id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
5,201 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8 | 1 | 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... | 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 $(6)(68)+34+34+34a+34b$ , du... | 197 |
5,202 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8 | 2 | 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... | 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 $... | 197 |
5,203 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8 | 3 | 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... | 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... | 197 |
5,204 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8 | 4 | 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... | 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 $C$ . R... | 197 |
5,205 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8 | 5 | 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... | 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 connectin... | 197 |
5,206 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_9 | 1 | 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. | 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 $y = -\frac{1}{m}x.$ Bec... | 480 |
5,207 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_9 | 2 | 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. | 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 ... | 480 |
5,208 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_9 | 4 | 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. | 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 |
5,209 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10 | 1 | 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 $m + n$ | 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... | 113 |
5,210 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10 | 2 | 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 $m + n$ | 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 |
5,211 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10 | 3 | 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 $m + n$ | 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 $\... | 113 |
5,212 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10 | 4 | 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 $m + n$ | 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... | 113 |
5,213 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_10 | 5 | 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 ... | 113 |
5,214 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_12 | 1 | Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\] | 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... | 385 |
5,215 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_12 | 2 | Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\] | 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 ever... | 385 |
5,216 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_12 | 3 | Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\] | 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 $\lef... | 385 |
5,217 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_13 | 1 | 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 $p^{2}$ | 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\... | 110 |
5,218 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_13 | 2 | 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 $p^{2}$ | 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 \(\boxed{110}\). | 110 |
5,219 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_13 | 3 | 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 $p^{2}$ | 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}... | 110 |
5,220 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_14 | 1 | 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 $m$ $n$ , ... | 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\\... | 104 |
5,221 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_14 | 2 | 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 $m$ $n$ , ... | [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--c... | 104 |
5,222 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_14 | 3 | 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 $m$ $n$ , ... | 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 fi... | 104 |
5,223 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_14 | 4 | 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 $m$ $n$ , ... | 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... | 104 |
5,224 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_14 | 5 | 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 $m$ $n$ , ... | 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 observ... | 104 |
5,225 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_15 | 1 | 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... | 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 $a$... | 721 |
5,226 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_15 | 2 | 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... | 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) \\
& = \le... | 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... | 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+... | 721 |
5,228 | 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 sa... | 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 $... | 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 woul... | 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... | 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 valu... | 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... | 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:
$3000a+300b+30c+3d+1111 \equiv 0 \pmod{7}$
A... | 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 eq... | 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 rad... | 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 int... | 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 ... | 902 |
5,238 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_9 | 3 | 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: | 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 be... | 902 |
5,239 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_10 | 1 | Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$ , circumradius $13$ , and inradius $6$ . Find $AB\cdot AC$ | 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 mi... | 468 |
5,240 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_10 | 2 | 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=\... | 468 |
5,241 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_11 | 1 | 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 - 10... | 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 poly... | 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 ... | 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... | 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... | 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 $12$ , w... | 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 $(... | 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)
& =... | 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}... | 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, $x + y \leq b - 1$
T... | 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... | 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... | 315 |
5,251 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15 | 2 | 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... | 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 ... | 315 |
5,252 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15 | 3 | 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... | 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.
Be... | 315 |
5,253 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 1 | 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 $m+n.$ | 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 fiv... | 191 |
5,254 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 2 | 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 $m+n.$ | 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... | 191 |
5,255 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 3 | 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 $m+n.$ | 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 ... | 191 |
5,256 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 4 | 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 $m+n.$ | 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 $5$ ... | 191 |
5,257 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 5 | 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 $m+n.$ | 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 constr... | 191 |
5,258 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 6 | 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 $m+n.$ | 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 ... | 191 |
5,259 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 7 | 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 $m+n.$ | 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 ... | 191 |
5,260 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 8 | 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 $m+n.$ | 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... | 191 |
5,261 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | 9 | 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 $m+n.$ | 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... | 191 |
5,262 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_3 | 1 | 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. Fin... | 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 int... | 607 |
5,263 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 2 | 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 $ABCD.$ | 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 perpen... | 106 |
5,264 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 3 | 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 $ABCD.$ | [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... | 106 |
5,265 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 4 | 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 $ABCD.$ | 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, ... | 106 |
5,266 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 5 | 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 $ABCD.$ | 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... | 106 |
5,267 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 8 | 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 $ABCD.$ | [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)... | 106 |
5,268 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 9 | 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 $ABCD.$ | 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... | 106 |
5,269 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | 10 | 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 $ABCD.$ | 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 $r \sin \theta$ . There... | 106 |
5,270 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6 | 4 | 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. ... | 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 b... | 51 |
5,271 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | 1 | 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 $ABCD.$ | 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},\overleftrigh... | 125 |
5,272 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | 2 | 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 $ABCD.$ | 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 $\ove... | 125 |
5,273 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | 3 | 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 $ABCD.$ | 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 ci... | 125 |
5,274 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | 4 | 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 $ABCD.$ | 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 $BC$ ... | 125 |
5,275 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | 5 | 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 $ABCD.$ | 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{O... | 125 |
5,276 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | 6 | 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 $ABCD.$ | 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. \[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\] \[PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.\] \[PE = \sqrt{RP^2 + RE^2} = 15 \implies E = E'.\] \[\... | 125 |
5,277 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9 | 1 | 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 $p(m) = p(2).$ | 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... | 738 |
5,278 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9 | 2 | 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 $p(m) = p(2).$ | $p(x)-p(2)$ 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 $|4+4m|\leq 20$ (whi... | 738 |
5,279 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | 1 | 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 $x$ .) | 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)... | 944 |
5,280 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | 2 | 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 $x$ .) | 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' &= \su... | 944 |
5,281 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | 3 | 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 $x$ .) | 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, no... | 944 |
5,282 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 1 | 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 $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | 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 i... | 235 |
5,283 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 2 | 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 ha... | 235 |
5,284 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 3 | 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 doe... | 235 |
5,285 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 4 | 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 $a_n$ 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 su... | 235 |
5,286 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 5 | 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 =... | 235 |
5,287 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 6 | 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 an... | 235 |
5,288 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 7 | 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 $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | 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... | 235 |
5,289 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | 8 | 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 tran... | 235 |
5,290 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | 1 | 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 ... | 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... | 125 |
5,291 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | 2 | 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 ... | 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 o... | 125 |
5,292 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | 3 | 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 ... | 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 bo... | 125 |
5,293 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | 4 | 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 ... | 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... | 125 |
5,294 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14 | 1 | 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) { ... | 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... | 608 |
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) { ... | 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 bac... | 608 |
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) { ... | 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, ... | 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... | 349 |
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, \b... | 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 ... | 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 ... | 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 ... | 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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