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 |
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5,401 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | 6 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | Let the midpoint of $BC$ be $M$ . Angle-chase and observe that $\Delta AMD~\Delta ABM~\Delta MCD$ . Let $BM=CM=a$ and $AM=x$ and $DM=y$ . As a result of this similarity, we write
\[\dfrac2a=\dfrac a3,\]
which gives $a=\sqrt 6$ . Similarly, we write
\[\dfrac2x=\dfrac x7\]
and
\[\dfrac3y=\dfrac y7\]
to get $x=\sqrt{14}$ ... | 180 |
5,402 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 | 1 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | Because $0 < x < 1$ , we have \begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^... | 220 |
5,403 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 | 2 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | Note that $2022 = 210\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ $70$ $42$ , or $30$ is $132 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$ , it can only be expressed as a sum in terms of onl... | 220 |
5,404 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_14 | 1 | For positive integers $a$ $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a,... | This should be $\lfloor \frac{1000}{c} \rfloor$ . The current function breaks when $c \mid 1000$ and $b \mid c$ . Take $c = 200$ and $b = 20$ . Then, we have $\lfloor \frac{999}{200} \rfloor = 4$ stamps of value 200, $\lfloor \frac{199}{20} \rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a co... | 188 |
5,405 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | 1 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexago... | First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Further... | 140 |
5,406 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | 2 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexago... | Denote by $O$ the center of $\Omega$ .
Denote by $r$ the radius of $\Omega$
We have $O_1$ $O_2$ $A$ $B$ $C$ $D$ are all on circle $\Omega$
Denote $\angle O_1 O O_2 = 2 \theta$ .
Denote $\angle O_1 O B = \alpha$ .
Denote $\angle O_2 O A = \beta$
Because $B$ and $C$ are on circles $\omega_1$ and $\Omega$ $BC$ is a perpen... | 140 |
5,407 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | 3 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexago... | Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \[B'O_2 = BO_1 = O_1 P = O_1 C,\] \[A'O_1 = AO_2 = O_2 P = O_2 D.\] We establish the equality of the arcs and conclude that the corresponding chords are equal \[\overset{\Large\frown} {CO_1} + \ove... | 140 |
5,408 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | 1 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] ... | Again, let the intersection of $AE$ and $BC$ be $G$ . By AA similarity, $\triangle AFG \sim \triangle CDG$ with a $\frac{7}{3}$ ratio. Define $x$ as $\frac{[CDG]}{9}$ . Because of similar triangles, $[AFG] = 49x$ . Using $ABCD$ , the area of the parallelogram is $33-18x$ . Using $AECF$ , the area of the parallelogram i... | 109 |
5,409 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | 2 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] ... | Let $P = AD \cap FC$ , and $K = AE \cap BC$ . Also let $AP = x$
$CK$ also has to be $x$ by parallelogram properties. Then $PD$ and $BK$ must be $11-x$ because the sum of the segments has to be $11$
We can easily solve for $PC$ by the Pythagorean Theorem: \begin{align*} DC^2 + PD^2 &= PC^2\\ 9 + (11-x)^2 &= PC^2 \end{al... | 109 |
5,410 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | 3 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] ... | Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$
Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\overline{CE}$ is $(x-11)^2+y^2=49.$
... | 109 |
5,411 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | 4 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] ... | Let the intersection of $AE$ and $BC$ be $G$ . It is useful to find $\tan(\angle DAE)$ , because $\tan(\angle DAE)=\frac{3}{BG}$ and $\frac{3}{\tan(\angle DAE)}=BG$ . From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = $33-3BG=33-\frac{9}{\tan(\angle DAE)}$
let $\angle CAD =... | 109 |
5,412 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4 | 1 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$ . Continuing this pattern until $21$ coins in the first pile, w... | 331 |
5,413 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4 | 2 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\geq b\geq c.$ By stars and bars, we have $\dbinom{65}{2}=2080$ to assign positive integer solutions to $a + b + c = 66.$ Now we need to subtract off the cases where it doesn't ... | 331 |
5,414 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4 | 3 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$ . Then, let $b = a + k_1$ , and $c = b + k_2$ , such that each $k_i \geq 1$ . The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66$ . This is simply the number of positive solutions to the equation $3x+2y+z = 66$ . Now, we take cases on $a... | 331 |
5,415 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | 1 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations \begin{align*} (s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\ x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\ x^2+y^2+(s-z)^2&=\left(10\sqr... | 192 |
5,416 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | 2 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\] Subtracting the fourth equation gives \begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\ x^2 + y^2... | 192 |
5,417 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | 3 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | Let $E$ be the vertex of the cube such that $ABED$ is a square.
Using the British Flag Theorem , we can easily show that \[PA^2 + PE^2 = PB^2 + PD^2\] and \[PA^2 + PG^2 = PC^2 + PE^2\] Hence, by adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2... | 192 |
5,418 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | 4 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | For all points $X$ in space, define the function $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$ . Then $f$ is linear; let $O=\frac{2A+G}{3}$ be the center of $\triangle BCD$ . Then since $f$ is linear, \begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\ \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\ri... | 192 |
5,419 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 1 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by ... | 567 |
5,420 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 2 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$ , and the length of the altitude from $B$ to $AC$ is also $10$ . Let the foot of this altitude be $F$ , and let the foot of the altitude from $A$ to $BC$ be denoted as $E$ . Then, $\Delta BCF \sim \Delta ACE$ . So, $\fra... | 567 |
5,421 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 3 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Make $AE$ perpendicular to $BC$ $AG$ perpendicular to $BD$ $AF$ perpendicular $DC$
It's obvious that $\triangle{AEB} \sim \triangle{AFD}$ . Let $EB=5x; AB=5y; DF=6x; AD=6y$ . Then make $BQ$ perpendicular to $DC$ , it's easy to get $BQ=18$
Since $AB$ parallel to $DC$ $\angle{ABG}=\angle{BDQ}$ , so $\triangle{ABG} \sim \... | 567 |
5,422 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 4 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let the foot of the altitude from $A$ to $BC$ be $P$ , to $CD$ be $Q$ , and to $BD$ be $R$
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$ . As $\angle QAB = 90^\circ$ , we have that $\angle QAR = 90^\ci... | 567 |
5,423 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 5 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$ , respectively.
Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$
Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadril... | 567 |
5,424 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 6 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $AD=BC=a$ . Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$ $F$ be the foot of the perpendicular from $A$ to line $BC$ , and $H$ be the foot of the perpendicular from $A$ to $DC$
Note that $\triangle CBG\sim\triangle CAF$ , and we get that $\frac{10}{15}=\frac{a}{AC}$ . Therefore, $... | 567 |
5,425 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 7 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Draw the distances in terms of $B$ , as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$ . As a result, let $AB=u$ , then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$ . The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$ . By angle subtraction, $\cos(180-\theta)=-\cos\theta$ . There... | 567 |
5,426 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 8 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | [asy] size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightangl... | 567 |
5,427 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$ .
Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$ . Denote $\theta = \angle{CBH}.$ It is clear t... | 567 |
5,428 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | 10 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $F$ be on $DC$ such that $AF \| DC$ . Let $G$ be on $BD$ such that $AG \| BD$
Let $m$ be the length of $AB$ . Let $n$ be the length of $AD$
The area of $\triangle ABD$ can be expressed in three ways: $\frac{1}{2}(15)(BC) = \frac{1}{2}(15)(n)$ $\frac{1}{2}(18)(m)$ , and $\frac{1}{2}(10)(BD)$
\[\frac{1}{2}(15)(n) = \... | 567 |
5,429 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | 1 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn... | Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively: \begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\t... | 301 |
5,430 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | 2 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn... | Let the brackets denote areas.
We find $[ABCD]$ in two different ways:
Equating the expressions for $[ABCD],$ we have \[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{... | 301 |
5,431 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | 3 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn... | We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09]* WLOG we ... | 301 |
5,432 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | 4 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn... | Solution
In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to $\cos \theta,$ where $\theta$ is the acute angle between the diagonals. \begin{align*} s &= A'B' + B'C' + C'D' + D'A' \\ &= (AB + BC + CD + DA)\cos \theta \\ &= (a + b + c + d)\cos \theta \\ &= 22\cos \the... | 301 |
5,433 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 1 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | Let $O_i$ and $r_i$ be the center and radius of $\omega_i$ , and let $O$ and $r$ be the center and radius of $\omega$
Since $\overline{AB}$ extends to an arc with arc $120^\circ$ , the distance from $O$ to $\overline{AB}$ is $r/2$ . Let $X=\overline{AB}\cap \overline{O_1O_2}$ . Consider $\triangle OO_1O_2$ . The line $... | 672 |
5,434 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 2 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | Let $O_{1}$ $O_{2}$ , and $O$ be the centers of $\omega_{1}$ $\omega_{2}$ , and $\omega$ with $r_{1}$ $r_{2}$ , and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$ . Let $x=O_{1}O_{2}$ and $O^{\prime}$ the or... | 672 |
5,435 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 4 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | Suppose we label the points as shown below [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0... | 672 |
5,436 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 5 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$ , and let $u=TO_{2}$ $v=TO_{1}$ $x=PT$ , where the length $O_... | 672 |
5,437 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 6 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$ . Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_... | 672 |
5,438 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 7 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | We are not given the radius of circle $w$ , but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$ , call it $W$ , and $w_2$ , call it $W_... | 672 |
5,439 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | 8 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$ $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$
Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$
Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\b... | 672 |
5,440 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | 1 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | We first claim that $n$ must be divisible by $42$ . Since $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \neq 0,1 \pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.
Then $\sigma(a^n)-1 = \su... | 125 |
5,441 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | 2 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | $n$ only needs to satisfy $\sigma(a^n)\equiv 1 \pmod{43}$ and $\sigma(a^n)\equiv 1 \pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . The following three statements are the same:
We ... | 125 |
5,442 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | 3 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | We perform casework on $a:$
Finally, the least such positive integer $n$ for all cases is \begin{align*} n&=\operatorname{lcm}(42,43,46,47) \\ &=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\ &=2\cdot3\cdot7\cdot23\cdot43\cdot47, \end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\boxed{125}.$ | 125 |
5,443 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | 4 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem , we get that $n \equiv 0 \pmod{42}$ and $n \equiv 0 \pmod{46}.$ Then, we can look at $a$ being a ... | 125 |
5,444 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15 | 1 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ | Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately:
Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\boxed{285}.$ | 285 |
5,445 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15 | 2 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ | Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$ . Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$ . Completing the square gives us $\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}$ ; this explains why the two parabolas int... | 285 |
5,446 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15 | 3 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ | Claim
Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$ . Suppose that these parabolas intersect at four points.
Then these points lie on the circle centered at point $(p_2, p_1)$ with rad... | 285 |
5,447 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | 1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=\boxed{550},$ which is the final answer. | 550 |
5,448 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | 2 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | For any palindrome $\underline{ABA},$ note that $\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ The average for $A$ is $5$ since $A$ can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ The average for $B$ is $4.5$ since $B$ is either $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ Therefore, the answer is $505 + 45 = \boxed{550}.... | 550 |
5,449 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | 3 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | For every three-digit palindrome $\underline{ABA}$ with $A\in\{1,2,3,4,5,6,7,8,9\}$ and $B\in\{0,1,2,3,4,5,6,7,8,9\},$ note that $\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \be... | 550 |
5,450 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | 4 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | We notice that a three-digit palindrome looks like this: $\underline{aba}.$
And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\times{10}=90$ three-digit palindromes.
We want to find the sum of these $90$ palindromes and divide it by $90$ to find the ari... | 550 |
5,451 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | 5 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$ | 550 |
5,452 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | 1 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ suc... | By angle chasing, we conclude that $\triangle AGF$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle, and $\triangle BED$ is a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle.
Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\triangle BED,$ we have $DE=\frac{840-x}{2}$ and... | 336 |
5,453 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | 2 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ suc... | We express the areas of $\triangle BED$ and $\triangle AFG$ in terms of $AF$ in order to solve for $AF.$
We let $x = AF.$ Because $\triangle AFG$ is isosceles and $\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$
Let the height of $\triangle ABC$ be $h$ and the height of $\triangle AEF$ be $h'.$ Then we have tha... | 336 |
5,454 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | 3 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ suc... | $\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}$ So, If $\Delta AFG$ is isosceles, it means that $AF = FG$
Let $AF = FG = AE = EF = x$
So, $[\Delta AFG] = \frac{1}{2} \cdot x^{2} \textup{sin} 120^{0} = \frac{\sqrt{3}}{4}x^{2}$
In $\Delta BED$ $BE = 840 - x$ , Hence $DE = \frac{840 - x}{2... | 336 |
5,455 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | 4 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ suc... | Since $\triangle AFG$ is isosceles, $AF = FG$ , and since $\triangle AEF$ is equilateral, $AF = EF$ . Thus, $EF = FG$ , and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from $E$ and $F$ to line $BC$ ; call the meeting points $P$ and $Q$ , respectively. $\triangle BEP$ is cl... | 336 |
5,456 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | 1 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\sqrt{n}\cdot i$ and $\overline{z}=m-\sqrt{n}\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\overline{z}.$
We know tha... | 330 |
5,457 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | 2 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | $(-20)^{3} + (-20)a + b = 0$ , hence $-20a + b = 8000$
Also, $(-21)^{3} + c(-21)^{2} + d = 0$ , hence $441c + d = 9261$
$m + i \sqrt{n}$ satisfies both $\Rightarrow$ we can put it in both equations and equate to 0.
In the first equation, we get $(m + i \sqrt{n})^{3} + a(m + i \sqrt{n}) + b = 0$ Simplifying this further... | 330 |
5,458 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | 3 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$ . After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$ , respectively. Since we know that if you substitute the root of a function back into the function, the output ... | 330 |
5,459 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | 4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$
Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$ , and $Q(x) = x^2 + (c-21)x + (441 - 21c)$
By t... | 330 |
5,460 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | 5 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | We plug -20 into the equation obtaining $(-20)^3-20a+b$ , likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$
Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$
By Vieta's, the sum of the roots in the first equation is $0$... | 330 |
5,461 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | 6 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | Since $m+i\sqrt{n}$ is a common root and all the coefficients are real, $m-i\sqrt{n}$ must be a common root, too.
Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients).
First, however, the product of the two common roots is: \begin{align*} &&&(x-m-... | 330 |
5,462 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 1 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$ rd side is between $6$ and $14$ , exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the ... | 736 |
5,463 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 2 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:
For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:
Case (1): The longest side has length $\boldsymbol{10,}$ so $... | 736 |
5,464 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 3 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | We have the diagram below.
[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("$A$",(0,0),SW); label("$B$",(1,2*sqrt(3)),N); label("$C$",(10,0),SE); label("$\theta$",(0,0),NE); label("$\alpha$",(1,2*sqrt(3)),SSE); label("$4$",(0,0)--(1,2*sqrt(3)),WNW); label("$10$",(0,0)--(10,0),... | 736 |
5,465 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 4 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | Note: Archimedes15 Solution which I added an answer
here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most ... | 736 |
5,466 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, ... | 736 |
5,467 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 6 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | Let a triangle in $\tau(s)$ be $ABC$ , where $AB = 4$ and $BC = 10$ . We will proceed with two cases:
Case 1: $\angle ABC$ is obtuse. If $\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$ ; therefore, the area of the triangle will fall in the r... | 736 |
5,468 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 7 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides $4$ and $10,$ when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$
The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10$ $AC\perp... | 736 |
5,469 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | 8 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | If $4$ and $10$ are the shortest sides and $\angle C$ is the included angle, then the area is \[\frac{4\cdot10\cdot\sin\angle C}{2} = 20\sin\angle C.\] Because $0\leq\sin\angle C\leq1$ , the maximum value of $20\sin\angle C$ is $20$ , so $s\leq20$
If $4$ is a shortest side and $10$ is the longest side, the length of th... | 736 |
5,470 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | 1 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$ . Substituting into the equation and factoring, we get that $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ , so therefore $A \subseteq B$ or $B \subseteq A$ . WLOG $A\subseteq B$ , then for each element there are $3$ possibilities, either it is in both $A$ and $B$ , it is in $B$ bu... | 454 |
5,471 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | 2 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | We denote $\Omega = \left\{ 1 , 2 , 3 , 4 , 5 \right\}$ .
We denote $X = A \cap B$ $Y = A \backslash \left( A \cap B \right)$ $Z = B \backslash \left( A \cap B \right)$ $W = \Omega \backslash \left( A \cup B \right)$
Therefore, $X \cup Y \cup Z \cup W = \Omega$ and the intersection of any two out of sets $X$ $Y$ $Z$ $W... | 454 |
5,472 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | 3 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | The answer is \begin{align*} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\ &=2(2+1)^5-(1+1)^5 \\ &=2(243)-32 \\ &=\boxed{454} ~MRENTHUSIASM | 454 |
5,473 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | 4 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | The answer is \begin{align*} &\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ &=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\bino... | 454 |
5,474 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | 5 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | Proceed with Solution 1 to get $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ . WLOG, assume $|A| = |A \cap B|$ . Thus, $A \subseteq B$
Since $A \subseteq B$ , if $|B| = n$ , there are $2^n$ possible sets $A$ , and there are also ${5 \choose n}$ ways of choosing such $B$
Therefore, the number of possible pairs of sets $(A,... | 454 |
5,475 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | 1 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | From the fourth equation we get $d=\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14$ . Hence $(abc)^2 - 14(abc)-120 = 0$ . Solving, we get $abc = -6$ or $abc = 20$ . From the first and second equation, we get $ab + bc + ca = ab-3c = -4... | 145 |
5,476 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | 2 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | Note that $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$ . Hence, $ab = 3c - 4$
Rewriting $abc+bcd+cda+dab = 14$ , we get $ab(c+d) + cd(a+b) = 14$ .
Substitute $ab = 3c - 4$ and solving, we get \[3c^{2} - 4c - 4d - 14 = 0.\] We refer to this as Equation 1.
Note that $abcd = 30$ gives $(3c-4)cd = 30$ . So, ... | 145 |
5,477 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | 3 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.
Rearranging $(2)$ and solving for $c,$ we have \begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \begin{align*} ab\left(\... | 145 |
5,478 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | 4 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: \[abc+d(ab+bc+ca)=14\] Then we plug in equation 2 to receive $abc-4d=14$ . By equation 4 we get $abc=\frac{30}{d}$ . Plugging in, we get $\frac{30}{d}-4d=14$ . Multiply by $d$ on both sides to get the quadratic e... | 145 |
5,479 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | 5 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | Let the four equations from top to bottom be listed $(1)$ through $(4)$ respectively. Multiplying both sides of $(3)$ by $d$ and factoring some terms gives us $abcd + d^2(ab + ac + bc) = 14d$ . Substituting using equations $(4)$ and $(2)$ gives us $30 -4 d^2 = 14d$ , and solving gives us $d = -5$ or $d = \frac{3}{2}$ .... | 145 |
5,480 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_9 | 1 | Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$ | This solution refers to the Remarks section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1... | 295 |
5,481 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10 | 1 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with ... | This solution refers to the Diagram section.
As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$ ) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is th... | 335 |
5,482 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10 | 2 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with ... | The centers of the three spheres form a $49$ $49$ $72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$ , which is $36$ away from the midpoint $A$ of the $72$ si... | 335 |
5,483 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10 | 3 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with ... | This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((10,-3,-40)); triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T... | 335 |
5,484 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11 | 1 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $... | Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$
In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain... | 258 |
5,485 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11 | 2 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $... | We know right away that $42\not\in S$ and $84\not\in S$ as stated in Solution 1.
To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible set... | 258 |
5,486 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11 | 3 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $... | In a solution that satisfies these constraints, the multiple of 6 must be adjacent to multiple of 7. The other two numbers must be on either side.
WLOG assume the set is $\{a,6j,7k,b\}$ . The student with numbers $a$ $6j$ , and $7k$ can think the set is $\{a-1, a,6j,7k\}$ or $\{a,6j,7k,b\}$ , and the students with numb... | 258 |
5,487 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11 | 4 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $... | Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$ . If one of the values in $S$ is $3$ or $4 \pmod{7}$ , observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \pmod{7}$ and $S$ contains a $0 \pmod{7}$ , then the values of $S$ must be $(b-3, b-2, b-1... | 258 |
5,488 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_13 | 1 | Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$ | Recall that $1000$ divides this expression if $8$ and $125$ both divide it. It should be fairly obvious that $n \geq 3$ ; so we may break up the initial condition into two sub-conditions.
(1) $5^n \equiv n \pmod{8}$ . Notice that the square of any odd integer is $1$ modulo $8$ (proof by plugging in $1^2,3^2,5^2,7^2$ in... | 797 |
5,489 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_13 | 2 | Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$ | We have that $2^n + 5^n \equiv n\pmod{1000}$ , or $2^n + 5^n \equiv n \pmod{8}$ and $2^n + 5^n \equiv n \pmod{125}$ by CRT. It is easy to check $n < 3$ don't work, so we have that $n \geq 3$ . Then, $2^n \equiv 0 \pmod{8}$ and $5^n \equiv 0 \pmod{125}$ , so we just have $5^n \equiv n \pmod{8}$ and $2^n \equiv n \pmod{1... | 797 |
5,490 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 | 1 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ ... | In this solution, all angle measures are in degrees.
Let $M$ be the midpoint of $\overline{BC}$ so that $\overline{OM}\perp\overline{BC}$ and $A,G,M$ are collinear. Let $\angle ABC=13k,\angle BCA=2k$ and $\angle XOY=17k.$
Note that:
Together, we conclude that $\triangle OAM \sim \triangle OXY$ by AA, so $\angle AOM = \... | 592 |
5,491 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 | 2 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ ... | Let $M$ be the midpoint of $BC$ . Because $\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o$ $AXOG$ and $OMYG$ are cyclic, so $O$ is the center of the spiral similarity sending $AM$ to $XY$ , and $\angle{XOY}=\angle{AOM}$ . Because $\angle{AOM}=2\angle{BCA}+\angle{BAC}$ , it's easy to get $\frac{585}{7} \implies \b... | 592 |
5,492 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 | 3 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ ... | Firstly, let $M$ be the midpoint of $BC$ . Then, $\angle OMB = 90^o$ . Now, note that since $\angle OGX = \angle XAO = 90^o$ , quadrilateral $AGOX$ is cyclic. Also, because $\angle OMY + \angle OGY = 180^o$ $OMYG$ is also cyclic. Now, we define some variables: let $\alpha$ be the constant such that $\angle ABC = 13\alp... | 592 |
5,493 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 | 4 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ ... | [asy] /* Made by MRENTHUSIASM */ size(375); pair A, B, C, O, G, X, Y; A = origin; B = (1,0); C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); O = circumcenter(A,B,C); G = centroid(A,B,C); Y = intersectionpoint(G--G+(100,0),B--C); X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); pair O1=cir... | 592 |
5,494 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 | 5 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ ... | Extend $XA$ and meet line $CB$ at $P$ . Extend $AG$ to meet $BC$ at $F$ . Since $AF$ is the median from $A$ to $BC$ $A,G,F$ are collinear. Furthermore, $OF$ is perpendicular to $BC$
Draw the circumcircle of $\triangle{XPY}$ , as $OA\bot XP, OG\bot XY, OF\bot PY$ $A,G,F$ are collinear, $O$ lies on $(XYP)$ as $AGF$ is th... | 592 |
5,495 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 | 1 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n... | Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$ . Note that this formula also returns the correct value when $n=(k+1)^2$ , but not when $n=k^2$ . ... | 258 |
5,496 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 | 2 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n... | Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$ . If $k$ is odd, then $f(n)=g(n)$ . If $k$ is even, then \begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*} from which \begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*} Since $k$ is even, $m$ ... | 258 |
5,497 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 | 3 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n... | To begin, note that if $n$ is a perfect square, $f(n)=g(n)$ , so $f(n)/g(n)=1$ , so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$ , making the values of $f(n+k)$ and $g(n+k)$ integers. Usi... | 258 |
5,498 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 | 4 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n... | First of all, if $n$ is a perfect square, $f(n)=g(n)=\sqrt{n}$ and their quotient is $1.$ So, for the rest of this solution, assume $n$ is not a perfect square.
Let $a^2$ be the smallest perfect square greater than $n$ and let $b^2$ be the smallest perfect square greater than $n$ with the same parity as $n,$ and note t... | 258 |
5,499 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 | 5 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n... | Say the answer is in the form n^2-x, then x must be odd or else f(x) = g(x). Say y = n^2-x. f(y) = x+n, g(y) = 3n+2+x. Because f(y)/g(y) = 4*(an integer)/7*(an integer), f(y) is 4*(an integer) so n must be odd or else f(y) would be odd. Solving for x in terms of n gives integer x = (5/3)n+8/3 which means n is 2 mod 3,... | 258 |
5,500 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_1 | 1 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | [asy] size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, W); label("$E$", F, E); [/asy]
If w... | 547 |
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