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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8
| 1
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Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]
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Draw an altitude from both end circles of the diagram with the circles of radius one, and call the lengths you get drawing the altitudes of the circles down to $BC$ $a$ and $b$ . Now we have the length of side $BC$ of being $(2)(2022)+1+1+a+b$ . However, the side $BC$ can also be written as $(6)(68)+34+34+34a+34b$ , due to similar triangles from the second diagram. If we set the equations equal, we have $\frac{1190}{11} = a+b$ . Call the radius of the incircle $r$ , then we have the side BC to be $r(a+b)$ . We find $r$ as $\frac{4046+\frac{1190}{11}}{\frac{1190}{11}}$ , which simplifies to $\frac{10+((34)(11))}{10}$ ,so we have $\frac{192}{5}$ , which sums to $\boxed{197}$
| 197
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5,202
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8
| 2
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Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]
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Assume that $ABC$ is isosceles with $AB=AC$
If we let $P_1$ be the intersection of $BC$ and the leftmost of the eight circles of radius $34$ $N_1$ the center of the leftmost circle, and $M_1$ the intersection of the leftmost circle and $AB$ , and we do the same for the $2024$ circles of radius $1$ , naming the points $P_2$ $N_2$ , and $M_2$ , respectively, then we see that $BP_1N_1M_1\sim BP_2N_2M_2$ . The same goes for vertex $C$ , and the corresponding quadrilaterals are congruent.
Let $x=BP_2$ . We see that $BP_1=34x$ by similarity ratios (due to the radii). The corresponding figures on vertex $C$ are also these values. If we combine the distances of the figures, we see that $BC=2x+4046$ and $BC=68x+476$ , and solving this system, we find that $x=\frac{595}{11}$
If we consider that the incircle of $\triangle ABC$ is essentially the case of $1$ circle with $r$ radius (the inradius of $\triangle ABC$ , we can find that $BC=2rx$ . From $BC=2x+4046$ , we have:
$r=1+\frac{2023}{x}$
$=1+\frac{11\cdot2023}{595}$
$=1+\frac{187}{5}$
$=\frac{192}{5}$
Thus the answer is $192+5=\boxed{197}$
| 197
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5,203
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8
| 3
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Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]
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Let $x = \cot{\frac{B}{2}} + \cot{\frac{C}{2}}$ . By representing $BC$ in two ways, we have the following: \[34x + 7\cdot 34\cdot 2 = BC\] \[x + 2023 \cdot 2 = BC\]
Solving we find $x = \frac{1190}{11}$ .
Now draw the inradius, let it be $r$ . We find that $rx =BC$ , hence \[xr = x + 4046 \implies r-1 = \frac{11}{1190}\cdot 4046 = \frac{187}{5}.\] Thus \[r = \frac{192}{5} \implies \boxed{197}.\] ~AtharvNaphade
| 197
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5,204
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8
| 4
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Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]
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First, let the circle tangent to $AB$ and $BC$ be $O$ and the other circle that is tangent to $AC$ and $BC$ be $R$ . Let $x$ be the distance from the tangency point on line segment $BC$ of the circle $O$ to $B$ . Also, let $y$ be the distance of the tangency point of circle $R$ on the line segment $BC$ to point $C$ . Realize that we can let $n$ be the number of circles tangent to line segment $BC$ and $r$ be the corresponding radius of each of the circles. Also, the circles that are tangent to $BC$ are similar. So, we can build the equation $BC = (x+y+2(n-1)) \times r$ . Looking at the given information, we see that when $n=8$ $r=34$ , and when $n=2024$ $r=1$ , and we also want to find the radius $r$ in the case where $n=1$ . Using these facts, we can write the following equations:
$BC = (x+y+2(8-1)) \times 34 = (x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r$
We can find that $x+y = \frac{1190}{11}$ . Now, let $(x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r$
Substituting $x+y = \frac{1190}{11}$ in, we find that \[r = \frac{192}{5} \implies \boxed{197}.\]
| 197
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5,205
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8
| 5
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Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]
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Define $I, x_1, x_8, y_1, y_{2024}$ to be the incenter and centers of the first and last circles of the $8$ and $2024$ tangent circles to $BC,$ and define $r$ to be the inradius of triangle $\bigtriangleup ABC.$ We calculate $\overline{x_1x_8} = 34 \cdot 14$ and $\overline{y_1y_{2024}} = 1 \cdot 4046$ because connecting the center of the circles voids two extra radii.
We can easily see that $B, x_1, x_8,$ and $I$ are collinear, and the same follows for $C, y_1, y_2024,$ and $I$ (think angle bisectors).
We observe that triangles $\bigtriangleup I x_1 x_8$ and $\bigtriangleup I y_1 y_{2024}$ are similar, and therefore the ratio of the altitude to the base is the same, so we note
\[\frac{\text{altitude}}{\text{base}} = \frac{r-34}{34\cdot 14} = \frac{r-1}{1\cdot 4046}.\]
Solving yields $r = \frac{192}{5},$ so the answer is $192+5 = \boxed{197}.$
| 197
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5,206
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_9
| 1
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Let $A$ $B$ $C$ , and $D$ be point on the hyperbola $\frac{x^2}{20}- \frac{y^2}{24} = 1$ such that $ABCD$ is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than $BD^2$ for all such rhombi.
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A quadrilateral is a rhombus if and only if its two diagonals bisect each other and are perpendicular to each other. The first condition is automatically satisfied because of the hyperbola's symmetry about the origin. To satisfy the second condition, we set $BD$ as the line $y = mx$ and $AC$ as $y = -\frac{1}{m}x.$ Because the hyperbola has asymptotes of slopes $\pm \frac{\sqrt6}{\sqrt5},$ we have $m, -\frac{1}{m} \in \left(-\frac{\sqrt6}{\sqrt5}, \frac{\sqrt6}{\sqrt5}\right).$ This gives us $m^2 \in \left(\frac{5}{6}, \frac{6}{5}\right).$
Plugging $y = mx$ into the equation for the hyperbola yields $x^2 = \frac{120}{6-5m^2}$ and $y^2 = \frac{120m^2}{6-5m^2}.$ By symmetry of the hyperbola, we know that $\left(\frac{BD}{2}\right)^2 = x^2 + y^2,$ so we wish to find a lower bound for $x^2 + y^2 = 120\left(\frac{1+m^2}{6-5m^2}\right).$ This is equivalent to minimizing $\frac{1+m^2}{6-5m^2} = -\frac{1}{5} + \frac{11}{5(6-5m^2)}$ . It's then easy to see that this expression increases with $m^2,$ so we plug in $m^2 = \frac{5}{6}$ to get $x^2+y^2 > 120,$ giving $BD^2 > \boxed{480}.$
| 480
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5,207
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_9
| 2
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Let $A$ $B$ $C$ , and $D$ be point on the hyperbola $\frac{x^2}{20}- \frac{y^2}{24} = 1$ such that $ABCD$ is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than $BD^2$ for all such rhombi.
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Assume $AC$ is the asymptope of the hyperbola, $BD$ in that case is the smallest. The expression of $BD$ is $y=-\sqrt{\frac{5}{6}}x$ . Thus, we could get $\frac{x^2}{20}-\frac{y^2}{24}=1\implies x^2=\frac{720}{11}$ . The desired value is $4\cdot \frac{11}{6}x^2=480$ . This case wouldn't achieve, so all $BD^2$ would be greater than $\boxed{480}$
| 480
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5,208
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_9
| 4
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Let $A$ $B$ $C$ , and $D$ be point on the hyperbola $\frac{x^2}{20}- \frac{y^2}{24} = 1$ such that $ABCD$ is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than $BD^2$ for all such rhombi.
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The only "numbers" provided in this problem are $24$ and $20$ , so the answer must be a combination of some operations on these numbers. If you're lucky, you could figure the most likely option is $24\cdot 20$ , as this yields $\boxed{480}$ and seems like a plausible answer for this question.
| 480
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5,209
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10
| 1
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Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ $BC=9$ , and $AC=10$ $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$
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From the tangency condition we have $\let\angle BCD = \let\angle CBD = \let\angle A$ . With LoC we have $\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}$ and $\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}$ . Then, $CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}$ . Using LoC we can find $AD$ $AD^2 = AC^2 + CD^2 - 2(AC)(CD)\cos(A+C) = 10^2+(\frac{225}{22})^2 + 2(10)\frac{225}{22}\cos(B) = 100 + \frac{225^2}{22^2} + 2(10)\frac{225}{22}*\frac{1}{15} = \frac{5^4*13^2}{484}$ . Thus, $AD = \frac{5^2*13}{22}$ . By Power of a Point, $DP*AD = CD^2$ so $DP*\frac{5^2*13}{22} = (\frac{225}{22})^2$ which gives $DP = \frac{5^2*9^2}{13*22}$ . Finally, we have $AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}$
| 113
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5,210
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10
| 2
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Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ $BC=9$ , and $AC=10$ $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$
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We know $AP$ is the symmedian, which implies $\triangle{ABP}\sim \triangle{AMC}$ where $M$ is the midpoint of $BC$ . By Appolonius theorem, $AM=\frac{13}{2}$ . Thus, we have $\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}$
| 113
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5,211
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10
| 3
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Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ $BC=9$ , and $AC=10$ $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$
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Extend sides $\overline{AB}$ and $\overline{AC}$ to points $E$ and $F$ , respectively, such that $B$ and $C$ are the feet of the altitudes in $\triangle AEF$ . Denote the feet of the altitude from $A$ to $\overline{EF}$ as $X$ , and let $H$ denote the orthocenter of $\triangle AEF$ . Call $M$ the midpoint of segment $\overline{EF}$ . By the Three Tangents Lemma, we have that $MB$ and $MC$ are both tangents to $(ABC)$ $\implies$ $M = D$ , and since $M$ is the midpoint of $\overline{EF}$ $MF = MB$ . Additionally, by angle chasing, we get that: \[\angle ABC \cong \angle AHC \cong \angle EHX\] Also, \[\angle EHX = 90 ^\circ - \angle HEF = 90 ^\circ - (90 ^\circ - \angle AFE) = \angle AFE\] Furthermore, \[AB = AF \cdot \cos(A)\] From this, we see that $\triangle ABC \sim \triangle AFE$ with a scale factor of $\cos(A)$ . By the Law of Cosines, \[\cos(A) = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}\] Thus, we can find that the side lengths of $\triangle AEF$ are $\frac{250}{11}, \frac{125}{11}, \frac{225}{11}$ . Then, by Stewart's theorem, $AM = \frac{13 \cdot 25}{22}$ . By Power of a Point, \[\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MP}\] \[\frac{225}{22} \cdot \frac{225}{22} = \overline{MP} \cdot \frac{13 \cdot 25}{22} \implies \overline{MP} = \frac{225 \cdot 9}{22 \cdot 13}\] Thus, \[AP = AM - MP = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}\] Therefore, the answer is $\boxed{113}$
| 113
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5,212
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10
| 4
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Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ $BC=9$ , and $AC=10$ $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$
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Connect lines $\overline{PB}$ and $\overline{PC}$ . From the angle by tanget formula, we have $\angle PBD = \angle DAB$ . Therefore by AA similarity, $\triangle PBD \sim \triangle BAD$ . Let $\overline{BP} = x$ . Using ratios, we have \[\frac{x}{5}=\frac{BD}{AD}.\] Similarly, using angle by tangent, we have $\angle PCD = \angle DAC$ , and by AA similarity, $\triangle CPD \sim \triangle ACD$ . By ratios, we have \[\frac{PC}{10}=\frac{CD}{AD}.\] However, because $\overline{BD}=\overline{CD}$ , we have \[\frac{x}{5}=\frac{PC}{10},\] so $\overline{PC}=2x.$ Now using Law of Cosines on $\angle BAC$ in triangle $\triangle ABC$ , we have \[9^2=5^2+10^2-100\cos(\angle BAC).\] Solving, we find $\cos(\angle BAC)=\frac{11}{25}$ . Now we can solve for $x$ . Using Law of Cosines on $\triangle BPC,$ we have
\begin{align*}
81&=x^2+4x^2-4x^2\cos(180-\angle BAC) \\
&= 5x^2+4x^2\cos(BAC). \\
\end{align*}
Solving, we get $x=\frac{45}{13}.$ Now we have a system of equations using Law of Cosines on $\triangle BPA$ and $\triangle CPA$ \[AP^2=5^2+\left(\frac{45}{13}\right)^2 -(10) \left(\frac{45}{13} \right)\cos(ABP)\] \[AP^2=10^2+4 \left(\frac{45}{13} \right)^2 + (40) \left(\frac{45}{13} \right)\cos(ABP).\]
Solving, we find $\overline{AP}=\frac{100}{13}$ , so our desired answer is $100+13=\boxed{113}$
| 113
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5,213
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10
| 5
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Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ $BC=9$ , and $AC=10$ $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$
|
Following from the law of cosines, we can easily get $\cos A = \frac{11}{25}$ $\cos B = \frac{1}{15}$ $\cos C = \frac{13}{15}$
Hence, $\sin A = \frac{6 \sqrt{14}}{25}$ $\cos 2C = \frac{113}{225}$ $\sin 2C = \frac{52 \sqrt{14}}{225}$ .
Thus, $\cos \left( A + 2C \right) = - \frac{5}{9}$
Denote by $R$ the circumradius of $\triangle ABC$ .
In $\triangle ABC$ , following from the law of sines, we have $R = \frac{BC}{2 \sin A} = \frac{75}{4 \sqrt{14}}$
Because $BD$ and $CD$ are tangents to the circumcircle $ABC$ $\triangle OBD \cong \triangle OCD$ and $\angle OBD = 90^\circ$ .
Thus, $OD = \frac{OB}{\cos \angle BOD} = \frac{R}{\cos A}$
In $\triangle AOD$ , we have $OA = R$ and $\angle AOD = \angle BOD + \angle AOB = A + 2C$ .
Thus, following from the law of cosines, we have
\begin{align*}
AD & = \sqrt{OA^2 + OD^2 - 2 OA \cdot OD \cos \angle AOD} \\
& = \frac{26 \sqrt{14}}{33} R.
\end{align*}
Following from the law of cosines,
\begin{align*}
\cos \angle OAD & = \frac{AD^2 + OA^2 - OD^2}{2 AD \cdot OA} \\
& = \frac{8 \sqrt{14}}{39} .
\end{align*}
Therefore,
\begin{align*}
AP & = 2 OA \cos \angle OAD \\
& = \frac{100}{13} .
\end{align*}
Therefore, the answer is $100 + 13 = \boxed{113}$
| 113
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5,214
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_12
| 1
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Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]
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If we graph $4g(f(x))$ , we see it forms a sawtooth graph that oscillates between $0$ and $1$ (for values of $x$ between $-1$ and $1$ , which is true because the arguments are between $-1$ and $1$ ). Thus by precariously drawing the graph of the two functions in the square bounded by $(0,0)$ $(0,1)$ $(1,1)$ , and $(1,0)$ , and hand-counting each of the intersections, we get $\boxed{385}$
| 385
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5,215
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_12
| 2
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Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]
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We will denote $h(x)=4g(f(x))$ for simplicity. Denote $p(x)$ as the first equation and $q(y)$ as the graph of the second. We notice that both $f(x)$ and $g(x)$ oscillate between 0 and 1. The intersections are thus all in the square $(0,0)$ $(0,1)$ $(1,1)$ , and $(1,0)$ . Every $p(x)$ wave going up and down crosses every $q(y)$ wave. Now, we need to find the number of times each wave touches 0 and 1.
We notice that $h(x)=0$ occurs at $x=-\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}$ , and $h(x)=1$ occurs at $x=-1, -\frac{1}{2}, 0,\frac{1}{2},1$ . A sinusoid passes through each point twice during each period, but it only passes through the extrema once. $p(x)$ has 1 period between 0 and 1, giving 8 solutions for $p(x)=0$ and 9 solutions for $p(x)=1$ , or 16 up and down waves. $q(y)$ has 1.5 periods, giving 12 solutions for $q(y)=0$ and 13 solutions for $q(y)=1$ , or 24 up and down waves. This amounts to $16\cdot24=384$ intersections.
However, we have to be very careful when counting around $(1, 1)$ . At this point, $q(y)$ has an infinite downwards slope and $p(x)$ is slanted, giving us an extra intersection; thus, we need to add 1 to our answer to get $\boxed{385}$
| 385
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5,216
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_12
| 3
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Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]
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We can easily see that only $x, y \in \left[0,1 \right]$ may satisfy both functions.
We call function $y = 4g \left( f \left( \sin \left( 2 \pi x \right) \right) \right)$ as Function 1 and function $x = 4g \left( f \left( \cos \left( 3 \pi y \right) \right) \right)$ as Function 2.
For Function 1, in each interval $\left[ \frac{i}{4} , \frac{i+1}{4} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ , Function 1's value oscillates between 0 and 1. It attains 1 at $x = \frac{i}{4}$ $\frac{i+1}{4}$ and another point between these two.
Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1.
So the graph of this function in this interval consists of 4 monotonic pieces.
For Function 2, in each interval $\left[ \frac{i}{6} , \frac{i+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 5 \right\}$ , Function 2's value oscillates between 0 and 1. It attains 1 at $y = \frac{i}{6}$ $\frac{i+1}{6}$ and another point between these two.
Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1.
So the graph of this function in this interval consists of 4 monotonic curves.
Consider any region $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ and $j \in \left\{0, 1, \cdots , 5 \right\}$ but $\left( i, j \right) \neq \left( 3, 5 \right)$ .
Both functions have four monotonic pieces.
Because Function 1's each monotonic piece can take any value in $\left[ \frac{j}{6} , \frac{j+1}{6} \right]$ and Function 2' each monotonic piece can take any value in $\left[ \frac{i}{4} , \frac{i+1}{4} \right]$ , Function 1's each monotonic piece intersects with Function 2's each monotonic piece.
Therefore, in the interval $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ , the number of intersecting points is $4 \cdot 4 = 16$
Next, we prove that if an intersecting point is on a line $x = \frac{i}{4}$ for $i \in \left\{ 0, 1, \cdots, 4 \right\}$ , then this point must be $\left( 1, 1 \right)$
For $x = \frac{i}{4}$ , Function 1 attains value 1.
For Function 2, if $y = 1$ , then $x = 1$ .
Therefore, the intersecting point is $\left( 1, 1 \right)$
Similarly, we can prove that if an intersecting point is on a line $y = \frac{i}{6}$ for $i \in \left\{ 0, 1, \cdots, 6 \right\}$ , then this point must be $\left( 1, 1 \right)$
Therefore, in each region $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ and $j \in \left\{0, 1, \cdots , 5 \right\}$ but $\left( i, j \right) \neq \left( 3, 5 \right)$ , all 16 intersecting points are interior.
That is, no two regions share any common intersecting point.
Next, we study region $\left[ \frac{3}{4} , 1 \right] \times \left[ \frac{5}{6} , 1 \right]$ .
Consider any pair of monotonic pieces, where one is from Function 1 and one is from Function 2, except the pair of two monotonic pieces from two functions that attain $\left( 1 , 1 \right)$ .
Two pieces in each pair intersects at an interior point on the region.
So the number of intersecting points is $4 \cdot 4 - 1 = 15$
Finally, we compute the number intersection points of two functions' monotonic pieces that both attain $\left( 1, 1 \right)$
One trivial intersection point is $\left( 1, 1 \right)$ .
Now, we study whether they intersect at another point.
Define $x = 1 - x'$ and $y = 1 - y'$ .
Thus, for positive and sufficiently small $x'$ and $y'$ , Function 1 is reduced to \[ y' = 4 \sin 2 \pi x' \hspace{1cm} (1) \] and Function 2 is reduced to \[ x' = 4 \left( 1 - \cos 3 \pi y' \right) . \hspace{1cm} (2) \]
Now, we study whether there is a non-zero solution.
Because we consider sufficiently small $x'$ and $y'$ , to get an intuition and quick estimate, we do approximations of the above equations.
Equation (1) is approximated as \[ y' = 4 \cdot 2 \pi x' \] and Equation (2) is approximated as \[ x' = 2 \left( 3 \pi y' \right)^2 \]
To solve these equations, we get $x' = \frac{1}{8^2 \cdot 18 \pi^4}$ and $y' = \frac{1}{8 \cdot 18 \pi^3}$ .
Therefore, two functions' two monotonic pieces that attain $\left( 1, 1 \right)$ have two intersecting points.
Putting all analysis above, the total number of intersecting points is $16 \cdot 4 \cdot 6 + 1 = \boxed{385}$
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_13
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Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$ . Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$
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If \(p=2\), then \(4\mid n^4+1\) for some integer \(n\). But \(\left(n^2\right)^2\equiv0\) or \(1\pmod4\), so it is impossible. Thus \(p\) is an odd prime.
For integer \(n\) such that \(p^2\mid n^4+1\), we have \(p\mid n^4+1\), hence \(p\nmid n^4-1\), but \(p\mid n^8-1\). By Fermat's Little Theorem , \(p\mid n^{p-1}-1\), so
\begin{equation*}
p\mid\gcd\left(n^{p-1}-1,n^8-1\right)=n^{\gcd(p-1,8)}-1.
\end{equation*}
Here, \(\gcd(p-1,8)\) mustn't be divide into \(4\) or otherwise \(p\mid n^{\gcd(p-1,8)}-1\mid n^4-1\), which contradicts. So \(\gcd(p-1,8)=8\), and so \(8\mid p-1\). The smallest such prime is clearly \(p=17=2\times8+1\).
So we have to find the smallest positive integer \(m\) such that \(17\mid m^4+1\). We first find the remainder of \(m\) divided by \(17\) by doing
\begin{array}{|c|cccccccccccccccc|}
\hline
\vphantom{\tfrac11}x\bmod{17}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\\hline
\vphantom{\dfrac11}\left(x^4\right)^2+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\\hline
\end{array}
So \(m\equiv\pm2\), \(\pm8\pmod{17}\). If \(m\equiv2\pmod{17}\), let \(m=17k+2\), by the binomial theorem,
\begin{align*}
0&\equiv(17k+2)^4+1\equiv\mathrm {4\choose 1}(17k)(2)^3+2^4+1=17(1+32k)\pmod{17^2}\\[3pt]
\implies0&\equiv1+32k\equiv1-2k\pmod{17}.
\end{align*}
So the smallest possible \(k=9\), and \(m=155\).
If \(m\equiv-2\pmod{17}\), let \(m=17k-2\), by the binomial theorem,
\begin{align*}
0&\equiv(17k-2)^4+1\equiv\mathrm {4\choose 1}(17k)(-2)^3+2^4+1=17(1-32k)\pmod{17^2}\\[3pt]
\implies0&\equiv1-32k\equiv1+2k\pmod{17}.
\end{align*}
So the smallest possible \(k=8\), and \(m=134\).
If \(m\equiv8\pmod{17}\), let \(m=17k+8\), by the binomial theorem,
\begin{align*}
0&\equiv(17k+8)^4+1\equiv\mathrm {4\choose 1}(17k)(8)^3+8^4+1=17(241+2048k)\pmod{17^2}\\[3pt]
\implies0&\equiv241+2048k\equiv3+8k\pmod{17}.
\end{align*}
So the smallest possible \(k=6\), and \(m=110\).
If \(m\equiv-8\pmod{17}\), let \(m=17k-8\), by the binomial theorem,
\begin{align*}
0&\equiv(17k-8)^4+1\equiv\mathrm {4\choose 1}(17k)(-8)^3+8^4+1=17(241-2048k)\pmod{17^2}\\[3pt]
\implies0&\equiv241+2048k\equiv3+9k\pmod{17}.
\end{align*}
So the smallest possible \(k=11\), and \(m=179\).
In conclusion, the smallest possible \(m\) is \(\boxed{110}\).
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Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$ . Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$
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We work in the ring \(\mathbb Z/289\mathbb Z\) and use the formula \[\sqrt[4]{-1}=\pm\sqrt{\frac12}\pm\sqrt{-\frac12}.\] Since \(-\frac12=144\), the expression becomes \(\pm12\pm12i\), and it is easily calculated via Hensel that \(i=38\), thus giving an answer of \(\boxed{110}\).
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_13
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Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$ . Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$
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Note that $n^4 + 1 \equiv 0 \pmod{p}$ means $\text{ord}_{p}(n) = 8 \mid p-1.$ The smallest prime that does this is $17$ and $2^4 + 1 = 17$ for example. Now let $g$ be a primitive root of $17^2.$ The satisfying $n$ are of the form, $g^{\frac{p(p-1)}{8}}, g^{3\frac{p(p-1)}{8}}, g^{5\frac{p(p-1)}{8}}, g^{7\frac{p(p-1)}{8}}.$ So if we find one such $n$ , then all $n$ are $n, n^3, n^5, n^7.$ Consider the $2$ from before. Note $17^2 \mid 2^{4 \cdot 17} + 1$ by LTE. Hence the possible $n$ are, $2^{17}, 2^{51}, 2^{85}, 2^{119}.$ Some modular arithmetic yields that $2^{51} \equiv \boxed{110}$ is the least value.
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_14
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Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$
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Notice that \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal
\begin{equation*}
\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.
\end{equation*}
Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is
\begin{equation*}
h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.
\end{equation*}
Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for cones,
\begin{align*}
\frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\
&=\frac13Sr\cdot4.
\end{align*}
Hence, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).
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Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$
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[asy] import three; currentprojection = orthographic(1,1,1); triple O = (0,0,0); triple A = (0,2,0); triple B = (0,0,1); triple C = (3,0,0); triple D = (3,2,1); triple E = (3,2,0); triple F = (0,2,1); triple G = (3,0,1); draw(A--B--C--cycle, red); draw(A--B--D--cycle, red); draw(A--C--D--cycle, red); draw(B--C--D--cycle, red); draw(E--A--O--C--cycle); draw(D--F--B--G--cycle); draw(O--B); draw(A--F); draw(E--D); draw(C--G); label("$O$", O, SW); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, S); label("$D$", D, NE); label("$E$", E, SE); label("$F$", F, NW); label("$G$", G, NE); [/asy]
Inscribe tetrahedron $ABCD$ in an rectangular prism as shown above.
By the Pythagorean theorem, we note
\[OA^2 + OB^2 = AB^2 = 41,\] \[OA^2 + OC^2 = AC^2 = 80, \text{and}\] \[OB^2 + OC^2 = BC^2 = 89.\]
Solving yields $OA = 4, OB = 5,$ and $OC = 8.$
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of $ABCD.$ We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of $ABCD.$
We know that the distance from all $4$ faces must be the same, so we only need to find the distance from the center to plane $ABC$
Let $O = (0,0,0), A = (4,0,0), B = (0,5,0),$ and $C = (0,0,8).$ We obtain that the plane of $ABC$ can be marked as $\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,$ or $10x + 8y + 5z - 40 = 0,$ and the center of the prism is $(2,\frac{5}{2},4).$
Using the Point-to-Plane distance formula, our distance is
\[d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.\]
Our answer is $20 + 21 + 63 = \boxed{104}.$
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Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$
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We use the formula for the volume of iscoceles tetrahedron. $V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}$
Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find \[\cos{\angle ACB} = \frac{80 + 89 - 41}{2\sqrt{80\cdot 89}}= \frac{16}{9\sqrt{5}}.\]
From this, we find \[\sin{\angle ACB} = \sqrt{1-\cos^2{\angle ACB}} = \sqrt{1 - \frac{256}{405}} = \sqrt{\frac{149}{405}}\] and can find the area of $\triangle ABC$ as \[A = \frac{1}{2} \sqrt{89\cdot 80}\cdot \sin{\angle ACB} = 6\sqrt{21}.\]
Let $R$ be the distance we want to find. By taking the sum of (equal) volumes \[[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,\] We have \[V = \frac{4AR}{3}.\] Plugging in and simplifying, we get $R = \frac{20\sqrt{21}}{63}$ for an answer of $20 + 21 + 63 = \boxed{104}$
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Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$
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Let $AH$ be perpendicular to $BCD$ that meets this plane at point $H$ .
Let $AP$ $AQ$ , and $AR$ be heights to lines $BC$ $CD$ , and $BD$ with feet $P$ $Q$ , and $R$ , respectively.
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as $A$ , is $A = 6 \sqrt{21}$
Hence, by using this area, we can compute $AP$ $AQ$ and $AR$ .
We have $AP = \frac{2 A}{BC} = \frac{2A}{\sqrt{89}}$ $AQ = \frac{2 A}{CD} = \frac{2A}{\sqrt{41}}$ , and $AR = \frac{2 A}{BC} = \frac{2A}{\sqrt{80}}$
Because $AH \perp BCD$ , we have $AH \perp BC$ . Recall that $AP \perp BC$ .
Hence, $BC \perp APH$ . Hence, $BC \perp HP$
Analogously, $CD \perp HQ$ and $BD \perp HR$
We introduce a function $\epsilon \left( l \right)$ for $\triangle BCD$ that is equal to 1 (resp. -1) if point $H$ and the opposite vertex of side $l$ are on the same side (resp. opposite sides) of side $l$
The area of $\triangle BCD$ is
\begin{align*}
A & = \epsilon_{BC} {\rm Area} \ \triangle HBC
+ \epsilon_{CD} {\rm Area} \ \triangle HCD
+ \epsilon_{BD} {\rm Area} \ \triangle HBD \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot HP
+ \frac{1}{2} \epsilon_{CD} CD \cdot HQ +
\frac{1}{2} \epsilon_{BD} BD \cdot HR \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2}
+ \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\
& \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1)
\end{align*}
Denote $B = 2A$ .
The above equation can be organized as
\begin{align*}
B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
+ \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\
& \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
This can be further reorganized as
\begin{align*}
B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
& = \epsilon_{CD} \sqrt{B^2 - 41 AH^2}
+ \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
Taking squares on both sides and reorganizing terms, we get
\begin{align*}
& 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\
& = \epsilon_{CD} \epsilon_{BD}
\sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} .
\end{align*}
Taking squares on both sides and reorganizing terms, we get \[ - \epsilon_{BC} 2 B \sqrt{B^2 - 89 AH^2} = - 2 B^2 + 189 AH^2 . \]
Taking squares on both sides, we finally get
\begin{align*}
AH & = \frac{20B}{189} \\
& = \frac{40A}{189}.
\end{align*}
Now, we plug this solution to Equation (1). We can see that $\epsilon_{BC} = -1$ $\epsilon_{CD} = \epsilon_{BD} = 1$ .
This indicates that $H$ is out of $\triangle BCD$ . To be specific, $H$ and $D$ are on opposite sides of $BC$ $H$ and $C$ are on the same side of $BD$ , and $H$ and $B$ are on the same side of $CD$
Now, we compute the volume of the tetrahedron $ABCD$ , denoted as $V$ . We have $V = \frac{1}{3} A \cdot AH = \frac{40 A^2}{3 \cdot 189}$
Denote by $r$ the inradius of the inscribed sphere in $ABCD$ .
Denote by $I$ the incenter.
Thus, the volume of $ABCD$ can be alternatively calculated as
\begin{align*}
V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\
& = \frac{1}{3} r \cdot 4A .
\end{align*}
From our two methods to compute the volume of $ABCD$ and equating them, we get
\begin{align*}
r & = \frac{10A}{189} \\
& = \frac{20 \sqrt{21}}{63} .
\end{align*}
Therefore, the answer is $20 + 21 + 63 = \boxed{104}$
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Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$
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Consider the following construction of the tetrahedron. Place $AB$ on the floor. Construct an isosceles vertical triangle with $AB$ as its base and $M$ as the top vertex. Place $CD$ on the top vertex parallel to the ground with midpoint $M.$ Observe that $CD$ can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project $AB$ onto the plane of $CD$ , let the minor angle $\theta$ be this discrepancy.
By Median formula or Stewart's theorem, $AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.$ Consequently the area of $\triangle AMB$ is $\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.$ Note the altitude $8$ is also the distance between the parallel planes containing $AB$ and $CD.$
By Distance Formula, \begin{align*} (\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \\ (\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \\ \implies CD \cos{\theta} \sqrt{41} &= 9 \\ \sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}. \end{align*} Then the volume of the tetrahedron is given by $\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.$
The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from $I$ w.r.t each of the faces. If $r$ is the inradius, i.e the distance to the faces, then $\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])$ must the volume. Each face has the same area by SSS congruence, and by Heron's it is $\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.$
Therefore the answer is, $\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.$
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_15
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Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23$ . Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}$ . The value of $r^2$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
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Observe that the "worst" possible box is one of the maximum possible length.
By symmetry, the height and the width are the same in this antioptimal box. (If the height and width weren't the same, the extra difference between them could be used to make the length longer.) Thus, let the width and height be of length $a$ and the length be $L$
We're given that the volume is $23$ ; thus, $a^2L=23$ . We're also given that the surface area is $54=2\cdot27$ ; thus, $a^2+2aL=27$
From the first equation, we can get $L=\dfrac{23}{a^2}$ . We do a bunch of algebra:
\begin{align*}
L&=\dfrac{23}{a^2} \\
27&=a^2+2aL \\
&=a^2+2a\left(\dfrac{23}{a^2}\right) \\
&=a^2+\dfrac{46}a \\
27a&=a^3+46 \\
a^3-27a+46&=0. \\
\end{align*}
We can use the Rational Root Theorem and test a few values. It turns out that $a=2$ works. We use synthetic division to divide by $a-2$
Asdf.png
As we expect, the remainder is $0$ , and we are left with the polynomial $x^2+2x-23$ . We can now simply use the quadratic formula and find that the remaining roots are $\dfrac{-2\pm\sqrt{4-4(-23)}}2=\dfrac{-2\pm\sqrt{96}}2=\dfrac{-2\pm4\sqrt{6}}2=-1\pm2\sqrt6$ . We want the smallest $a$ to maximize $L$ , and it turns out that $a=2$ is in fact the smallest root. Thus, we let $a=2$ . Substituting this into $L=\dfrac{23}{a^2}$ , we find that $L=\dfrac{23}4$ . However, this is not our answer! This is simply the length of the box; we want the radius of the sphere enclosing it. We know that the diameter of the sphere is the diagonal of the box, and the 3D Pythagorean Theorem can give us the space diagonal. Applying it, we find that the diagonal has length $\sqrt{2^2+2^2+\left(\dfrac{23}4\right)^2}=\sqrt{8+\dfrac{529}{16}}=\sqrt{\dfrac{128+529}{16}}=\dfrac{\sqrt{657}}4$ . This is the diameter; we halve it to find the radius, $\dfrac{\sqrt{657}}8$ . We then square this and end up with $\dfrac{657}{64}$ , giving us an answer of $657+64=\boxed{721}$
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| 2
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Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23$ . Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}$ . The value of $r^2$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
Denote by $x$ $y$ $z$ the length, width, and height of a rectangular box.
We have
\begin{align*}
xy + yz + zx & = \frac{54}{2} \hspace{1cm} (1) \\
xyz & = 23 \hspace{1cm} (2)
\end{align*}
We have
\begin{align*}
4 r^2 & = x^2 + y^2 + z^2 \\
& = \left( x + y + z \right)^2 - 2 \cdot \left( xy + yz + zx \right) \\
& = \left( x + y + z \right)^2 - 54 .
\end{align*}
Therefore, we solve the following constrained optimization problem:
\begin{align*}
\max_{x,y,z} \ & x + y + z \\
\mbox{subject to } & (1), (2)
\end{align*}
First, we prove that an optimal solution must have at least two out of $x$ $y$ $z$ that are the same.
Denote by $\lambda$ and $\eta$ lagrangian multipliers of constraints (1) and (2), respectively.
Consider the following Lagrangian:
\begin{align*}
\max_{x,y,z, \lambda, \eta} & x + y + z + \lambda \left( xy + yz + zx - 27 \right)
+ \eta \left( xyz - 23 \right) .
\end{align*}
Taking first-order-condition with respect to $x$ $y$ $z$ , respectively, we get
\begin{align*}
1 + \lambda \left( y + z \right) + \eta yz & = 0 \hspace{1cm} (3) \\
1 + \lambda \left( z + x \right) + \eta zx & = 0 \hspace{1cm} (4) \\
1 + \lambda \left( x + y \right) + \eta xy & = 0 \hspace{1cm} (5)
\end{align*}
Suppose there is an optimal solution with $x$ $y$ $z$ that are all distinct.
Taking $(4)-(3)$ , we get \[ \left( x - y \right) \left( \lambda + \eta z \right) = 0 . \]
Because $x \neq y$ , we have \[ \lambda + \eta z = 0 \hspace{1cm} (6) \]
Analogously, we have
\begin{align*}
\lambda + \eta x & = 0 \hspace{1cm} (7)
\end{align*}
Taking $(6) - (7)$ , we get $\eta \left( z - x \right) = 0$ .
Because $z \neq x$ , we have $\eta = 0$ . Plugging this into (6), we get $\lambda = 0$
However, the solution that $\lambda = \eta = 0$ is a contradiction with (3).
Therefore, in an optimal solution, we cannot have $x$ $y$ , and $z$ to be all distinct.
W.L.O.G, in our remaining analysis, we assume an optimal solution satisfies $y = z$
Therefore, we need to solve the following two-variable optimization problem:
\begin{align*}
\max_{x,y} \ & x + 2y \\
\mbox{subject to } & 2 xy + y^2 = 27 \\
& xy^2 = 23
\end{align*}
Replacing $x$ with $y$ by using the constraint $xy^2 = 23$ , we solve the following single-variable optimization problem:
\begin{align*}
\max_y \ & \frac{23}{y^2} + 2y \hspace{1cm} (8) \\
\mbox{subject to } & \frac{46}{y} + y^2 = 27 \hspace{1cm} (9)
\end{align*}
By solving (9), we get $y = 2$ and $-1 + 2 \sqrt{6}$
Plugging $y = 2$ into (8), we get $\frac{23}{y^2} + 2y = \frac{39}{4}$
Plugging $y = -1 + 2 \sqrt{6}$ into (8), we get $\frac{23}{y^2} + 2y = \frac{96 \sqrt{6} - 21}{23}$
We have $\frac{96 \sqrt{6} - 21}{23} < \frac{39}{4}$ .
Therefore, the maximum value of $x + y + z$ is $\frac{39}{4}$
Therefore,
\begin{align*}
r^2 & = \frac{1}{4} \left( \left( x + y + z \right)^2 - 54 \right) \\
& = \frac{1}{4} \left( \left( \frac{39}{4} \right)^2 - 54 \right) \\
& = \frac{657}{64} .
\end{align*}
Therefore, the answer is $657 + 64 = \boxed{721}$
| 721
|
5,227
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_15
| 3
|
Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23$ . Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}$ . The value of $r^2$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
First, let's list the conditions:
Denote by $l$ $w$ $h$ the length, width, and height of a rectangular box.
\[lwh=23\] \begin{align*}
2(lw+wh+hl)&=54\\
lw+wh+hl&=27.
\end{align*}
Applying the Pythagorean theorem, we can establish that
\begin{align*}
(2r)^2&=(l^2+w^2+h^2)\\
4r^2&=(l^2+w^2+h^2)\\
4r^2&=(l+w+h)^2-2(lw+wh+hl)\\
4r^2&=(l+w+h)^2-54.
\end{align*}
We can spot Vieta's formula hidden inside this equation and call this $m$ . Now we have three equations:
\[lwh=23\] \[lw+wh+hl)=27\] \[l+w+h=m\]
Let there be a cubic equation. $x^3+bx^2+cx+d=0$ . Its roots are $l$ $w$ and $h$ . We can use our formulas from before to derive $c$ and $d$
\[-b=l+w+h=m\]
\[c=lw+wh+lh=27\]
\[-d=lwh=23\]
We can now rewrite the equation from before:
$x^3-mx^2+27x-23=0$
To find the maximum $r$ we need the maximum $m$ . This only occurs when this equation has double roots illustrated with graph below.
WLOG we can set $h=w$
Thus:
\[lw+w^2+wl=27\] \[lw^2=23\]
We can substitute $l$ and form a depressed cubic equation with $w$ .
\begin{align*}
lw^2&=23\\
l&=\frac{23}{w^2}\\
2\left(\frac{23}{w^2}\right)w+w^2&=27\\
\frac{46}{w}+w^2&=27\\
w^2+\frac{46}{w}-27&=0\\
w^3 -27w+46&=0.
\end{align*}
Based on Rational Root Theorem the possible rational roots are $\pm1, \pm2, \pm23$
A quick test reveals that $2$ is a root of the equation. Comparing coefficients we can factorize the equation into:
$(w-2)(w^2+2w-23)=0$
Besides $2$ , we derive another positive root using the quadratic formula, $2\sqrt{6}-1$ But to maximize the $m$ we need to pick the smaller $w$ , which is $2$
Substituting this into $l=\frac{23}{w^2}$ , we find that $l=\dfrac{23}4$
Applying it to our equation above:
\begin{align*}
4r^2&=(l+w+h)^2-54\\
4r^2&=(l+2w)^2-54\\
4r^2&=\left(\dfrac{23}4+2\cdot2\right)^2-54\\
4r^2&=\left(\dfrac{39}4\right)^2-54\\
4r^2&=\left(\dfrac{1521}{16}\right)-54\\
4r^2&=\left(\dfrac{657}{16}\right)\\
r^2&=\left(\dfrac{657}{64}\right).
\end{align*} $657+64=\boxed{721}$
| 721
|
5,228
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_2
| 1
|
A list of positive integers has the following properties:
$\bullet$ The sum of the items in the list is $30$
$\bullet$ The unique mode of the list is $9$
$\bullet$ The median of the list is a positive integer that does not appear in the list itself.
Find the sum of the squares of all the items in the list.
|
The third condition implies that the list's size must be an even number, as if it were an odd number, the median of hte list would surely appear in the list itself.
Therefore, we can casework on what even numbers work.
Say the size is 2. Clearly, this doesn't work as the only list would be $\{9, 9\}$ , which doesn't satisfy condition 1.
If the size is 4, then we can have two $9$ s, and a remaining sum of $12$ . Since the other two values in the list must be distinct, and their sum must equal $30-18=12$ , we have that the two numbers are in the form $a$ and $12-a$ . Note that we cannot have both values greater than $9$ , and we cannot have only one value greater than $9$ , because this would make the median $9$ , which violates condition 3. Since the median of the list is a positive integer, this means that the greater of $a$ and $12-a$ must be an odd number. The only valid solution to this is $a=5$ . Thus, our answer is $5^2+7^2+9^2+9^2 = \boxed{236}$ . ~akliu
| 236
|
5,229
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_2
| 2
|
A list of positive integers has the following properties:
$\bullet$ The sum of the items in the list is $30$
$\bullet$ The unique mode of the list is $9$
$\bullet$ The median of the list is a positive integer that does not appear in the list itself.
Find the sum of the squares of all the items in the list.
|
If there were an odd number of elements, the median would be in the set. Thus, we start with 4 elements. For 9 to be the mode, there must be 2 9s. For 9 to not be the median, either both numbers are greater than 9, or both numbers are less than 9. Clearly, both numbers must be less. From here, the numbers are clearly $(5,7,9,9)$ , and we add their squares to get $\boxed{236}$ -westwoodmonster
| 236
|
5,230
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_2
| 3
|
A list of positive integers has the following properties:
$\bullet$ The sum of the items in the list is $30$
$\bullet$ The unique mode of the list is $9$
$\bullet$ The median of the list is a positive integer that does not appear in the list itself.
Find the sum of the squares of all the items in the list.
|
Since the median is not in the list, there must not be an odd number of elements. Suppose the list has two elements. To meet the mode condition, both must equal $9$ , but this does not satisfy the other conditions.
Next, suppose the list has six elements. If there were at least three $9$ s, then the other elements would sum to at most $30-27=3$ . Since the elements are positive integers, this can only be achieved with the set $\{1,1,1,9,9,9\}$ , which violates the unique mode condition. Therefore, there must be exactly two $9$ s, and the other four elements must be distinct to satisfy the unique mode condition. Two sets of four unique positive integers add to $12$ $\{1,2,3,6\}$ and $\{1,2,4,5\}$ . Neither can act as the remaining four elements since both possibilities violate the constraint that the median is an integer.
Next, suppose the list had at least eight elements. For the sake of contradiction, suppose the third-largest element was at least $9$ . Then, since every element is a positive integer, the minimum sum would be $1+1+1+1+1+9+9+9>30$ . So, to satisfy the unique mode condition, there must be exactly two $9$ s, and the other elements must be distinct. But then the minimum sum is $1+2+3+4+5+6+9+9>30$ , so the sum constraint can never be satisfied. From these deductions, we conclude that the list has exactly four elements.
Note that no element can appear three times in the list, or else the middle-two-largest elements would be equal, violating the condition that the median is not in the list. Therefore, to satisfy the unique mode condition, the list contains two $9$ s and two other distinct integers that add to $30-18=12$ . Five sets of two unique positive integers add to $12$ $\{1,11\}$ $\{2,10\}$ $\{3,9\}$ $\{4,8\}$ , and $\{5,7\}$ . The first four options violate the median condition (either they make the median one of the list elements, or they make the median a non-integer). Thus, the set must be $\{5,7,9,9\}$ , and the sum of the squares of these elements is $25+49+81+81=\boxed{236}$
| 236
|
5,231
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_6
| 2
|
Alice chooses a set $A$ of positive integers. Then Bob lists all finite nonempty sets $B$ of positive integers with the property that the maximum element of $B$ belongs to $A$ . Bob's list has 2024 sets. Find the sum of the elements of A.
|
Let $A = \left\{ a_1, a_2, \cdots, a_n \right\}$ with $a_1 < a_2 < \cdots < a_n$
If the maximum element of $B$ is $a_i$ for some $i \in \left\{ 1, 2, \cdots , n \right\}$ , then each element in $\left\{ 1, 2, \cdots, a_i- 1 \right\}$ can be either in $B$ or not in $B$ .
Therefore, the number of such sets $B$ is $2^{a_i - 1}$
Therefore, the total number of sets $B$ is
\begin{align*}
\sum_{i=1}^n 2^{a_i - 1} & = 2024 .
\end{align*}
Thus
\begin{align*}
\sum_{i=1}^n 2^{a_i} & = 4048 .
\end{align*}
Now, the problem becomes writing 4048 in base 2, say, $4048 = \left( \cdots b_2b_1b_0 \right)_2$ .
We have $A = \left\{ j \geq 1: b_j = 1 \right\}$
We have $4048 = \left( 111,111,010,000 \right)_2$ .
Therefore, $A = \left\{ 4, 6, 7, 8, 9, 10, 11 \right\}$ .
Therefore, the sum of all elements in $A$ is $\boxed{55}$
| 55
|
5,232
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_7
| 1
|
Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$ , the resulting number is divisible by $7$ . Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$ . Find $Q+R$
|
We note that by changing a digit to $1$ for the number $\overline{abcd}$ , we are subtracting the number by either $1000(a-1)$ $100(b-1)$ $10(c-1)$ , or $d-1$ . Thus, $1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}$ . We can casework on $a$ backwards, finding the maximum value.
(Note that computing $1000 \equiv 6 \pmod{7}, 100 \equiv 2 \pmod{7}, 10 \equiv 3 \pmod{7}$ greatly simplifies computation).
Applying casework on $a$ , we can eventually obtain a working value of $\overline{abcd} = 5694 \implies \boxed{699}$ . ~akliu
| 699
|
5,233
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_7
| 2
|
Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$ , the resulting number is divisible by $7$ . Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$ . Find $Q+R$
|
Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Reducing, we get
$6+2b+3c+d \equiv 0 \pmod{7}$ $(1)$
$6a+2+3c+d \equiv 0 \pmod{7}$ $(2)$
$6a+2b+3+d \equiv 0 \pmod{7}$ $(3)$
$6a+2b+3c+1 \equiv 0 \pmod{7}$ $(4)$
Subtracting $(2)-(1), (3)-(2), (4)-(3), (4)-(1)$ , we get:
$3a-b \equiv 2 \pmod{7}$
$2b-3c \equiv 6 \pmod{7}$
$3c-d \equiv 2 \pmod{7}$
$6a-d \equiv 5 \pmod{7}$
For the largest 4 digit number, we test values for a starting with 9. When a is 9, b is 4, c is 3, and d is 7. However, when switching the digits with 1, we quickly notice this doesnt work. Once we get to a=5, we get b=6,c=9,and d=4. Adding 694 with 5, we get $\boxed{699}$ -westwoodmonster
| 699
|
5,234
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_7
| 3
|
Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$ , the resulting number is divisible by $7$ . Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$ . Find $Q+R$
|
Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Add the equations together, we get:
$3000a+300b+30c+3d+1111 \equiv 0 \pmod{7}$
And since the remainder of 1111 divided by 7 is 5, we get:
$3abcd \equiv 2 \pmod{7}$
Which gives us:
$abcd \equiv 3 \pmod{7}$
And since we know that changing each digit into 1 will make abcd divisible by 7, we get that $d-1$ $10c-10$ $100b-100$ , and $1000a-1000$ all have a remainder of 3 when divided by 7. Thus, we get $a=5$ $b=6$ $c=9$ , and $d=4$ . Thus, we get 5694 as abcd, and the answer is $694+5=\boxed{699}$
| 699
|
5,235
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_7
| 4
|
Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$ , the resulting number is divisible by $7$ . Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$ . Find $Q+R$
|
Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Then, we let x, y, z, t be the smallest whole number satisfying the following equations:
$1000a \equiv x \pmod{7}$
$100b \equiv y \pmod{7}$
$10a \equiv z \pmod{7}$
$d \equiv t \pmod{7}$
Since 1000, 100, 10, and 1 have a remainder of 6, 2, 3, and 1 when divided by 7, we can get the equations of:
(1): $6+y+z+t \equiv 0 \pmod{7}$
(2): $x+2+z+t \equiv 0 \pmod{7}$
(3): $x+y+3+t \equiv 0 \pmod{7}$
(4): $x+y+z+1 \equiv 0 \pmod{7}$
Add (1), (2), (3) together, we get:
$2x+2y+2z+3t+11 \equiv 0 \pmod{7}$
We can transform this equation to:
$2(x+y+z+1)+3t+9 \equiv 0 \pmod{7}$
Since, according to (4), $x+y+z+1$ has a remainder of 0 when divided by 7, we get:
$3t+9 \equiv 0 \pmod{7}$
And because t is 0 to 6 due to it being a remainder when divided by 7, we use casework and determine that t is 4.
Using the same methods of simplification, we get that x=2, y=5, and z=6, which means that 1000a, 100b, 10c, and d has a remainder of 2, 5, 6, and 4, respectively. Since a, b, c, and d is the largest possible number between 0 to 9, we use casework to determine the answer is a=5, b=6, c=9, and d=4, which gives us an answer of $5+694=\boxed{699}$
| 699
|
5,236
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_8
| 1
|
Torus $T$ is the surface produced by revolving a circle with radius $3$ around an axis in the plane of the circle that is a distance $6$ from the center of the circle (so like a donut). Let $S$ be a sphere with a radius $11$ . When $T$ rests on the inside of $S$ , it is internally tangent to $S$ along a circle with radius $r_i$ , and when $T$ rests on the outside of $S$ , it is externally tangent to $S$ along a circle with radius $r_o$ . The difference $r_i-r_o$ can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] unitsize(0.3 inch); draw(ellipse((0,0), 3, 1.75)); draw((-1.2,0.1)..(-0.8,-0.03)..(-0.4,-0.11)..(0,-0.15)..(0.4,-0.11)..(0.8,-0.03)..(1.2,0.1)); draw((-1,0.04)..(-0.5,0.12)..(0,0.16)..(0.5,0.12)..(1,0.04)); draw((0,2.4)--(0,-0.15)); draw((0,-0.15)--(0,-1.75), dashed); draw((0,-1.75)--(0,-2.25)); draw(ellipse((2,0), 1, 0.9)); draw((2.03,-0.02)--(2.9,-0.4)); [/asy]
|
First, let's consider a section $\mathcal{P}$ of the solids, along the axis.
By some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\mathcal{P}$ we took crosses one of the equator of the sphere.
Here I drew two graphs, the first one is the case when $T$ is internally tangent to $S$
[asy] unitsize(0.35cm); pair O = (0, 0); real r1 = 11; real r2 = 3; draw(circle(O, r1)); pair A = O + (0, -r1); pair B = O + (0, r1); draw(A--B); pair C = O + (0, -1.25*r1); pair D = O + (0, 1.25*r1); draw(C--D, dashed); dot(O); pair E = (2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)); pair F = (0, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)); pair G = (-r2 * O + r1 * E) / (r1 - r2); pair H = (-r2 * O + r1 * F) / (r1 - r2); draw(circle(E, r2)); draw(circle((-2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)), r2)); draw(O--G, dashed); draw(F--E, dashed); draw(G--H, dashed); label("$O$", O, SW); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NW); label("$D$", D, SW); label("$E_i$", E, NE); label("$F_i$", F, W); label("$G_i$", G, SE); label("$H_i$", H, W); label("$r_i$", 0.5 * H + 0.5 * G, NE); label("$3$", 0.5 * E + 0.5 * G, NE); label("$11$", 0.5 * O + 0.5 * G, NE); [/asy]
and the second one is when $T$ is externally tangent to $S$
[asy] unitsize(0.35cm); pair O = (0, 0); real r1 = 11; real r2 = 3; draw(circle(O, r1)); pair A = O + (0, -r1); pair B = O + (0, r1); draw(A--B); pair C = O + (0, -1.25*(r1 + r2)); pair D = O + (0, 1.25*r1); draw(C--D, dashed); dot(O); pair E = (2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)); pair F = (0, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)); pair G = (r2 * O + r1 * E) / (r1 + r2); pair H = (r2 * O + r1 * F) / (r1 + r2); draw(circle(E, r2)); draw(circle((-2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)), r2)); draw(O--E, dashed); draw(F--E, dashed); draw(G--H, dashed); label("$O$", O, SW); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NW); label("$D$", D, SW); label("$E_o$", E, NE); label("$F_o$", F, SW); label("$G_o$", G, S); label("$H_o$", H, W); label("$r_o$", 0.5 * H + 0.5 * G, NE); label("$3$", 0.5 * E + 0.5 * G, NE); label("$11$", 0.5 * O + 0.5 * G, NE); [/asy]
For both graphs, point $O$ is the center of sphere $S$ , and points $A$ and $B$ are the intersections of the sphere and the axis. Point $E$ (ignoring the subscripts) is one of the circle centers of the intersection of torus $T$ with section $\mathcal{P}$ . Point $G$ (again, ignoring the subscripts) is one of the tangents between the torus $T$ and sphere $S$ on section $\mathcal{P}$ $EF\bot CD$ $HG\bot CD$
And then, we can start our calculation.
In both cases, we know $\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}$
Hence, in the case of internal tangent, $\frac{E_iF_i}{OE_i} =\frac{G_iH_i}{OG_i}\Longrightarrow \frac{6}{11-3} =\frac{r_i}{11}\Longrightarrow r_i=\frac{33}{4}$
In the case of external tangent, $\frac{E_oF_o}{OE_o} =\frac{G_oH_o}{OG_o}\Longrightarrow \frac{6}{11+3} =\frac{r_o}{11}\Longrightarrow r_o=\frac{33}{7}$
Thereby, $r_i-r_o=\frac{33}{4}-\frac{33}{7}=\frac{99}{28}$ . And there goes the answer, $99+28=\boxed{127}$
| 127
|
5,237
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_9
| 2
|
There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that:
|
Note that the answer is equivalent to the number of ways to choose rows and columns that the white chips occupy, as once those are chosen, there is only one way to place the chips, and every way to place the chips corresponds to a set of rows and columns occupied by the white pieces.
If the white pieces occupy none of the rows, then because they don't appear on the board, they will not occupy any of the columns. Similar logic can be applied to show that if white pieces occupy all of the rows, they will also occupy all of the columns.
The number of sets of rows and columns that white can occupy are $2^{5} - 2 = 30$ each, accounting for the empty and full set.
So, including the board with 25 white pieces and the board with 25 black pieces, the answer is $30^{2}+2 = \boxed{902}$
| 902
|
5,238
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_9
| 3
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There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that:
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Case 1: All chips on the grid have the same color.
In this case, all cells are occupied with chips with the same color.
Therefore, the number of configurations in this case is 2.
Case 2: Both black and white chips are on the grid.
Observation 1: Each colored chips must occupy at least one column and one row.
This is because, for each given color, there must be at least one chip. Therefore, all chips placed in the cells that are in the same row or the same column with this given chip must have the same color with this chip.
Observation 2: Each colored chips occupy at most 4 rows and 4 columns.
This directly follows from Observation 1.
Observation 3: For each color, if all chips in this color occupy columns with $x$ -coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$ -coordinates $\left\{ y_1, \cdots, y_n \right\}$ , then every cell $\left( x, y \right)$ with $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$ is occupied by a chip with the same color.
This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.
If there is any cell in this region that is empty, then it violates Condition 3.
Observation 4: For each color, if all chips in this color occupy columns with $x$ -coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$ -coordinates $\left\{ y_1, \cdots, y_n \right\}$ , then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$ , or $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is empty.
This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.
Observation 5: For each color, if all chips in this color occupy columns with $x$ -coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$ -coordinates $\left\{ y_1, \cdots, y_n \right\}$ , then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is occupied by chips with the different color.
This follows from Condition 3.
By using the above observations, the number of feasible configurations in this case is given by
\begin{align*}
\sum_{n=1}^4 \sum_{m=1}^4 \binom{5}{n} \binom{5}{m}
& = \left( \sum_{n=1}^4 \binom{5}{n} \right)
\left( \sum_{m=1}^4 \binom{5}{m} \right) \\
& = \left( 2^5 - 2 \right)^2 \\
& = 900 .
\end{align*}
Putting all cases together, the total number of feasible configurations is $2 + 900 = \boxed{902}$
| 902
|
5,239
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_10
| 1
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Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$ , circumradius $13$ , and inradius $6$ . Find $AB\cdot AC$
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By Euler's formula $OI^{2}=R(R-2r)$ , we have $OI^{2}=13(13-12)=13$ . Thus, by the Pythagorean theorem, $AI^{2}=13^{2}-13=156$ . Let $AI\cap(ABC)=M$ ; notice $\triangle AOM$ is isosceles and $\overline{OI}\perp\overline{AM}$ which is enough to imply that $I$ is the midpoint of $\overline{AM}$ , and $M$ itself is the midpoint of $II_{a}$ where $I_{a}$ is the $A$ -excenter of $\triangle ABC$ . Therefore, $AI=IM=MI_{a}=\sqrt{156}$ and \[AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.\]
| 468
|
5,240
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_10
| 2
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Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$ , circumradius $13$ , and inradius $6$ . Find $AB\cdot AC$
|
Denote $AB=a, AC=b, BC=c$ . By the given condition, $\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6$ , where $A$ is the area of $\triangle{ABC}$
Moreover, since $OI\bot AI$ , the second intersection of the line $AI$ and $(ABC)$ is the reflection of $A$ about $I$ , denote that as $D$ . By the incenter-excenter lemma, $DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c$
Thus, we have $\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c$ . Now, we have $\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}$
| 468
|
5,241
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_11
| 1
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Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
\begin{equation*}
a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
\end{equation*}
|
$ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000$
Note $(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0$ . Thus, $a/b/c=100$ . There are $201$ cases for each but we need to subtract $2$ for $(100,100,100)$ . The answer is $\boxed{601}$
| 601
|
5,242
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_11
| 2
|
Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
\begin{equation*}
a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
\end{equation*}
|
We have
\begin{align*}
& a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \\
& = ab \left( a + b \right) + bc \left( b + c \right) + ca \left( c + a \right) \\
& = ab \left( 300 - c \right) + bc \left( 300 - a \right) + ca \left( 300 - b \right) \\
& = 300 \left( ab + bc + ca \right) - 3 abc \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 10^4 \left( a + b + c \right) + 10^6
\right) \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 2 \cdot 10^6
\right) \\
& = 6 \cdot 10^6 .
\end{align*}
The first and the fifth equalities follow from the condition that $a+b+c = 300$
Therefore,
\[
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .
\]
Case 1: Exactly one out of $a - 100$ $b - 100$ $c - 100$ is equal to 0.
Step 1: We choose which term is equal to 0. The number ways is 3.
Step 2: For the other two terms that are not 0, we count the number of feasible solutions.
W.L.O.G, we assume we choose $a - 100 = 0$ in Step 1. In this step, we determine $b$ and $c$
Recall $a + b + c = 300$ . Thus, $b + c = 200$ .
Because $b$ and $c$ are nonnegative integers and $b - 100 \neq 0$ and $c - 100 \neq 0$ , the number of solutions is 200.
Following from the rule of product, the number of solutions in this case is $3 \cdot 200 = 600$
Case 2: At least two out of $a - 100$ $b - 100$ $c - 100$ are equal to 0.
Because $a + b + c = 300$ , we must have $a = b = c = 100$
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of solutions is $600 + 1 = \boxed{601}$
| 601
|
5,243
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_11
| 3
|
Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
\begin{equation*}
a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
\end{equation*}
|
We will use Vieta's formulas to solve this problem. We assume $a + b + c = 300$ $ab + bc + ca = m$ , and $abc = n$ . Thus $a$ $b$ $c$ are the three roots of a cubic polynomial $f(x)$
We note that $300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n$ , which simplifies to $100m - 2000000 = n$
Our polynomial $f(x)$ is therefore equal to $x^3 - 300x^2 + mx - (100m - 2000000)$ . Note that $f(100) = 0$ , and by polynomial division we obtain $f(x) = (x - 100)(x^2 - 200x - (m-20000))$
We now notice that the solutions to the quadratic equation above are $x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}$ , and that by changing the value of $m$ we can let the roots of the equation be any pair of two integers which sum to $200$ . Thus any triple in the form $(100, 100 - x, 100 + x)$ where $x$ is an integer between $0$ and $100$ satisfies the conditions.
Now to count the possible solutions, we note that when $x \ne 100$ , the three roots are distinct; thus there are $3! = 6$ ways to order the three roots. As we can choose $x$ from $0$ to $99$ , there are $100 \cdot 3! = 600$ triples in this case. When $x = 100$ , all three roots are equal to $100$ , and there is only one triple in this case.
In total, there are thus $\boxed{601}$ distinct triples.
| 601
|
5,244
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_11
| 4
|
Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
\begin{equation*}
a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
\end{equation*}
|
Let's define $a=100+x$ $b=100+y$ $c=100+z$ . Then we have $x+y+z=0$ and $6000000 = \sum a^2(b+c)$
$= \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) = \sum (20000 - 10000 x + x(40000-x^2))$
$= \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3$ , so we get $x^3 + y^3 + z^3 = 0$ . Then from $x+y+z = 0$ , we can find $0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz$ , which means that one of $a$ $b$ $c$ must be 0. There are 201 solutions for each of $a=0$ $b=0$ and $c=0$ , and subtract the overcounting of 2 for solution $(200, 200, 200)$ , the final result is $201 \times 3 - 2 = \boxed{601}$
| 601
|
5,245
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_12
| 3
|
Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point \(C\) on \(\overline{AB}\), distinct from \(A\) and \(B\), that does not belong to any segment from \(\mathcal{F}\) other than \(\overline{AB}\). Then \(OC^2=\tfrac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).
|
The equation of line $AB$ is \[ y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (1) \]
The position of line $PQ$ can be characterized by $\angle QPO$ , denoted as $\theta$ .
Thus, the equation of line $PQ$ is
\[ y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) \]
Solving (1) and (2), the $x$ -coordinate of the intersecting point of lines $AB$ and $PQ$ satisfies the following equation:
\[ \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} + \frac{x}{\cos \theta} = 1 . \hspace{1cm} (1) \]
We denote the L.H.S. as $f \left( \theta; x \right)$
We observe that $f \left( 60^\circ ; x \right) = 1$ for all $x$ .
Therefore, the point $C$ that this problem asks us to find can be equivalently stated in the following way:
We interpret Equation (1) as a parameterized equation that $x$ is a tuning parameter and $\theta$ is a variable that shall be solved and expressed in terms of $x$ .
In Equation (1), there exists a unique $x \in \left( 0, 1 \right)$ , denoted as $x_C$ $x$ -coordinate of point $C$ ), such that the only solution is $\theta = 60^\circ$ . For all other $x \in \left( 0, 1 \right) \backslash \{ x_C \}$ , there are more than one solutions with one solution $\theta = 60^\circ$ and at least another solution.
Given that function $f \left( \theta ; x \right)$ is differentiable, the above condition is equivalent to the first-order-condition \[ \frac{\partial f \left( \theta ; x_C \right) }{\partial \theta} \bigg|_{\theta = 60^\circ} = 0 . \]
Calculating derivatives in this equation, we get \[ - \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} + x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} = 0. \]
By solving this equation, we get \[ x_C = \frac{1}{8} . \]
Plugging this into Equation (1), we get the $y$ -coordinate of point $C$ \[ y_C = \frac{3 \sqrt{3}}{8} . \]
Therefore,
\begin{align*}
OC^2 & = x_C^2 + y_C^2 \\
& = \frac{7}{16} .
\end{align*}
Therefore, the answer is $7 + 16 = \boxed{23}$
| 23
|
5,246
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_13
| 1
|
Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when \[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\] is divided by 1000.
|
\[\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)\]
Now, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten product from $0$ to $12$ , we see that both instances of $\omega^k$ cycle through each of the 13th roots. Then, our answer is:
\[((1 + i)^{13} - 1)(1 - i)^{13} - 1)\]
\[= (-64(1 + i) - 1)(-64(1 - i) - 1)\]
\[= (65 + 64i)(65 - 64i)\]
\[= 65^2 + 64^2\]
\[= 8\boxed{321}\]
| 321
|
5,247
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_13
| 2
|
Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when \[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\] is divided by 1000.
|
To find $\prod_{k=0}^{12} (2 - 2w^k + w^{2k})$ , where $w\neq1$ and $w^{13}=1$ , rewrite this is as
$(r-w)(s-w)(r-w^2)(s-w^2)...(r-w^{12})(s-w^{12})$ where $r$ and $s$ are the roots of the quadratic $x^2-2x+2=0$
Grouping the $r$ 's and $s$ 's results in $\frac{r^{13}-1}{r-1} \cdot\frac{s^{13}-1}{s-1}$
the denomiator $(r-1)(s-1)=1$ by vietas.
the numerator $(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321$ by newtons sums
so the answer is $\boxed{321}$
| 321
|
5,248
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_13
| 3
|
Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when \[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\] is divided by 1000.
|
Denote $r_j = e^{\frac{i 2 \pi j}{13}}$ for $j \in \left\{ 0, 1, \cdots , 12 \right\}$
Thus, for $\omega \neq 1$ $\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)$ is a permutation of $\left( r_0, r_1, \cdots, r_{12} \right)$
We have
\begin{align*}\
\Pi_{k = 0}^{12} \left( 2 - 2 \omega^k + \omega^{2k} \right)
& = \Pi_{k=0}^{12} \left( 1 + i - \omega^k \right)
\left( 1 - i - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - \omega^k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) \\
& = \left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\right)
\left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right)
\right) . \hspace{1cm} (1)
\end{align*}
The third equality follows from the above permutation property.
Note that $r_0, r_1, \cdots , r_{12}$ are all zeros of the polynomial $z^{13} - 1$ .
Thus, \[ z^{13} - 1 = \Pi_{k=0}^{12} \left( z - r_k \right) . \]
Plugging this into Equation (1), we get
\begin{align*}
(1)
& = \left( \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^{13} - 1 \right)
\left( \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^{13} - 1 \right) \\
& = \left( - 2^{13/2} e^{i \frac{\pi}{4}} - 1 \right)
\left( - 2^{13/2} e^{-i \frac{\pi}{4}} - 1 \right) \\
& = 2^{13} + 1 + 2^{13/2} \cdot 2 \cos \frac{\pi}{4} \\
& = 2^{13} + 1 + 2^7 \\
& = 8321 .
\end{align*}
Therefore, the answer is $\boxed{321}$
| 321
|
5,249
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_14
| 1
|
Let \(b\ge 2\) be an integer. Call a positive integer \(n\) \(b\text-\textit{eautiful}\) if it has exactly two digits when expressed in base \(b\) and these two digits sum to \(\sqrt n\). For example, \(81\) is \(13\text-\textit{eautiful}\) because \(81 = \underline{6} \ \underline{3}_{13} \) and \(6 + 3 = \sqrt{81}\). Find the least integer \(b\ge 2\) for which there are more than ten \(b\text-\textit{eautiful}\) integers.
|
We write the base- $b$ two-digit integer as $\left( xy \right)_b$ .
Thus, this number satisfies \[ \left( x + y \right)^2 = b x + y \] with $x \in \left\{ 1, 2, \cdots , b-1 \right\}$ and $y \in \left\{ 0, 1, \cdots , b - 1 \right\}$
The above conditions imply $\left( x + y \right)^2 < b^2$ . Thus, $x + y \leq b - 1$
The above equation can be reorganized as \[ \left( x + y \right) \left( x + y - 1 \right) = \left( b - 1 \right) x . \]
Denote $z = x + y$ and $b' = b - 1$ .
Thus, we have \[ z \left( z - 1 \right) = b' x , \hspace{1cm} (1) \] where $z \in \left\{ 2, 3, \cdots , b' \right\}$ and $x \in \left\{ 1, 2, \cdots , b' \right\}$
Next, for each $b'$ , we solve Equation (1).
We write $b'$ in the prime factorization form as $b' = \Pi_{i=1}^n p_i^{k_i}$ .
Let $\left(A, \bar A \right)$ be any ordered partition of $\left\{ 1, 2, \cdots , n \right\}$ (we allow one set to be empty).
Denote $P_A = \Pi_{i \in A} p_i^{k_i}$ and $P_{\bar A} = \Pi_{i \in \bar A} p_i^{k_i}$
Because ${\rm gcd} \left( z, z-1 \right) = 1$ , there must exist such an ordered partition, such that $P_A | z$ and $P_{\bar A} | z-1$
Next, we prove that for each ordered partition $\left( A, \bar A \right)$ , if a solution of $z$ exists, then it must be unique.
Suppose there are two solutions of $z$ under partition $\left( A, \bar A \right)$ $z_1 = c_1 P_A$ $z_1 - 1 = d_1 P_{\bar A}$ , and $z_2 = c_2 P_A$ $z_2 - 1 = d_2 P_{\bar A}$ .
W.L.O.G., assume $c_1 < c_2$ .
Hence, we have \[ \left( c_2 - c_1 \right) P_A = \left( d_2 - d_1 \right) P_{\bar A} . \]
Because ${\rm gcd} \left( P_A, P_{\bar A} \right) = 1$ and $c_1 < c_2$ , there exists a positive integer $m$ , such that $c_2 = c_1 + m P_{\bar A}$ and $d_2 = d_1 + m P_A$ .
Thus,
\begin{align*}
z_2 & = z_1 + m P_A P_{\bar A} \\
& = z_1 + m b' \\
& > b' .
\end{align*}
However, recall $z_2 \leq b'$ . We get a contradiction.
Therefore, under each ordered partition for $b'$ , the solution of $z$ is unique.
Note that if $b'$ has $n$ distinct prime factors, the number of ordered partitions is $2^n$ .
Therefore, to find a $b'$ such that the number of solutions of $z$ is more than 10, the smallest $n$ is 4.
With $n = 4$ , the smallest number is $2 \cdot 3 \cdot 5 \cdot 7 = 210$ .
Now, we set $b' = 210$ and check whether the number of solutions of $z$ under this $b'$ is more than 10.
We can easily see that all ordered partitions (except $A = \emptyset$ ) guarantee feasible solutions of $z$ .
Therefore, we have found a valid $b'$ .
Therefore, $b = b' + 1 = \boxed{211}$
| 211
|
5,250
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15
| 1
|
Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy]
|
By Furaken
There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1, and those whose sides are not (Case 2).
For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangle's sides coincide with these segments as shown in the diagram. [asy] real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); pair B1 = (0.5,0.5); pair B2 = (-0.5,0.5); pair B3 = (-0.5,-0.5); pair B4 = (0.5,-0.5); dot(B1); dot(B2); dot(B3); dot(B4); draw(A1--A5--A7--A11--cycle); draw(A2--A4--A8--A10--cycle); draw(A3--A9); draw(A6--A12); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); label("$B_1$", B1, SW); label("$B_2$", B2, SE); label("$B_3$", B3, NE); label("$B_4$", B4, NW); [/asy] We use inclusion-exclusion for this. There are 30 valid rectangles contained in $A_1A_5A_7A_{11}$ , as well as 30 in $A_2A_4A_8A_{10}$ . However, the 9 rectangles contained in $B_1B_2B_3B_4$ have been counted twice, so we subtract 9 and we have 51 rectangles in the diagram. Multiplying by 3, we get 153 rectangles for Case 1.
For Case 2, we have this diagram. To be honest, you can count the rectangles here in whatever way you like. [asy] real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); draw(A1--A6--A7--A12--cycle); draw(A3--A4--A9--A10--cycle); draw(A2--A5--A8--A11--cycle); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); [/asy] There are 36 rectangles contained within $A_2A_5A_8A_{11}$ , and 18 that use points outside $A_2A_5A_8A_{11}$ . So we get a total of $3(36+18)=162$ rectangles for Case 2.
Adding the two cases together, we get the answer $\boxed{315}$
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15
| 2
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Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy]
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Using the same diagram as Solution 1, we can get the number of rectangles from Case 1 by adding the number of rectangles of $A_2$ $A_8$ $A_8$ $A_{10}$ and $A_1$ $A_5$ $A_7$ $A_{11}$ and then subtracting the overlaps,
\[\binom{5}{2}\binom{3}{2} + \binom{5}{2}\binom{3}{2} - \binom{3}{2}\binom{3}{2}\] \[=51\]
We multiply this by 3 to get the total number of rectangles for Case 1, which is 153.
For Case 2, we can first get the total number of rectangles from $A_2A_3A_4A_5A_8A_9A_{10}A_{11}$ then add $A_1A_6A_7A_{12}$ and subtract by the overlaps, \[\binom{4}{2}\binom{4}{2} + \binom{6}{2} - \binom{4}{2} + \binom{6}{2} - \binom{4}{2}\] \[= 54\] Multiply that by 3 and add it to Case 1 to get $\boxed{315}$
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https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15
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Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy]
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We put the dodecagon in the right position that there exists a side whose slope is 0.
Note that finding a rectangle is equivalent to finding two pairs of lines, such that two lines in each pair are parallel and lines from different pairs are perpendicular.
Now, we use this property to count the number of rectangles.
Because two pairs of lines that form a rectangle are perpendicular, we only need to use the slope of one pair, denoted as $k$ , to determine the direction of the rectangle.
The slope of the other pair is thus $- \frac{1}{k}$ . To avoid overcounting, we do casework analysis by defining each case in term of $0 \leq k < \infty$ only (we make a convention that if $k = 0$ , then $- \frac{1}{k} = \infty$ ).
In our counting, we will frequently quantify the distance between two vertices of the regular dodecagon.
To characterize this in a straightforward way, we simply measure the number of vertices (on the minor arc) between our measured two vertices. For instance, two vertices on a side has distance 0. Distances between two vertices that are diagonals can be 1, 2, 3, 4, 5.
Case 1: $k = 0, \tan 30^\circ, \tan 60^\circ$
We only count for $k = 0$ . The number of solutions for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Consider $k = 0$ .
We need to find a pair of horizontal segments and a pair of vertical segments to form a rectangle.
For $k = 0$ , the length of each horizontal segment can only be 0, 2, 4.
Denote by $2i$ the shorter length of two parallel horizontal segments.
Given $i$ , the number of pairs of two parallel horizontal segments is $1 + 2 \left( 4 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 2}{2}$
Therefore, for $k = 0$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^2 \left( 1 + 2 \left( 4 - 2 i \right) \right)
\binom{2i + 2}{2}
& = 54 .
\end{align*}
The number of rectangles for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Therefore, the total number of rectangles in this case is $54 \cdot 3 = 162$
Case 2: $k = \tan 15^\circ$ $\tan 45^\circ$ $\tan 75^\circ$
The number of rectangles under all these $k$ s are the same.
So we only count for $k = \tan 15^\circ$
For $k = \tan 15^\circ$ , the length of each segment can only be 1, 3, 5.
However, there is only one segment with length 5.
So this cannot be the shorter length of two parallel segments with slope $\tan 15^\circ$
Denote by $2i + 1$ the shorter length of two parallel segments with slope $\tan 15^\circ$ .
Given $i$ , the number of pairs of two parallel segments is $1 + 2 \left( 3 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 3}{2}$
Therefore, for $k = \tan 15^\circ$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^1 \left( 1 + 2 \left( 3 - 2 i \right) \right)
\binom{2i + 3}{2}
& = 51 .
\end{align*}
The number of rectangles for $k = \tan 45^\circ$ and $\tan 75^\circ$ are the same.
Therefore, the total number of rectangles in this case is $51 \cdot 3 = 153$
Putting all cases together, the total number of rectangles is $162 + 153 = \boxed{315}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions.
First, there are $\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions.
Together, the requested probability is \[\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{21\cdot(10\cdot8\cdot6\cdot4\cdot2)}{14\cdot13\cdot12\cdot11\cdot10} = \frac{48}{143},\] from which the answer is $48+143 = \boxed{191}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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We can simply just loop through each of the men and find the probability that the person opposite from him is a woman.
Start by sitting down the $1$ st man. The probability that the person opposite to him is a woman is $\frac{9}{13}$ since out of the $13$ people who can sit opposite to him, $9$ can be a woman. With the $2$ nd man, we can use the same logic: there are $11$ people who can sit opposite to him, but only $8$ of them are a woman, so the probability is $\frac{8}{11}.$ We use the same logic for the $3$ rd, $4$ th and $5$ th men to get probabilities of $\frac{7}{9}$ $\frac{6}{7}$ and $\frac{5}{5},$ respectively.
Multiplying these probabilities, we get a final answer of \[\frac{9}{13}\cdot\frac{8}{11}\cdot\frac{7}{9}\cdot\frac{6}{7}\cdot\frac{5}{5}=\frac{48}{143}\longrightarrow\boxed{191}.\]
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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This problem is equivalent to solving for the probability that no man is standing diametrically opposite to another man. We can simply just construct this.
We first place the $1$ st man anywhere on the circle, now we have to place the $2$ nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of $\frac{12}{13}$ because there are $13$ available spots, and $12$ of them are not opposite to the first man.
We do the same thing for the $3$ rd man, finding a spot for him such that he is not opposite to the other $2$ men, which would happen with a probability of $\frac{10}{12}$ using similar logic. Doing this for the $4$ th and $5$ th men, we get probabilities of $\frac{8}{11}$ and $\frac{6}{10}$ respectively.
Multiplying these probabilities, we get, \[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\]
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange $5$ men in a circle and the number of ways to arrange $9$ women in a circle, are constants.) The total number of ways to arrange $5$ M's and $9$ W's is $\binom{14}{5} = 2002.$
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly $2$ of the pairs of diametrically opposite positions must be occupied by $2$ W's. There are $\binom{7}{2} = 21$ ways to choose these $2$ pairs. For the remaining $5$ pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in $2^{5} = 32$ ways. Therefore, there are $21*32 = 672$ valid arrangements.
Therefore, the probability that an arrangement is valid is $\frac{672}{2002} = \frac{48}{143}$ for an answer of $\boxed{191}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
| 5
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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To start off, we calculate the total amount of ways to organize all $14$ people irrespective of any constraints. This is simply ${14\choose5} = 2002$ , because we just count how many ways we can place all $5$ men in any of the $14$ slots.
Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle unable to hold any men. This means that placing down one man will effectively take away $2$ spots.
There are $14$ possible slots the first man can be placed. Once that man was placed, the next man only have $12$ possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has $10$ possible slots. The fourth man has $8$ possible slots, and the fifth man has $6$ possible slots.
This means the number of ways you can place all $5$ men down is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ . However, since the men are all indistinct from each other, you also have to divide that value by $5! = 120$ , since there are $120$ ways to arrange the $5$ men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: $\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6}{5!} = 672$ . The women simply fill in the rest of the available slots in each arrangement of men.
Thus, the final probability is $\frac{672}{2002} = \frac{48}{143}$ , meaning the answer is $48 + 143 = \boxed{191}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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We will first assign seats to the men. The first man can be placed in any of the $14$ slots. The second man can be placed in any of the remaining $13$ seats, except for the one diametrically opposite to the first man. So, there are $13 - 1 = 12$ ways to seat him. With a similar argument, the third man can be seated in $10$ ways, the fourth man in $8$ ways and the last man in $6$ ways.
So, the total number of ways to arrange the men is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$
The women go to the remaining $9$ spots. Note that since none of the seats diametrically opposite to the men is occupied, each man is opposite a woman. The number of ways to arrange the women is therefore, simply $9!$ , meaning that the total number of ways to arrange the people with restrictions is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!$ In general, there are $14!$ ways to arrange the people without restrictions. So, the probability is \[\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!}{14!} = \frac{8 \cdot 6}{13 \cdot 11} = \frac{48}{143}.\] The answer is $48 + 143 = \boxed{191}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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First pin one man on one seat (to ensure no rotate situations). Then there are $13!$ arrangements.
Because $5$ men must have women at their opposite side, we consider the $2$ nd man and the woman opposite as one group and name it $P_2.$ There are $4$ groups, $P_1, P_2, P_3, P_4$ except the first man pinned on the same point. And for the rest $4$ women, name them $P_5$ and $P_6.$ First to order $P_1, P_2, P_3, P_4, P_5, P_6,$ there are $6!$ ways. For the $1$ st man, there are $9$ women to choose, $8$ for the $2$ nd, $\ldots,$ $5$ for the $5$ th, and then for the $2$ women pairs $3$ and $1.$ Because every $2$ person in the group have chance to change their position, there are $2^6$ possibilities.
So the possibility is \[P=\frac{6!\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 3\cdot 1 \cdot 2^6}{13!}=\frac{48}{143}.\]
The answer is $48+143=\boxed{191}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
| 8
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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We get around the condition that each man can't be opposite to another man by simply considering all $7$ diagonals, and choosing $5$ where there will be a single man. For each diagonal, the man can go on either side, and there are $\binom{14}{5}$ ways to arrange the men and the women in total. Thus our answer is $\frac{\binom{7}{5}\cdot 2^5}{\binom{14}{5}} = \frac{48}{143}.$ We get $48 + 143 = \boxed{191}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
| 9
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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We can find the probability of one arrangement occurring, and multiply it by the total number of arrangements.
The probability of a man being in any specific position is $\frac{5}{14}.$ The probability of a woman being across from him is $\frac{9}{13}.$ The probability of a man being in any valid position is now $\frac{4}{12},$ and the probability of a woman being across from him is $\frac{8}{11},$ and so forth. We stop when there are no more men left. Multiplying these probabilities together,
\[P(\mathrm{One\ successful\ outcome})=\frac{5}{14}\cdot \frac{9}{13}\cdot \frac{4}{12}\cdot \frac{8}{11}\cdot \frac{3}{10}\cdot \frac{7}{9}\cdot \frac{2}{8}\cdot \frac{1}{6}\cdot \frac{6}{7}\cdot \frac{5}{5} = \frac{1}{2002}.\] To find the total number of successful outcomes, we consider the diagonals; the total number of diagonals to be made is $\binom75$ , since there are $7$ total diagonals, and we want to choose $5$ of them to connect a man to a woman. For each of these diagonals, the man can be on either side of the diagonal.
It follows that there are $2$ possibilities for each diagonal (man on one side, woman on the other, and vice versa). There are $5$ diagonals with a man and a woman, so there are $2^5$ different ways for these diagonals to appear.
There are $\binom75$ successful diagonals, and for each of these diagonals, there are $2^5$ ways to seat the men and the women, there are $\binom75$ $\cdot 2^5$ successful outcomes.
Recall that \[P(\mathrm{Any\ successful\ outcome})=P(\mathrm{One\ successful\ outcome})\cdot P(\mathrm{Total\ number\ of\ successful\ outcomes}).\] Therefore, \[P(\mathrm{Any\ successful\ outcome}) = \frac{1}{2002}\cdot 2^5\cdot \binom75 = \frac{1}{2002}\cdot 2^5\cdot 21 = \frac{2^5\cdot 21}{2002} = \frac{48}{143}.\] The requested sum is $48+143=\boxed{191}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_3
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A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.
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In this solution, let $\boldsymbol{n}$ -line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$ -line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:
It follows that the $2$ -line points account for $780-9-24-50-90=\boxed{607}$ pairs of lines, where each pair intersect at a single point.
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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By the Inscribed Angle Theorem , we conclude that $\triangle PAC$ and $\triangle PBD$ are right triangles.
Let the brackets denote areas. We are given that \begin{alignat*}{8} 2[PAC] &= PA \cdot PC &&= 56, \\ 2[PBD] &= PB \cdot PD &&= 90. \end{alignat*} Let $O$ be the center of the circle, $X$ be the foot of the perpendicular from $P$ to $\overline{AC},$ and $Y$ be the foot of the perpendicular from $P$ to $\overline{BD},$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); [/asy] Let $d$ be the diameter of $\odot O.$ It follows that \begin{alignat*}{8} 2[PAC] &= d\cdot PX &&= 56, \\ 2[PBD] &= d\cdot PY &&= 90. \end{alignat*} Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have \[PX^2+PY^2=PO^2.\] We rewrite this equation in terms of $d:$ \[\left(\frac{56}{d}\right)^2+\left(\frac{90}{d}\right)^2=\left(\frac d2\right)^2,\] from which $d^2=212.$ Therefore, we get \[[ABCD] = \frac{d^2}{2} = \boxed{106}.\] ~MRENTHUSIASM
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); [/asy] Let the center of the circle be $O$ , and the radius of the circle be $r$ . Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$ , its area is $\dfrac{1}{2}(2r)(2r) = 2r^2$ . Since $AC$ and $BD$ are diameters of the circle, $\triangle APC$ and $\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitudes to $AC$ and $BD$ , respectively. We have \[[\triangle APC] = \frac{1}{2}(PA)(PC) = \frac{1}{2}(PX)(AC),\] so $PX = \dfrac{(PA)(PC)}{AC} = \dfrac{28}{r}$ . Similarly, \[[\triangle BPD] = \frac{1}{2}(PB)(PD) = \frac{1}{2}(PY)(PB),\] so $PY = \dfrac{(PB)(PD)}{BD} = \dfrac{45}{r}$ . Since $\triangle APX \sim \triangle PCX,$ \[\frac{AX}{PX} = \frac{PX}{CX}\] \[\frac{AO - XO}{PX} = \frac{PX}{OC + XO}.\] But $PXOY$ is a rectangle, so $PY = XO$ , and our equation becomes \[\frac{r - PY}{PX} = \frac{PX}{r + PY}.\] Cross multiplying and rearranging gives us $r^2 = PX^2 + PY^2 = \left(\dfrac{28}{r}\right)^2 + \left(\dfrac{45}{r}\right)^2$ , which rearranges to $r^4 = 2809$ . Therefore $[ABCD] = 2r^2 = \boxed{106}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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Drop a height from point $P$ to line $\overline{AC}$ and line $\overline{BC}$ . Call these two points to be $X$ and $Y$ , respectively. Notice that the intersection of the diagonals of $\square ABCD$ meets at a right angle at the center of the circumcircle, call this intersection point $O$
Since $OXPY$ is a rectangle, $OX$ is the distance from $P$ to line $\overline{BD}$ . We know that $\tan{\angle{POX}} = \frac{PX}{XO} = \frac{28}{45}$ by triangle area and given information. Then, notice that the measure of $\angle{OCP}$ is half of $\angle{XOP}$
Using the half-angle formula for tangent,
\begin{align*} \frac{(2 \cdot \tan{\angle{OCP}})}{(1-\tan^2{\angle{OCP}})} = \tan{\angle{POX}} = \frac{28}{45} \\ 14\tan^2{\angle{OCP}} + 45\tan{\angle{OCP}} - 14 = 0 \end{align*}
Solving the equation above, we get that $\tan{\angle{OCP}} = -7/2$ or $2/7$ . Since this value must be positive, we pick $\frac{2}{7}$ . Then, $\frac{PA}{PC} = 2/7$ (since $\triangle CAP$ is a right triangle with line $\overline{AC}$ the diameter of the circumcircle) and $PA * PC = 56$ . Solving we get $PA = 4$ $PC = 14$ , giving us a diagonal of length $\sqrt{212}$ and area $\boxed{106}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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Denote by $x$ the half length of each side of the square.
We put the square to the coordinate plane, with $A = \left( x, x \right)$ $B = \left( - x , x \right)$ $C = \left( - x , - x \right)$ $D = \left( x , - x \right)$
The radius of the circumcircle of $ABCD$ is $\sqrt{2} x$ .
Denote by $\theta$ the argument of point $P$ on the circle.
Thus, the coordinates of $P$ are $P = \left( \sqrt{2} x \cos \theta , \sqrt{2} x \sin \theta \right)$
Thus, the equations $PA \cdot PC = 56$ and $PB \cdot PD = 90$ can be written as \begin{align*} \sqrt{\left( \sqrt{2} x \cos \theta - x \right)^2 + \left( \sqrt{2} x \sin \theta - x \right)^2} \cdot \sqrt{\left( \sqrt{2} x \cos \theta + x \right)^2 + \left( \sqrt{2} x \sin \theta + x \right)^2} & = 56 \\ \sqrt{\left( \sqrt{2} x \cos \theta + x \right)^2 + \left( \sqrt{2} x \sin \theta - x \right)^2} \cdot \sqrt{\left( \sqrt{2} x \cos \theta - x \right)^2 + \left( \sqrt{2} x \sin \theta + x \right)^2} & = 90 \end{align*}
These equations can be reformulated as \begin{align*} x^4 \left( 4 - 2 \sqrt{2} \left( \cos \theta + \sin \theta \right) \right) \left( 4 + 2 \sqrt{2} \left( \cos \theta + \sin \theta \right) \right) & = 56^2 \\ x^4 \left( 4 + 2 \sqrt{2} \left( \cos \theta - \sin \theta \right) \right) \left( 4 - 2 \sqrt{2} \left( \cos \theta - \sin \theta \right) \right) & = 90^2 \end{align*}
These equations can be reformulated as \begin{align*} 2 x^4 \left( 1 - 2 \cos \theta \sin \theta \right) & = 28^2 \hspace{1cm} (1) \\ 2 x^4 \left( 1 + 2 \cos \theta \sin \theta \right) & = 45^2 \hspace{1cm} (2) \end{align*}
Taking $\frac{(1)}{(2)}$ , by solving the equation, we get \[ 2 \cos \theta \sin \theta = \frac{45^2 - 28^2}{45^2 + 28^2} . \hspace{1cm} (3) \]
Plugging (3) into (1), we get \begin{align*} {\rm Area} \ ABCD & = \left( 2 x \right)^2 \\ & = 4 \sqrt{\frac{28^2}{2 \left( 1 - 2 \cos \theta \sin \theta \right)}} \\ & = 2 \sqrt{45^2 + 28^2} \\ & = 2 \cdot 53 \\ & = \boxed{106}
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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[asy] pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,"$A$",NW); label(B,"$B$",NE); label(C,"$C$",SE); label(D,"$D$",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,"$P$", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); [/asy] Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$ . Let the center of the circle be $(0,0)$ .
The two equations would thus become: \[\left(\left(a+\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a-\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=56^2\] \[\left(\left(a-\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a+\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=90^2\] Now, let $m=\left(a+\dfrac{k}2\right)^2$ $n=\left(a-\dfrac{k}2\right)^2$ $o=\left(b+\dfrac{k}2\right)^2$ , and $p=\left(b-\dfrac{k}2\right)^2$ . Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$ . Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\cdot146$ . Substituting back in and expanding, we have $2ak\cdot-2bk=34\cdot146$ , so $abk^2=-17\cdot73$ . We now have one of our terms we need ( $k^2$ ). Therefore, we only need to find $ab$ to find $k^2$ .
We now write the equation of the circle, which point $P$ satisfies: \[a^2+b^2=\left(\dfrac{k\sqrt{2}}{2}\right)^2=\dfrac{k^2}2\] We can expand the second equation, yielding \[\left(a^2+b^2+\dfrac{k^2}2+(ak+bk)\right)\left(a^2+b^2+\dfrac{k^2}2-(ak+bk)\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.\] Now, with difference of squares, we get $k^4-k^2\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$ . We can add $2abk^2=-17\cdot73\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$ . We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\left(k^2-\dfrac{k^2}2\right)=5618$ . We can combine like terms to get $k^2\cdot\dfrac{k^2}2=5618$ , so $(k^2)^2=11236$ .
Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$ , so we can factor $11236=2^2\cdot2809$ . With some testing with approximations and last-digit methods, we can find that $53^2=2809$ . Therefore, taking the square root, we find that $k^2$ , the area of square $ABCD$ , is $2\cdot53=\boxed{106}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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WLOG, let $P$ be on minor arc $AD.$ Draw in $AP$ $BP$ $CP$ $DP$ and let $\angle ABP = x.$ We can see, by the inscribed angle theorem, that $\angle APB = \angle ACB = 45$ , and $\angle CPD = \angle CAD = 45.$ Then, $\angle PAB = 135-x$ $\angle PCD = \angle PAD = (135-x)-90 = 45-x$ , and $\angle PDC = 90+x.$ Letting $(PA, PB, PC, PD, AB) = (a,b,c,d,s)$ , we can use the law of sines on triangles $PAB$ and $PCD$ to get \[s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(135-x)} = \frac{c}{\sin(90+x)} = \frac{d}{\sin(45-x)}.\] Making all the angles in the above equation acute gives \[s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(45+x)} = \frac{c}{\sin(90-x)} = \frac{d}{\sin(45-x)}.\]
Note that we are looking for $s^{2}.$ We are given that $ac = 56$ and $bd = 90.$ This means that $s^{2}\sin(x)\sin(90-x) = 28$ and $s^{2}\sin(45+x)\sin(45-x) = 45.$ However, \[\sin(x)\sin(90-x) = \sin(x)\cos(x) = \frac{\sin(2x)}{2}\] and \[\sin(45+x)\sin(45-x) = \frac{(\cos(x) + \sin(x))(\cos(x) - \sin(x))}{2} = \frac{\cos^{2}(x) - \sin^{2}(x)}{2} = \frac{\cos(2x)}{2}.\] Therefore, $s^{2}\sin(2x) = 56$ and $s^{2}\cos(2x) = 90.$ Therefore, by the Pythagorean Identity, \[s^{2} = \sqrt{(s^{2}\sin(2x))^{2} + (s^{2}\cos(2x))^{2}} = \sqrt{56^{2} + 90^{2}} = \boxed{106}.\]
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
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Similar to Solution 6, let $P$ be on minor arc $\overarc {AB}$ $r$ and $O$ be the radius and center of the circumcircle respectively, and $\theta = \angle AOP$ . Since $\triangle APC$ is a right triangle, $PA \cdot PC$ equals the hypotenuse, $2r$ , times its altitude, which can be represented as $r \sin \theta$ . Therefore, $2r^2 \sin \theta = 56$ . Applying similar logic to $\triangle BPD$ , we get $2r^2 \sin (90^\circ - \theta) = 2r^2 \cos \theta = 90$
Dividing the two equations, we have \begin{align*} \frac{\sin \theta}{\cos \theta} &= \frac{56}{90} \\ 56 \cos \theta &= 90 \sin \theta \\ (56 \cos \theta)^2 &= (90 \sin \theta)^2. \end{align*} Adding $(56 \sin \theta)^2$ to both sides allows us to get rid of $\cos \theta$ \begin{align*} (56 \cos \theta)^2 + (56 \sin \theta)^2 &= (90 \sin \theta)^2 + (56 \sin \theta)^2 \\ 56^2 &= (90^2 + 56^2)(\sin \theta)^2 \\ \frac{56^2}{90^2 + 56^2} &= (\sin \theta)^2 \\ \frac{28}{53} &= \sin \theta. \end{align*} Therefore, we have $2r^2\left(\frac{28}{53}\right) = 56$ , and since the area of the square can be represented as $2r^2$ , the answer is $56 \cdot \frac{53}{28} = \boxed{106}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6
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Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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Denote $E_n$ the expected number of cards Alice guesses correctly given $n$ red cards and $n$ black cards. We want to find $E_3$
Alice has a $\frac{1}{2}$ chance of guessing the first card. WLOG assume the first card color is red. For the next card, Alice has a $\frac{3}{5}$ chance of guessing the card if she chooses black; if they guess right, there's one less red and black card, so the expected number of cards Alice guesses from here is $E_2$ . If Alice does not guess correctly (which occurs with probability $\frac{2}{5}$ ), this means that there's 3 black cards and 1 red card left, so Alice should guess black next with a $\frac{3}{4}$ chance of being right. If Alice is wrong (with probability $\frac{1}{4}$ ), there are only 3 black cards left, so Alice can guess these with certainty; if Alice is right, there are 2 blacks and 1 red left, so Alice should again guess black. If Alice is right (with probability $\frac{2}{3}$ ), there is now 1 black and red card each, so the expected number of cards guessed is $E_1$ ; if she is wrong (with probability $\frac{1}{3}$ ), there are 2 black cards left, so Alice can guess these with certainty.
Summing this up into a formula: \[E_3 = \frac{1}{2} + \frac{3}{5} \left(1 + E_2 \right) + \frac{2}{5} \left( \frac{1}{4}(3) + \frac{3}{4}\left(1 + \frac{2}{3}\left(1 + E_1 \right) + \frac{1}{3}(2)\right) \right)\]
We can apply similar logic to compute $E_2$ and get \[E_2 = \frac{1}{2} + \frac{2}{3}(1 + E_1) + \frac{1}{3}(2)\]
To compute $E_1$ , we know that Alice can guess the last card with certainty, and there's a $\frac{1}{2}$ chance they get the first card as well, so $E_1 = \frac{3}{2}$
Thus, $E_2 = \frac{17}{6}$ , and after long computation, we get $E_3 = \frac{41}{10}$ . The requested answer is $41 + 10 = \boxed{51}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8
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Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
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This solution refers to the Diagram section.
Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrightarrow{AB},$ and $\overleftrightarrow{BC},$ respectively, such that $\overline{RT}$ intersects $\odot O$ at $P$ and $Q.$
We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; fill(R--T--Z--X--cycle,cyan); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); draw(P--R^^P--S^^P--T,red+dashed); draw(O--X^^O--Y^^O--Z); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$Y$",Y,1.5*dir(-90),linewidth(4.5)); dot("$Z$",Z,1.5*dir(-45),linewidth(4.5)); dot("$Q$",Q,1.5*dir(60),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy] Note that $\angle RXZ = \angle TZX = 90^\circ$ by the properties of tangents, so $RTZX$ is a rectangle. It follows that the diameter of $\odot O$ is $XZ = RT = 25.$
Let $x=PQ$ and $y=RX=TZ.$ We apply the Power of a Point Theorem to $R$ and $T:$ \begin{align*} y^2 &= 9(9+x), \\ y^2 &= 16(16-x). \end{align*} We solve this system of equations to get $x=7$ and $y=12.$ Alternatively, we can find these results by the symmetry on rectangle $RTZX$ and semicircle $\widehat{XPZ}.$
We extend $\overline{SP}$ beyond $P$ to intersect $\odot O$ and $\overleftrightarrow{CD}$ at $E$ and $F,$ respectively, where $E\neq P.$ So, we have $EF=SP=5$ and $PE=25-SP-EF=15.$ On the other hand, we have $PX=15$ by the Pythagorean Theorem on right $\triangle PRX.$ Together, we conclude that $E=X.$ Therefore, points $S,P,$ and $X$ must be collinear.
Let $G$ be the foot of the perpendicular from $D$ to $\overline{AB}.$ Note that $\overline{DG}\parallel\overline{XP},$ as shown below: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; G = foot(D,A,B); fill(D--A--G--cycle,green); fill(P--R--X--cycle,yellow); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); draw(P--R,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$G$",G,1.5*dir(-90),linewidth(4.5)); draw(P--X,MidArrow(0.3cm,Fill(red))); draw(G--D,MidArrow(0.3cm,Fill(red))); label("$9$",midpoint(P--R),dir(A-D),red); label("$12$",midpoint(R--X),dir(135),red); label("$15$",midpoint(X--P),dir(0),red); label("$25$",midpoint(G--D),dir(0),red); [/asy] As $\angle PRX = \angle AGD = 90^\circ$ and $\angle PXR = \angle ADG$ by the AA Similarity, we conclude that $\triangle PRX \sim \triangle AGD.$ The ratio of similitude is \[\frac{PX}{AD} = \frac{RX}{GD}.\] We get $\frac{15}{AD} = \frac{12}{25},$ from which $AD = \frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4AD = \boxed{125}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8
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Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
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This solution refers to the Diagram section.
Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let $H$ be the foot of the perpendicular from $P$ to $\overleftrightarrow{CD},$ $M$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{HS},$ and $N$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{RT}.$
We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, H, M, N; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T); fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(90),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$H$",H,1.5*dir(90),linewidth(4.5)); dot("$M$",M,1.5*dir(180),linewidth(4.5)); dot("$N$",N,1.5*dir(15),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy] Note that the diameter of $\odot O$ is $HS=RT=25,$ so $OP=\frac{25}{2}.$ It follows that:
Since $\overline{MO}\parallel\overline{AB}$ and $\overline{ON}\parallel\overline{DA},$ we conclude that $\angle A = \angle MON.$ We apply the Sine of a Sum Formula: \begin{align*} \sin\angle A &= \sin\angle MON \\ &= \sin(\angle MOP + \angle PON) \\ &= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\ &= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\ &= \frac{4}{5}. \end{align*} Note that \[\sin\angle A = \frac{HS}{DA},\] from which $\frac{4}{5} = \frac{25}{DA}.$ We solve this equation to get $DA=\frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4DA = \boxed{125}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8
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Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
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Label the points of the rhombus to be $X$ $Y$ $Z$ , and $W$ and the center of the incircle to be $O$ so that $9$ $5$ , and $16$ are the distances from point $P$ to side $ZW$ , side $WX$ , and $XY$ respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus $XYZW$ is $25$ and circle $O$ has radius $\frac{25}{2}$
Call the feet of the altitudes from $P$ to side $ZW$ , side $WX$ , and side $XY$ to be $A$ $B$ , and $C$ respectively. Additionally, call the feet of the altitudes from $O$ to side $ZW$ , side $WX$ , and side $XY$ to be $D$ $E$ , and $F$ respectively.
Draw a line segment from $P$ to $\overline{OD}$ so that it is perpendicular to $\overline{OD}$ . Notice that this segment length is equal to $AD$ and is $\sqrt{\left(\frac{25}{2}\right)^2-\left(\frac{7}{2}\right)^2}=12$ by Pythagorean Theorem.
Similarly, perform the same operations with perpendicular from $P$ to $\overline{OE}$ to get $BE=10$
By equal tangents, $WD=WE$ . Now, label the length of segment $WA=n$ and $WB=n+2$
Using Pythagorean Theorem again, we get
\begin{align*} WA^2+PA^2&=WB^2+PB^2 \\ n^2+9^2&=(n+2)^2+5^2 \\ n&=13. \end{align*}
Which also gives us $\tan{\angle{OWX}}=\frac{1}{2}$ and $OW=\frac{25\sqrt{5}}{2}$
Since the diagonals of the rhombus intersect at $O$ and are angle bisectors and are also perpendicular to each other, we can get that
\begin{align*} \frac{OX}{OW}&=\tan{\angle{OWX}} \\ OX&=\frac{25\sqrt{5}}{4} \\ WX^2&=OW^2+OX^2 \\ WX&=\frac{125}{4} \\ 4WX&=\boxed{125}
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8
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Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
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Denote by $O$ the center of $ABCD$ .
We drop an altitude from $O$ to $AB$ that meets $AB$ at point $H$ .
We drop altitudes from $P$ to $AB$ and $AD$ that meet $AB$ and $AD$ at $E$ and $F$ , respectively.
We denote $\theta = \angle BAC$ .
We denote the side length of $ABCD$ as $d$
Because the distances from $P$ to $BC$ and $AD$ are $16$ and $9$ , respectively, and $BC \parallel AD$ , the distance between each pair of two parallel sides of $ABCD$ is $16 + 9 = 25$ .
Thus, $OH = \frac{25}{2}$ and $d \sin \theta = 25$
We have \begin{align*} \angle BOH & = 90^\circ - \angle HBO \\ & = 90^\circ - \angle HBD \\ & = 90^\circ - \frac{180^\circ - \angle C}{2} \\ & = 90^\circ - \frac{180^\circ - \theta}{2} \\ & = \frac{\theta}{2} . \end{align*}
Thus, $BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}$
In $FAEP$ , we have $\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0$ .
Thus, \[ AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)} - PF i . \]
Taking the imaginary part of this equation and plugging $EP = 5$ and $PF = 9$ into this equation, we get \[ AE = \frac{9 + 5 \cos \theta}{\sin \theta} . \]
We have \begin{align*} OP^2 & = \left( OH - EP \right)^2 + \left( AH - AE \right)^2 \\ & = \left( \frac{25}{2} - 5 \right)^2 + \left( d - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) \\ & = \left( \frac{15}{2} \right)^2 + \left( \frac{25}{\sin \theta} - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) . \hspace{1cm} (\bigstar) \end{align*}
Because $P$ is on the incircle of $ABCD$ $OP = \frac{25}{2}$ . Plugging this into $(\bigstar)$ , we get the following equation \[ 20 \sin \theta - 15 \cos \theta = 7 . \]
By solving this equation, we get $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$ .
Therefore, $d = \frac{25}{\sin \theta} = \frac{125}{4}$
Therefore, the perimeter of $ABCD$ is $4d = \boxed{125}$
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| 5
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Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
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The center of the incircle is $O.$ Denote the points in which the incircle meets $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ and $\overline{DA}$ as $W,$ $X,$ $Y,$ and $Z,$ respectively. Next, also denote the base of the perpendicular from $P$ to $\overline{AB},$ $\overline{AD},$ $\overline{OW},$ and $\overline{OZ}$ as $M,$ $N,$ $S,$ and $T,$ respectively.
We can easily see that the radius of the circle is $\frac{25}{2}.$ Using this and Pythagorus on right $\triangle OSP$ and $\triangle OTP,$ we find that $MW = PS = 10$ and $NZ = PT = 12.$
Since $AW = AZ$ by properties of circle tangents, we can deduce by the above information that $AM = AN+2.$ Doing Pythagorus on right $\triangle AMP$ and $\triangle ANP$ we find that $a^2 = b^2 + 56$ (because $a^2+25=b^2+81.$ ) From solving the $2$ just derived equations, we find that $AM=15$ and $AN=13.$
Next, we use Pythagorus on right $\triangle AOB$ (we can see it's right because of properties of rhombuses.) We get \[AB^2 = AO^2 + BO^2.\] We know $AB = AW + WB = 25 + WB.$ By Pythagorus on $\triangle AWO$ and $\triangle BWO,$ we also know $AO^2 = 25^2+\left(\frac{25}{2}\right)^2$ and $BO^2=WB^2+\left(\frac{25}{2}\right)^2.$ Substituting these in, we have \[25^2 + 50WB + WB^2 = 25^2+\left(\frac{25}{2}\right)^2+\left(\frac{25}{2}\right)^2+WB^2.\] Solving for $WB,$ we get $WB = \frac{25}{4}.$ Now we find that each side of the rhombus $=AB=25+\frac{25}{4}=\frac{125}{4}.$ The perimeter of the rhombus would be that times $4.$ Our final answer is \[\frac{125}{4}\cdot4=\boxed{125}.\]
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5,276
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8
| 6
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Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
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Notation is shown on diagram, $RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.$ $RT = 9 + 16 = 25 = FG$ as hights of rhombus. \[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\] \[PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.\] \[PE = \sqrt{RP^2 + RE^2} = 15 \implies E = E'.\] \[\sin \alpha = \frac {RE}{PE} = \frac {GF}{AD} \implies AD = \frac {15 \cdot 25}{12} = \frac {125}{4}.\] The perimeter of $ABCD$ is $\frac{125}{4}\cdot4=\boxed{125}.$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9
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Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$
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Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$ . Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$ \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$ ; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$
$\textbf{Case 1:}$
There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a " $\pm$ " there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case.
$\textbf{Case 2:}$
$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$
Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \begin{align*} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.
However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.
This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$ ) we get $18\cdot41=\boxed{738}$ as our final answer.
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9
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Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$
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$p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$ , with $m\neq 2$
In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$ , with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$ ), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).
In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$ , and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $328$ polynomials.
The total is $\boxed{738}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10
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There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)
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Define $\left\{ x \right\} = x - \left\lfloor x \right\rfloor$
First, we bound $U$
We establish an upper bound of $U$ . We have \begin{align*} U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\ & = \frac{1}{5} \sum_{n=1}^{2023} n^2 - \frac{a}{5} \sum_{n=1}^{2023} n \\ & = \frac{1012 \cdot 2023}{5} \left( 1349 - a \right) \\ & \triangleq UB . \end{align*}
We establish a lower bound of $U$ . We have \begin{align*} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = \sum_{n=1}^{2023} \frac{n^2 - na}{5} - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} . \end{align*}
We notice that if $5 | n$ , then $\frac{n^2 - na}{5} \in \Bbb Z$ .
Thus, \begin{align*} U & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\ & = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\ & = UB - 1619 \\ & \triangleq LB . \end{align*}
Because $U \in \left[ - 1000, 1000 \right]$ and $UB - LB = 1619 < \left( 1000 - \left( - 1000 \right) \right)$ , we must have either $UB \in \left[ - 1000, 1000 \right]$ or $LB \in \left[ - 1000, 1000 \right]$
For $UB \in \left[ - 1000, 1000 \right]$ , we get a unique $a = 1349$ .
For $LB \in \left[ - 1000, 1000 \right]$ , there is no feasible $a$
Therefore, $a = 1349$ . Thus $UB = 0$
Next, we compute $U$
Let $n = 5 q + r$ , where $r = {\rm Rem} \ \left( n, 5 \right)$
We have \begin{align*} \left\{ \frac{n^2 - na}{5} \right\} & = \left\{ \frac{\left( 5 q + r \right)^2 - \left( 5 q + r \right)\left( 1350 - 1 \right)}{5} \right\} \\ & = \left\{ 5 q^2 + 2 q r - \left( 5 q + r \right) 270 + q + \frac{r^2 + r}{5} \right\} \\ & = \left\{\frac{r^2 + r}{5} \right\} \\ & = \left\{ \begin{array}{ll} 0 & \mbox{ if } r = 0, 4 \\ \frac{2}{5} & \mbox{ if } r = 1, 3 \\ \frac{1}{5} & \mbox{ if } r = 2 \end{array} \right. . \end{align*}
Therefore, \begin{align*} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{q=0}^{404} \sum_{r=0}^4 \left\{\frac{r^2 + r}{5} \right\} + \left\{ \frac{0^2 - 0 \cdot a}{5} \right\} + \left\{ \frac{2024^2 - 2024a}{5} \right\} \\ & = - \sum_{q=0}^{404} \left( 0 + 0 + \frac{2}{5} + \frac{2}{5} + \frac{1}{5} \right) + 0 + 0 \\ & = - 405 . \end{align*}
Therefore, $a + U = 1349 - 405 = \boxed{944}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10
| 2
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There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)
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We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$ . Since for any real number $x$ $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$ , we have $U \le U' \le U + 2023$ . Now, since $-1000 \le U \le 1000$ , we have $-1000 \le U' \le 3023$
Now, we can solve for $U'$ in terms of $a$ . We have: \begin{align*} U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\ &= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\ &= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\ &= \frac{2023(2024)(4047-3a)}{30} \\ \end{align*} So, we have $U' = \frac{2023(2024)(4047-3a)}{30}$ , and $-1000 \le U' \le 3023$ , so we have $-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023$ , or $-30000 \le 2023(2024)(4047-3a) \le 90690$ . Now, $2023 \cdot 2024$ is much bigger than $90690$ or $30000$ , and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$ , or $a = 1349$ , and $U' = 0$
Now, we can find $U - U'$ . We have: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\ &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \end{align*}. Now, if $n^2-1349n \equiv 0 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0$ , and if $n^2-1349n \equiv 1 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}$ , and so on. Testing with $n \equiv 0,1,2,3,4, \text{ (mod 5)}$ , we get $n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$ , we get: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\ &= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot t0 \\ &= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\ &= -405 \end{align*} Since $U' = 0$ , this gives $U = -405$ , and we have $a + U = 1349-405 = \boxed{944}$
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10
| 3
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There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)
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We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$ ) from the numerator of $\frac{n^2-na}{5}$ . For example, $\left\lfloor \frac{7}{5} \right\rfloor = \frac{7-2}{5} = 1$
Note that the congruence of $n^2 - na$ (mod $5$ ) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$ ) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$ ).
For example, if $a \equiv 1$ (mod $5$ ), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \equiv 1,2,3,4,0$ (mod $5$ ). Let the sum of these elements be $s$ . Note that for each “loop” of the numerator (mod $5$ ), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$ ) throughout the sum, as $5 \cdot 404=2020$ . Then
$U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2} -404s \right)}{5}$
Where $n=2023$ . Note that since $5 \cdot 404=2020$ , this is an approximation for $U$ because the equation disregards the remainder (mod $5$ ) when $n=2021, 2022$ , and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$ , so we can proceed with this approximation to solve for $a$
Factoring our approximation gives $U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$
We set $a= \frac{(2n+1)}{3} = 1349$ to make $\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$ , accordingly minimizing $|U|$ , yielding $U \approx \frac{-404s}{5}$
If $a$ increases or decreases by $1$ , then $U$ changes by $\frac {(n)(n+1)}{2 \cdot 5} = \frac {2023 \cdot 2024}{10}$ which clearly breaks the inequality on $U$ . Therefore $a=1349 \equiv 4$ (mod $5$ ) giving the set of remainders $(2,1,2,0,0)$ , so $s=5$ and our approximation yields $U \approx -404$ . However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \frac{(2+1+2)}{5} = - 405$ , giving an answer of $1349-405= \boxed{944}$
| 944
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5,282
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 1
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
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Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\]
We see that $f(1) = 2$ $f(2) = 3$ , and $f(n) = f(n-1) + f(n-2)$ . This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$ th Fibonacci number.
This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}$
| 235
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5,283
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 2
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
|
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
$\textbf{Case 1: (1,2)}$
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us $8 \choose {0}$ . We can also pick one number, giving us $7 \choose {1}$ because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us $6 \choose {2}$ . This pattern continues for each amount of numbers, yielding $5 \choose {3}$ for 3 numbers and $4 \choose {4}$ for four numbers. Adding these up, we have $8 \choose {0}$ $7 \choose {1}$ $6 \choose {2}$ $5 \choose {3}$ $4 \choose {4}$ $\textbf{34}$
$\textbf{Case 2: (2,3)}$
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have $7 \choose {0}$ $6 \choose {1}$ $5 \choose {2}$ $4 \choose {3}$ $\textbf{21}$
$\textbf{Case 3: (3,4)}$
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say $2 \choose {0}$ for choosing no numbers and $1 \choose {1}$ for choosing 1, leaving us with $2 \choose {0}$ $1 \choose {1}$ = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have $6 \choose {0}$ $5 \choose {1}$ $4 \choose {2}$ $3 \choose {3}$ = 13. With 2 choices on one side and 13 choices on the other side, we have $2\cdot13$ $\textbf{26}$ combinations in all.
$\textbf{Case 4: (4,5)}$
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have $3 \choose {0}$ $2 \choose {1}$ = 3 choices (1, 2, or neither) and for the numbers 6-10 we have $5 \choose {0}$ $4 \choose {1}$ $3 \choose {2}$ = 8 choices. With 3 choices on one side and 8 choices on the other side, we have $3\cdot8$ $\textbf{24}$ combinations in all.
$\textbf{Case 5: (5,6)}$
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices and for the numbers 5-10 we have the same $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices. $5\cdot5$ $\textbf{25}$ combinations in all.
$\textbf{Rest of the cases}$
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = $\boxed{235}$
| 235
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5,284
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 3
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
|
Denote by $N_1 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by $N_0 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers.
Denote by $a$ the smallest number in set $S$
First, we compute $N_1 \left( m \right)$
Consider $m \geq 3$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 1 \right)$
Case 2: A subset contains $a$ but does not contain $a + 1$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 2 \right)$
Case 3: A subset contains $a$ and $a + 1$
To have exactly one pair of consecutive integers, this subset cannot have $a + 2$ , and cannot have consecutive integers in $\left\{ a+3, a+4, \cdots , a + m - 1 \right\}$
Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \left( m - 3 \right)$
Therefore, for $m \geq 3$ \[N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .\]
For $m = 1$ , we have $N_1 \left( 1 \right) = 0$ .
For $m = 2$ , we have $N_1 \left( 2 \right) = 1$
Second, we compute $N_0 \left( m \right)$
Consider $m \geq 2$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has no consecutive integers is $N_0 \left( m - 1 \right)$
Case 2: A subset contains $a$
To avoid having consecutive integers, the subset cannot have $a + 1$
Thus, the number of subsets that has no consecutive integers is $N_0 \left( m - 2 \right)$
Therefore, for $m \geq 2$ \[N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right) .\]
For $m = 0$ , we have $N_0 \left( 0 \right) = 1$ .
For $m = 1$ , we have $N_0 \left( 1 \right) = 2$
By solving the recursive equations above, we get $N_1 \left( 10 \right) = \boxed{235}$
| 235
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5,285
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 4
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
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Let $a_n$ be the number of subsets of the set $\{1,2,3,\ldots,n\}$ such that there exists exactly 1 pair of consecutive elements.
Let $b_n$ be the number of subsets of the set $\{1, 2, 3\ldots, n\}$ such that there doesn't exist any pair of consecutive elements.
First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either:
1. $\{n-1, n\}\subseteq S,$ 2. $n\not\in S$ , or
3. $n-1 \not\in S$ and $n\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, \[a_n = a_{n-1} + a_{n-2} + b_{n-3}.\] Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either:
1. $n \not\in T,$ or
2. $n \in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, \[b_n = b_{n-1} + b_{n-2}.\] Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$ th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values.
\[\begin{array}{r|l} n & a_n \\ \hline 1 & 0 \\ 2 & 1\\ 3 & 2\\ 4 & 2 + 1 + F_2 = 5\\5 & 5 + 2 + F_3 = 10\\6 & 10 + 5 + F_4 = 20 \\ 7 & 20 + 10 + F_5 = 38\\ 8 & 38 + 20 + F_6 = 71 \\ 9& 71 + 38 + F_7 = 130\\ 10 & 130 + 71 + F_8 = 235 \end{array}\] Our answer is $a_{10} = \boxed{235}.$ ~AtharvNaphade
| 235
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5,286
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 5
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
|
Note: This is a very common stars and bars application.
Casework on number of terms, let the number of terms be $n$ . We can come up with a generalized formula for the number of subsets with n terms.
Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$
Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$ . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$ $1 + d_2 + ... + d_{n-1} = k$
To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$ $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers.
Finally, we add in $D_0$ , the distance between 0 and the first term of the set, and $D_n$ , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by $D_0$ , k, and $D_N$ \[D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11\] \[D_0 + D_2 + D_3 + ... + D_n = 12 - n\] There are n positive terms , by Balls and Urns , there are ${(12-n)-1 \choose (n)-1} = {11-n \choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence $(n-1){11-n \choose n-1}$ for $n > 2$
Case 1: Two terms $n = 2$ , so $(2-1){11-2 \choose 2-1} = 1{9 \choose 1} = 9$
Case 2: Three terms $n = 3$ , so $(3-1){11-3 \choose 3-1} = 2{8 \choose 2} = 56$
Case 3: Four terms $n = 4$ , so $(4-1){11-4 \choose 4-1} = 3{7 \choose 3} = 105$
Case 4: Five terms $n = 5$ , so $(5-1){11-5 \choose 5-1} = 4{6 \choose 4} = 60$
Case 5: Six terms $n = 6$ , so $(6-1){11-6 \choose 6-1} = 5{5 \choose 5} = 5$
We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=\boxed{235}$
| 235
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5,287
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 6
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
|
Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.)
Now, notice that if we take $i$ and $i+1$ as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\{1, 2, ... , i-2 \}$ is consecutive with any element of the set $\{ i+3, i+4, ... , 10 \}$
Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has $i-2$ consecutive integer elements and the second set has $8-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$
Summing this over for all $1 \leq i \leq 9$ yields \[\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \cdot 34 + 1 \cdot 21 + 2 \cdot 13 + 3 \cdot 8) + 5^2 = 2 \cdot 105+25 = \boxed{235}.\]
| 235
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5,288
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 7
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
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The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$ : 1 triomino ( $RGG$ ), $n$ dominoes ( $RG$ ), and $8-2n$ monominoes ( $R$ ). The $G$ spots cover the members of the subset.
The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
$R + RG + RGG =$ Polyominoes $\rightarrow$ Orderings
$0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5$
$2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60$
$4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105$
$6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56$
$8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9$
The sum is $\boxed{235}$
| 235
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5,289
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
| 8
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
|
Let $dp(i)$ be the number of subsets of a set $i$ consecutive integers such that the maximum value in the set is $i$ and there exists exactly one pair of consecutive integers. Define $dp2(i)$ similarly, but without any pair of consecutive integers. The base cases are $dp2(1)=dp2(2)=1$ $dp(1)=0$ , and $dp(2)=1$
The transitions are:
\[dp2(i)=\sum_{j=1}^{i-2}(dp(j))+1\]
\[dp(i)=dp2(i-1)+\sum_{j=1}^{i-2}dp(j)\]
Note that $dp2$ is the Fibonacci numbers.
\[\begin{array}{rcl} i & dp & dp2\\ \hline 1 & 0 & 1 \\ 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 3 & 3 \\ 5 & 5 & 5 \\ 6 & 10 & 8 \\ 7 & 18 & 13 \\ 8 & 33 & 21 \\ 9 & 59 & 34 \\ 10 & 105 & 55 \\ \end{array}\]
Summing over $dp$ yields $1+1+3+5+10+18+33+59+105=\boxed{235}$
| 235
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5,290
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13
| 1
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Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]
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Denote $\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}$ .
Denote by $d$ the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space.
We put the bottom face on the $x-O-y$ plane.
For this bottom face, we put a vertex with an acute angle $2 \alpha$ at the origin, denoted as $O$ .
For two edges that are on the bottom face and meet at $O$ , we put one edge on the positive side of the $x$ -axis. The endpoint is denoted as $A$ . Hence, $A = \left( d , 0 , 0 \right)$ .
We put the other edge in the first quadrant of the $x-O-y$ plane. The endpoint is denoted as $B$ . Hence, $B = \left( d \cos 2 \alpha , d \sin 2 \alpha , 0 \right)$
For the third edge that has one endpoint $O$ , we denote by $C$ its second endpoint.
We denote $C = \left( u , v , w \right)$ .
Without loss of generality, we set $w > 0$ .
Hence, \[ u^2 + v^2 + w^2 = d^2 . \hspace{1cm} (1) \]
We have \begin{align*} \cos \angle AOC & = \frac{\overrightarrow{OA} \cdot \overrightarrow{OC}}{|OA| \cdot |OC|} \\ & = \frac{u}{d} , \hspace{1cm} (2) \end{align*} and \begin{align*} \cos \angle BOC & = \frac{\overrightarrow{OB} \cdot \overrightarrow{OC}}{|OB| \cdot |OC|} \\ & = \frac{u \cos 2 \alpha + v \sin 2 \alpha}{d} . \hspace{1cm} (3) \end{align*}
Case 1: $\angle AOC = \angle BOC = 2 \alpha$ or $2 \left( 90^\circ - \alpha \right)$
By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \pm d \cos 2 \alpha \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} \\ & = \pm d \cos 2 \alpha \tan \alpha . \end{align*}
Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} . \hspace{1cm} (4) \end{align*}
Case 2: $\angle AOC = 2 \alpha$ and $\angle BOC = 2 \left( 90^\circ - \alpha \right)$ , or $\angle BOC = 2 \alpha$ and $\angle AOC = 2 \left( 90^\circ - \alpha \right)$
By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \mp d \cos 2 \alpha \frac{1 + \cos 2 \alpha}{\sin 2 \alpha} \\ & = \mp d \cos 2 \alpha \cot \alpha . \end{align*}
Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} . \hspace{1cm} (5) \end{align*}
We notice that $(4) > (5)$ . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is \begin{align*} \frac{(4)}{(5)} & = \frac{\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha}} {\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha}} \\ & = \sqrt{\frac{\tan^2 2 \alpha - \tan^2 \alpha}{\tan^2 2 \alpha - \cot^2 \alpha}} . \end{align*}
Recall that $\tan \alpha = \frac{\sqrt{21}}{\sqrt{31}}$ .
Thus, $\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{\sqrt{21 \cdot 31}}{5}$ .
Plugging this into the equation above, we get \begin{align*} \frac{(4)}{(5)} & = \frac{63}{62}. \end{align*}
Therefore, the answer is $63 + 62 = \boxed{125}$
| 125
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5,291
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13
| 2
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Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]
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Let one of the vertices be at the origin and the three adjacent vertices be $u$ $v$ , and $w$ . For one of the parallelepipeds, the three diagonals involving the origin have length $\sqrt {21}$ . Hence, $(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=21$ and $(u-v)\cdot (u-v)=u\cdot u+v\cdot v-2u\cdot v=31$ . Since all of $u$ $v$ , and $w$ have equal length, $u\cdot u=13$ $v\cdot v=13$ , and $u\cdot v=-2.5$ . Symmetrically, $w\cdot w=13$ $u\cdot w=-2.5$ , and $v\cdot w=-2.5$ . Hence the volume of the parallelepiped is given by $\sqrt{\operatorname{det}\begin{pmatrix}13&-2.5&-2.5\\-2.5&13&-2.5\\-2.5&-2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}15.5&-15.5&0\\-2.5&13&-2.5\\0&-15.5&15.5\end{pmatrix}}=\sqrt{15.5^2\operatorname\det\begin{pmatrix}1&-1&0\\-2.5&13&-2.5\\0&-1&1\end{pmatrix}}=\sqrt{15.5^2\cdot 8}$
For the other parallelepiped, the three diagonals involving the origin are of length $\sqrt{31}$ and the volume is $\sqrt{\operatorname{det}\begin{pmatrix}13&2.5&2.5\\2.5&13&2.5\\2.5&2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}10.5&-10.5&0\\2.5&13&2.5\\0&-10.5&10.5\end{pmatrix}}=\sqrt{10.5^2\operatorname\det\begin{pmatrix}1&-1&0\\2.5&13&2.5\\0&-1&1\end{pmatrix}}=\sqrt{10.5^2\cdot 18}$
Consequently, the answer is $\sqrt\frac{10.5^2\cdot 18}{15.5^2\cdot 8}=\frac{63}{62}$ , giving $\boxed{125}$
| 125
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5,292
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13
| 3
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Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]
|
Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting $P$ , and the one with three obtuse angles meeting $P'$
The area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.
Denote the point with three acute angles meeting at it in $P$ as $A$ , and its neighbors $B$ $C$ , and $D$ . Similarly, denote the point with three obtuse angles meeting at it in $P'$ as $A'$ , and its neighbors $B'$ $C'$ , and $D'$
We have the following equations:
\[\textrm{Height of }P\textrm{ from }ACD = \frac{\textrm{Vol}(ABCD) \cdot 3}{[ACD]},\] \[\textrm{Height of }P'\textrm{ from }A'C'D' = \frac{\textrm{Vol}(A'B'C'D') \cdot 3}{[A'C'D']}.\]
However, $ACD$ and $A'C'D'$ are both half the area of a rhombus with diagonals $\sqrt{31}$ and $\sqrt{21}$ , so our ratio is really
\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')}.\]
Because the diagonals of all of the faces are $\sqrt{31}$ and $\sqrt{21}$ , each edge of the parallelepipeds is $\sqrt{13}$ by the Pythagorean theorem.
We have $AB = AC = AD = \sqrt{13}$ , and $BC = CD = BD = \sqrt{21}$ . When we drop a perpendicular to the centroid of $BCD$ from $A$ (let's call this point $O$ ), we have $BO = \frac{\sqrt{21}}{\sqrt{3}} = \sqrt{7}$ . Thus,
\[AB^2 - BO^2 = AO^2\] \[13 - 7 = AO^2 = 6\] \[AO = \sqrt{6}.\]
The area of base $BCD$ is $\frac{21\sqrt{3}}{4}$ . Hence,
\[\textrm{Vol}(ABCD) = \frac{\sqrt{6}\cdot\frac{21\sqrt{3}}{4}}{3}\] \[= \frac{63\sqrt{2}}{12}.\]
We can apply a similar approach to $A'B'C'D'$
$A'B' = A'C' = A'D' = \sqrt{13}$ , and $B'C' = C'D' = B'D' = \sqrt{31}$ . When we drop a perpendicular to the centroid of $B'C'D'$ from $A'$ (let's call this point $O'$ ), we have $B'O' = \frac{\sqrt{31}}{\sqrt{3}} = \sqrt{\frac{31}{3}}$ . Thus,
\[A'B'^2 - B'O'^2 = A'O'^2\] \[13 - \frac{31}{3} = A'O'^2\] \[A'O' = \sqrt{8}{3} = \frac{2\sqrt{6}}{3}.\]
The area of base $B'C'D'$ is $\frac{31\sqrt{3}}{4}$ . Hence,
\[\textrm{Vol}(A'B'C'D') = \frac{\frac{2\sqrt{6}}{3}\cdot\frac{31\sqrt{3}}{4}}{3}\] \[= \frac{186\sqrt{2}}{36}\] \[= \frac{62\sqrt{2}}{12}.\]
Finally,
\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')} = \frac{\frac{63\sqrt{2}}{12}}{\frac{62\sqrt{2}}{12}} = \frac{63}{62}.\]
Our answer is $63 + 62 = \boxed{125}$
| 125
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5,293
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13
| 4
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Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]
|
Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
[asy] unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2; label("A",a,S); label("B",b,S); label("C",c,S); label("D",d,N); label("P",p,S); label("O",o,E); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); draw(d--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--c, dotted); draw(d--c, dotted); [/asy]
As illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$ . Denote the center of the base by $O$ . By symmetry, $P$ must be on $AO$ . Now we need to find $DP$
Apply Pythagorean theorem to $\triangle DPA$ we have \[DP^2 = AD^2 - AP^2.\]
Apply Pythagorean theorem to $\triangle DPC$ and then $\triangle CPO$ we have \[DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.\]
Combining the above two, we have \[AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.\]
Since $AD=\sqrt{13}$ $DC=\sqrt{21}$ $CO=\frac{\sqrt{21}}{2}$ $AO=\frac{\sqrt{31}}{2}$ , plug them into the above equation and solve for the only unknown variable $AP$ , we get $AP=\frac{5}{\sqrt{31}}.$
Thus the height \[DP = \sqrt{AD^2 - AP^2} = \sqrt{13 - \frac{25}{31}} = \sqrt{\frac{378}{31}}.\]
[asy] unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = 0.7*dir(80), e = c+d, p = 0.9*dir(10), o = (b+c)/2; label("A'",a,S); label("B'",b,S); label("C'",c,W); label("E'",e,N); label("P'",p,S); label("O'",o,W); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); draw(e--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--a, dotted); draw(e--a, dotted); [/asy]
For the other parallelepiped, using the same approach and drop a perpendicular from $E'$ onto the base at $P'$ . Similarly applying Pythagorean theorem to $\triangle E'P'C'$ $\triangle E'P'A'$ and $\triangle A'P'O'$ we have \[C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.\]
Plugging known values into the above equation and solve for the only unknown variable $C'P'$ , we get $C'P'=\frac{5}{\sqrt{21}}.$
Thus the height \[E'P' = \sqrt{C'E'^2 - C'P'^2} = \sqrt{13 - \frac{25}{21}} = \sqrt{\frac{248}{21}}.\]
The ratio between the two is therefore \[\frac{DP}{E'P'} = \frac{\sqrt{\frac{378}{31}}}{\sqrt{\frac{248}{21}}} = \sqrt{\frac{2\cdot3^3\cdot7}{31}\cdot\frac{3\cdot7}{2^3\cdot31}} = \frac{3^2\cdot7}{2\cdot31} = \frac{63}{62}\] , giving $\boxed{125}$
| 125
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5,294
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14
| 1
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The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$
|
This problem is, in essence, the following: A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an $x$ - or $y$ -coordinate of $12$ . At this point, if he tries to jump to a coordinate outside the square from $(0,0)$ to $(11,11)$ , he "wraps around" and ends up at an $x$ - or $y$ - coordinate of $0$ . How many ways are there for Felix to jump on every grid point in this square, so that he ends at $(0,0)$ ? This is consistent with the construction of the flat torus as $\mathbb Z^2/12\mathbb Z^2$ (2-dimensional modular arithmetic. $(\mathbb{Z}_{12})^2$
Moving on, define a $\textit{path}$ from point $A$ to point $B$ to be a sequence of "up"s and "right"s that takes Felix from $A$ to $B$ . The $\textit{distance}$ from $A$ to $B$ is the length of the shortest path from $A$ to $B$ . At the crux of this problem is the following consideration: The points $A_i=(i,12-i), i\in{0,...,11}$ are pairwise equidistant, each pair having distance of $12$ in both directions.
[asy] size(7cm); for (int x=0; x<12; ++x){ for (int y=0; y<12; ++y){ fill(circle((x,y),0.05));}} for (int i=0; i<12; ++i){ fill(circle((i,11-i),0.1),red);} pen p=green+dashed; path u=(3,8)--(4,8)--(4,9)--(4,10)--(4,11)--(5,11)--(5,11.5); path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4); draw(u,p); draw(v,p); pen p=blue+dashed; path u=(4,7)--(5,7)--(5,8)--(5,9)--(5,10)--(6,10)--(6,11)--(6,11.5); path v=(6,-0.5)--(6,0)--(7,0)--(7,1)--(7,2)--(7,3)--(8,3); draw(u,p); draw(v,p); [/asy]
A valid complete path then joins two $A_i$ 's, say $A_i$ and $A_j$ . In fact, a link between some $A_i$ and $A_j$ fully determines the rest of the cycle, as the path from $A_{i+1}$ must "hug" the path from $A_i$ , to ensure that there are no gaps. We therefore see that if $A_0$ leads to $A_k$ , then $A_i$ leads to $A_{i+k}$ . Only the values of $k$ relatively prime to $12$ result in solutions, though, because otherwise $A_0$ would only lead to $\{A_i:\exists n\in \mathbb Z:i\equiv kn\quad\text{mod 12}\}$ . The number of paths from $A_0$ to $A_k$ is ${12\choose k}$ , and so the answer is
\[{12\choose1}+{12\choose5}+{12\choose7}+{12\choose11}=1\boxed{608}.\]
| 608
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5,295
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14
| 2
|
The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$
|
This is more of a solution sketch and lacks rigorous proof for interim steps, but illustrates some key observations that lead to a simple solution.
Note that one can visualize this problem as walking on a $N \times N$ grid where the edges warp. Your goal is to have a single path across all nodes on the grid leading back to $(0,\ 0)$ . For convenience, any grid position are presumed to be in $\mod N$
Note that there are exactly two ways to reach node $(i,\ j)$ , namely $(i - 1,\ j)$ and $(i,\ j - 1)$
As a result, if a path includes a step from $(i,\ j)$ to $(i + 1,\ j)$ , there cannot be a step from $(i,\ j)$ to $(i,\ j + 1)$ . However, a valid solution must reach $(i,\ j + 1)$ , and the only valid step is from $(i - 1,\ j + 1)$
So a solution that includes a step from $(i,\ j)$ to $(i + 1,\ j)$ dictates a step from $(i - 1,\ j + 1)$ to $(i,\ j + 1)$ and by extension steps from $(i - a,\ j + a)$ to $(i - a + 1,\ j + a)$ . We observe the equivalent result for steps in the orthogonal direction.
This means that in constructing a valid solution, taking one step in fact dictates N steps, thus it's sufficient to count valid solutions with $N = a + b$ moves of going right $a$ times and $b$ times up the grid. The number of distinct solutions can be computed by permuting 2 kinds of indistinguishable objects $\binom{N}{a}$
Here we observe, without proof, that if $\gcd(a, b) \neq 1$ , then we will return to the origin prematurely. For $N = 12$ , we only want to count the number of solutions associated with $12 = 1 + 11 = 5 + 7 = 7 + 5 = 11 + 1$
(For those attempting a rigorous proof, note that $\gcd(a, b) = \gcd(a + b, b) = \gcd(N, b) = \gcd(N, a)$ ).
The total number of solutions, noting symmetry, is thus
\[2\cdot\left(\binom{12}{1} + \binom{12}{5}\right) = 1608\]
This yields $\boxed{608}$ as our desired answer.
| 608
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5,296
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14
| 3
|
The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$
|
Define a $12 \times 12$ matrix $X$ .
Each entry $x_{i, j}$ denotes the number of movements the longer hand moves, given that two hands jointly make $12 \left( i - 1 \right) + \left( j - 1 \right)$ movements.
Thus, the number of movements the shorter hand moves is $12 \left( i - 1 \right) + \left( j - 1 \right) - x_{i, j}$
Denote by $r_{i, j}$ the remainder of $x_{i, j}$ divided by 12.
Denote by $R$ this remainder matrix.
If two hands can return to their initial positions after 144 movements, then $r_{12, 12} = 0$ or 11.
Denote by $S_0$ (resp. $S_{11}$ ) the collection of feasible sequences of movements, such that $r_{12, 12} = 0$ (resp. $r_{12, 12} = 11$ ).
Define a function $f : S_0 \rightarrow S_{11}$ , such that for any $\left\{ x_{i,j} , \ \forall \ i, j \in \left\{ 1, 2, \cdots , 12 \right\} \right\} \in S_0$ , the functional value of the entry indexed as $\left( i, j \right)$ is $12 \left( i - 1 \right) + \left( j - 1 \right) - x_{i, j}$ .
Thus, function $f$ is bijective. This implies $| S_0 | = | S_{11} |$
In the rest of analysis, we count $| S_0 |$
We make the following observations:
\begin{enumerate}
\item $x_{1, 1} = 0$ and $12 | x_{12, 12}$
These follow from the definition of $S_0$
\item Each column of $R$ is a permutation of $\left\{ 0, 1, \cdots , 11 \right\}$
The reasoning is as follows. Suppose there exist $i < i'$ $j$ , such that $r_{i, j} = r_{i', j}$ . Then this entails that the positions of two hands after the $\left( 12 \left( i' - 1 \right) + \left( j - 1 \right) \right)$ th movement coincide with their positions after the $\left( 12 \left( i - 1 \right) + \left( j - 1 \right) \right)$ th movement.
\item For any $j \in \left\{ 1, 2 ,\cdots , 11 \right\}$ $x_{i, j+1} - x_{i, j}$ is equal to either 0 for all $i$ or 1 for all $i$
The reasoning is as follows. If this does not hold and the $j$ th column in $R$ is a permutation of $\left\{ 0, 1, \cdots , 12 \right\}$ , then the $j+1$ th column is no longer a permutation of $\left\{ 0, 1, \cdots , 12 \right\}$ . This leads to the infeasibility of the movements.
\item $x_{i+1, 1} = x_{i, 12}$ for any $i \in \left\{ 1, 2, \cdots , 11 \right\}$
This follows from the conditions that the $12$ th column in $R$ excluding $r_{12, 12}$ and the first column in $R$ excluding $x_{1, 1}$ are both permutations of $\left\{ 1, 2, \cdots , 11 \right\}$
\end{enumerate}
All observations jointly imply that $x_{i, 12} = i \cdot x_{1, 12}$ .
Thus, $\left\{ r_{1, 12}, r_{2, 12} , \cdots , r_{11, 12} \right\}$ is a permutation of $\left\{ 1, 2, \cdots , 11 \right\}$ .
Thus, $x_{1, 12}$ is relatively prime to 12.
Because $x_{1, 1} = 0$ and $x_{1, 12} - x_{1, 1} \leq 11$ , we have $x_{1, 12} = 1$ , 5, 7, or 11.
Recall that when we move from $x_{1, 1}$ to $x_{1, 12}$ , there are 11 steps of movements. Each movement has $x_{1, j+1} - x_{i, j} = 0$ or 1.
Thus, for each given $x_{1, 12}$ , the number of feasible movements from $x_{1, 1}$ to $x_{1, 12}$ is $\binom{11}{x_{1, 12}}$
Therefore, the total number of feasible movement sequences in this problem is \begin{align*} | S_0 | + | S_{11} | & = 2 | S_0 | \\ & = 2 \cdot \sum_{x_{1, 12} = 1, 5, 7, 11} \binom{11}{x_{1, 12}} \\ & = 2 \left( 11 + 462 + 330 + 1 \right) \\ & = 1608 . \end{align*}
Therefore, the answer is $\boxed{608}$
| 608
|
5,297
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_15
| 1
|
Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying
|
Assume that $z=a+bi$ . Then, \[z^3=(a^3-3ab^2)+(3a^2b-b^3)i\] Note that by the Triangle Inequality, \[|(a^3-3ab^2)-(3a^2b-b^3)|<p\implies |a^3+b^3-3ab^2-3a^2b|<a^2+b^2\] Thus, we know \[|a+b||a^2+b^2-4ab|<a^2+b^2\] Without loss of generality, assume $a>b$ (as otherwise, consider $i^3\overline z=b+ai$ ). If $|a/b|\geq 4$ , then \[17b^2\geq a^2+b^2>|a+b||a^2+b^2-4ab|\geq |b-4b||16b^2-16b^2+b^2|=3b^3\] `Thus, this means $b\leq\frac{17}3$ or $b\leq 5$ . Also note that the roots of $x^2-4x+1$ are $2\pm\sqrt 3$ , so thus if $b\geq 6$ \[2\sqrt 3b=(2(2-\sqrt 3)-4)b<a<4b\] Note that \[1000>p=a^2+b^2\geq 12b^2+b^2=13b^2\] so $b^2<81$ , and $b<9$ . If $b=8$ , then $16\sqrt 3\leq a\leq 32$ . Note that $\gcd(a,b)=1$ , and $a\not\equiv b\pmod 2$ , so $a=29$ or $31$ . However, then $5\mid a^2+b^2$ , absurd.
If $b=7$ , by similar logic, we have that $14\sqrt 3 <a< 28$ , so $b=26$ . However, once again, $5\mid a^2+b^2$ . If $b=6$ , by the same logic, $12\sqrt3<a<24$ , so $a=23$ , where we run into the same problem. Thus $b\leq 5$ indeed.
If $b=5$ , note that \[(a+5)(a^2+25-20a)<a^2+25\implies a<20\] We note that $p=5^2+18^2=349$ works. Thus, we just need to make sure that if $b\leq 4$ $a\leq 18$ . But this is easy, as \[p>(a+b)(a^2+b^2-4ab)\geq (4+18)(4^2+18^2-4\cdot 4\cdot 18)>1000\] absurd. Thus, the answer is $\boxed{349}$
| 349
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5,298
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_15
| 2
|
Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying
|
Denote $z = a + i b$ . Thus, $a^2 + b^2 = p$
Thus, \[z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) .\]
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have ${\rm Re} \left( z^3 \right) > 0$ and ${\rm Im} \left( z^3 \right) > 0$ .
Thus, \begin{align*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{align*}
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have the following triangle inequalities: \begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}
We notice that $| z^3 | = p^{3/2}$ , and ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ , and $| z^3 |$ form a right triangle. Thus, ${\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}$ .
Because $p > 1$ $p^{3/2} > p$ .
Therefore, (3) holds.
Conditions (4) and (5) can be written in the joint form as \[\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4)\]
We have \begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*} and $p = a^2 + b^2$
Thus, (5) can be written as \[\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| < a^2 + b^2 . \hspace{1cm} (6)\]
Therefore, we need to jointly solve (1), (2), (6).
From (1) and (2), we have either $a, b >0$ , or $a, b < 0$ .
In (6), by symmetry, without loss of generality, we assume $a, b > 0$
Thus, (1) and (2) are reduced to \[a > \sqrt{3} b . \hspace{1cm} (7)\]
Let $a = \lambda b$ . Plugging this into (6), we get \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) \end{align*}
Because $p= a^2 + b^2$ is a prime, $a$ and $b$ are relatively prime.
Therefore, we can use (7), (8), $a^2 + b^2 <1000$ , and $a$ and $b$ are relatively prime to solve the problem.
To facilitate efficient search, we apply the following criteria:
\begin{enumerate*}
\item To satisfy (7) and $a^2 + b^2 < 1000$ , we have $1 \leq b \leq 15$ .
In the outer layer, we search for $b$ in a decreasing order.
In the inner layer, for each given $b$ , we search for $a$ .
\item Given $b$ , we search for $a$ in the range $\sqrt{3} b < a < \sqrt{1000 - b^2}$ .
\item We can prove that for $b \geq 9$ , there is no feasible $a$ .
The proof is as follows.
For $b \geq 9$ , to satisfy $a^2 + b^2 < 1000$ , we have $a \leq 30$ .
Thus, $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Thus, the R.H.S. of (8) has the following upper bound \begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}
Hence, to satisfy (8), a necessary condition is \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{10}{27} . \end{align*}
However, this cannot be satisfied for $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Therefore, there is no feasible solution for $b \geq 9$ .
Therefore, we only need to consider $b \leq 8$
\item We eliminate $a$ that are not relatively prime to $b$
\item We use the following criteria to quickly eliminate $a$ that make $a^2 + b^2$ a composite number.
\item For the remaining $\left( b, a \right)$ , check whether (8) and the condition that $a^2 + b^2$ is prime are both satisfied.
The first feasible solution is $b = 5$ and $a = 18$ .
Thus, $a^2 + b^2 = 349$
\item For the remaining search, given $b$ , we only search for $a \geq \sqrt{349 - b^2}$
Following the above search criteria, we find the final answer as $b = 5$ and $a = 18$ .
Thus, the largest prime $p$ is $p = a^2 + b^2 = \boxed{349}$
| 349
|
5,299
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1
| 1
|
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.
|
In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\] We are given that \begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*} The second equation implies that $a=5d.$ Substituting this into the first equation, we get \begin{align*} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{align*} It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\boxed{220}.$
| 220
|
5,300
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1
| 2
|
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.
|
Let the terms in the sequence be defined as \[a_1, a_2, ..., a_6.\]
Since this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, \[\sum_{i=1}^6 a_i=3(a_1+a_6)=990.\] Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}.$
| 220
|
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