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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5,301 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | 1 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, P; A = orig... | 250 |
5,302 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | 2 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Let the common angle be $\theta$ . Note that $\angle PAC = 90^\circ-\theta$ , thus $\angle APC = 90^\circ$ . From there, we know that $AC = \frac{10}{\sin\theta}$
Note that $\angle ABP = 45^\circ-\theta$ , so from law of sines we have \[... | 250 |
5,303 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | 3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Do some angle chasing yielding:
We have $AC=\frac{10}{\sin\theta}$ since $\triangle APC$ is a right triangle. Since $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle, $AB=\frac{10}{\sin\theta}$ , and $BC=\frac{10\sqrt{2}}{\s... | 250 |
5,304 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | 4 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Notice that in triangle $PBC$ $\angle PBC + 45-\angle PCA = 45^\circ$ , so $\angle BPC = 135^\circ$ . Similar logic shows $\angle APC = 135^\circ$
Now, we see that $\triangle APB \sim \triangle BPC$ with ratio $1:\sqrt{2}$ (as $\triangle... | 250 |
5,305 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | 5 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB... | 250 |
5,306 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | 6 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Denote $\angle PCA = \theta$ . Then, by trig Ceva's: \begin{align*} \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\th... | 250 |
5,307 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | 1 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We first subtract the second equation from the first, noting that they both equal $60$ \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$
The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} from which it is apparent that $x=4$ and $x=11... | 273 |
5,308 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | 2 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We index these equations as (1), (2), and (3), respectively.
Taking $(1)-(2)$ , we get \[ \left( x - z \right) \left( y - 4 \right) = 0 . \]
Denote $x' = x - 4$ $y' = y - 4$ $z' = z - 4$ .
Thus, the above equation can be equivalently written as \[ \left( x' - z' \right) y' = 0 . \hspace{1cm} (1') \]
Similarly, by takin... | 273 |
5,309 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | 3 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We index these equations as (1), (2), and (3), respectively.
Using equation (1), we get $z = \frac{60 - xy}{4} = 15 - \frac{xy}{4}$ We need to solve for x, so we plug this value of z into equation (3) to get:
\[15x - \frac{x^2y}{4} - 4y = 60\] \[\frac{y}{4} * x^2 - 15x + (60 + 4y) = 0\]
We use the quadratic formula to ... | 273 |
5,310 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | 4 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | Since all three equations are in the form $\frac{K}{a} + 4a = 60$ where $K = xyz$ , we can rearrange this to see that $x$ $y$ , and $z$ all satisfy
\[ 4a^2 - 60a + K = 0. \]
Let this quadratic have roots $a_1$ and $a_2$ . Then, there are two cases to consider: two of $x$ $y$ $z$ are equal to $a_1$ and the third is equa... | 273 |
5,311 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | 5 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We index these equations as (1), (2), and (3) respectively (same as solution 2). There are two possible cases:
Case 1: $x = \pm 4$
In this case, we simply plug in $x = 4$ and $x = -4$ . We note that $x=4$ is a valid case.
Case 2: $x \neq \pm 4$
In this case, using equation (3), we get $y = 15 - \frac{xz}{4}$ . Plugging... | 273 |
5,312 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_5 | 1 | Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be... | Denote $r = \frac{a}{b}$ , where $\left( a, b \right) = 1$ .
We have $55 r = \frac{55a}{b}$ .
Suppose $\left( 55, b \right) = 1$ , then the sum of the numerator and the denominator of $55r$ is $55a + b$ .
This cannot be equal to the sum of the numerator and the denominator of $r$ $a + b$ .
Therefore, $\left( 55, b \rig... | 719 |
5,313 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | 1 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options... | 928 |
5,314 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | 2 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | First, we identify the geometric condition for the sentence ``four vertices colored the
same color are the four vertices of a rectangle Consider any four vertices on the dodecagon, $A$ $B$ $C$ $D$ .
Denote by $O$ the center of the dodecagon.
Because $OA = OB = OC$ $\angle OAB = \angle OBA$ and $\angle OBC = \angle OCB$... | 928 |
5,315 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | 3 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | Note that a rectangle is formed if and only if it's diagonals pass through the center of the dodecagon and the diagonal's endpoints are the same color.
Consider the $6$ diagonals that pass through the center. A rectangle is formed if the endpoints of some pair of them are all the same color. We can now perform casework... | 928 |
5,316 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | 4 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | \[\text{First, we notice that a rectangle is made from two pairs of vertices 1/2 turn away from each other.}\]
\[\textit{Note: The image is }\frac{\textit{280}}{\textit{841}}\approx\frac{\textit{1}}{\textit{3}}\textit{ size.}\]
\[\text{For there to be no rectangles, there can be at most one same-colored pair for each c... | 64 |
5,317 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 | 4 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | The product can be factored into $-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)$
where $r,s,t$ are the roots of the polynomial $x^3+x+1=0$
This is then $-(r^7-1)(s^7-1)(t^7-1)$ because $(r^7-1)$ and $(r-1)(r-w)(r-w^2)...(r-w^6)$ share the same roots.
To find $-(r^7-1)(s^7-1)(t^7-1)$
Noti... | 24 |
5,318 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_10 | 1 | Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\be... | We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then, t... | 144 |
5,319 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_10 | 2 | Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\be... | Let's carry out an archaeological study, that is, we will find the bones (the base) and "build up the meat."
1. Let "bones" of number $X$ be $X \pmod 3.$ Then the “skeleton” of the original table is
\[1 0 2 1 0 2\] \[2 1 0 2 1 0\] By condition, the table cannot have a column of two identical numbers (the difference of ... | 144 |
5,320 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_11 | 3 | Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | Firstly, there cannot be two subsets with cardinality 1, or they will not intersect.
If there is one subset $A$ with cardinality $1$ ; let the element in $A$ be $a$ , then there are $2^4=16$ subsets that do not include $a$ so they do not work. Every remaining subsets $S$ will have $a$ as an element so $S\cap{A}\geq1$ ,... | 81 |
5,321 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | 1 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | Because $M$ is the midpoint of $BC$ , following from the Stewart's theorem, $AM = 2 \sqrt{37}$
Because $A$ $B$ $C$ , and $P$ are concyclic, $\angle BPA = \angle C$ $\angle CPA = \angle B$
Denote $\theta = \angle PBQ$
In $\triangle BPQ$ , following from the law of sines, \[ \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\angle ... | 247 |
5,322 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | 3 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | It is clear that $BQCP$ is a parallelogram. By Stewart's Theorem, $AM=\sqrt{148}$ , POP on $M$ tells $PM=\frac{49}{\sqrt{148}}$
As $QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}$ leads to $\boxed{247}$ | 247 |
5,323 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | 4 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | First, note that by Law of Sines, $\frac{\sin(\angle{QBP})}{QP}=\frac{\sin(\angle{BPQ})}{BQ}$ and that $\frac{\sin(\angle{QCP})}{QP}=\frac{\sin(\angle{QPC})}{QP}$ . Equating the 2 expressions, you get that $\frac{\sin(\angle{BPQ})}{BQ}=\frac{\sin(\angle{QPC})}{QP}$ . Now drop the altitude from $A$ to $BC$ . As it is co... | 247 |
5,324 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | 5 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | We use the law of Cosine and get \[AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,\] \[AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies\] \[AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.\] We use the power of point $M$ with respect circumcircle $\triangle ABC$ and get \[AM \cdot MP = BM \cd... | 247 |
5,325 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_13 | 1 | Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | Denote $a_n = \sec^n A + \tan^n A$ .
For any $k$ , we have \begin{align*} a_n & = \sec^n A + \tan^n A \\ & = \left( \sec^{n-k} A + \tan^{n-k} A \right) \left( \sec^k A + \tan^k A \right) - \sec^{n-k} A \tan^k A - \tan^{n-k} A \sec^k A \\ & = a_{n-k} a_k - 2^k \sec^{n-k} A \cos^k A - 2^k \tan^{n-k} A \cot^k A \\ & = a_{... | 167 |
5,326 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_13 | 2 | Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | \[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\] \[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\] \[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\] \[= \left(\frac {\sqrt {17} ... | 167 |
5,327 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 | 1 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangl... | [asy] defaultpen(linewidth(0.6)); pair A = (0, 0), B = (5.656,2), C = (-2.828, 8), D = B+C, P = 0.875*C, Q = B+0.625*C, H = (-2.828, 0), G = (5.656, 0); pair P1 = (-5, 0), P2 = (10, 0); draw(A--B--D--C--A); filldraw(A--B--Q--P--cycle,rgb(0.35,0.7,0.9)); draw(C--H, dotted); draw(B--G, dotted); draw(P1--P2); dot("$A$",A,... | 751 |
5,328 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 | 2 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangl... | Denote $h(X)$ the distance from point $X$ to $\mathcal{P}, h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7, AB = a, AC = a \sqrt{2}.$
Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram. \[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC}{h(C... | 751 |
5,329 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 | 3 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangl... | We introduce a Cartesian coordinate system to the diagram.
We put the origin at $A$ . We let the $z$ -components of $B$ $C$ $D$ be positive.
We set the $x$ -axis in a direction such that $B$ is on the $x-O-z$ plane.
The coordinates of $A$ $B$ $C$ are $A = \left( 0, 0, 0 \right)$ $B = \left( x_B, 0 , 2 \right)$ $C = \le... | 751 |
5,330 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_15 | 1 | For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$ | Denote $a_n = 23 b_n$ .
Thus, for each $n$ , we need to find smallest positive integer $k_n$ , such that \[ 23 b_n = 2^n k_n + 1 . \]
Thus, we need to find smallest $k_n$ , such that \[ 2^n k_n \equiv - 1 \pmod{23} . \]
Now, we find the smallest $m$ , such that $2^m \equiv 1 \pmod{23}$ .
By Fermat's Theorem, we must ha... | 363 |
5,331 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | 1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let $R(x)=P(x)+Q(x).$ Since the $x^2$ -terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows... | 116 |
5,332 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | 2 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} for some constants $a,b,c$ and $d.$
We are given that \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53, &&(3)... | 116 |
5,333 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | 3 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53, &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} W... | 116 |
5,334 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | 4 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \begin{align*} 2(16)^2 + 16a + b &= 54, \\ 2(20)^2 + 20a + b &= 53. \end{align*} Hence, $a = -72.25$ and $b = 698.$
Similarly, \begin{align*}... | 116 |
5,335 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | 5 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Add the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$ . This equation must also pass through the two points $(16,54)$ and $(20,53)$
Let $m=a+c$ and $n=b+d$ . We then have two equations: \begin{align*} 108&=16m+n, \\ 106&=20m+n. \end{align*} We are trying to solve for $n=P(0)$ . ... | 116 |
5,336 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | 1 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\] Taking both sides modulo $71,$ we have \begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$
Therefore, the requested three-digit positive ... | 227 |
5,337 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | 2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we get $99a = 71b+8c$
Note that $99$ and $71$ are large numbers comparatively to $8$ , so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$ , which is a multiple of $4$ . So, if we multiply this by $2$ , it will be a multi... | 227 |
5,338 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | 3 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we get $99a = 71b+8c.$
We list a few multiples of $99$ out: \[99,198,297,396.\] Of course, $99$ can't be made of just $8$ 's. If we use one $71$ , we get a remainder of $28$ , which can't be made of $8$ 's either. So $99$ doesn't work. $198$ can't be made up of just $8$ 's. If we use one $71$ , ... | 227 |
5,339 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | 4 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we get $99a = 71b+8c$
We can see that $99$ is $28$ larger than $71$ , and we have an $8c$ . We can clearly see that $56$ is a multiple of $8$ , and any larger than $56$ would result in $c$ being larger than $9$ . Therefore, our only solution is $a = 2, b = 2, c = 7$ . Our answer is $\underline{a... | 227 |
5,340 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | 5 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we have that $99a = 71b + 8c$
Note that by the divisibility rule for $9$ , we have $a+b+c \equiv a \pmod{9}$ . Since $b$ and $c$ are base- $9$ digits, we can say that $b+c = 0$ or $b+c=9$ . The former possibility can be easily eliminated, and thus $b+c=9$ . Next, we write the equation from So... | 227 |
5,341 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 1 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | We have the following diagram: [asy] /* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(... | 242 |
5,342 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 2 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A... | 242 |
5,343 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoin... | 242 |
5,344 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 4 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint... | 242 |
5,345 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 5 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let t... | 242 |
5,346 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 6 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A... | 242 |
5,347 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 7 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | Let $PQ = x$ . Note that since $AP$ bisects $\angle{A}$ and $DP$ bisects $\angle{D}$ , we have \[\angle{APD} = 180^{\circ}-\tfrac12 \angle{A}-\tfrac12 \angle{D}=90^{\circ}.\] Let $\angle{ADP}=\theta$ . We have that $\angle{ADC} = 2\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$ . Notice that $DE=\tfrac{650-500}{... | 242 |
5,348 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | 8 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint... | 242 |
5,349 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4 | 1 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi... | 834 |
5,350 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4 | 2 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | First we recognize that $w = \operatorname{cis}(30^{\circ})$ and $z = \operatorname{cis}(120^{\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$ , respectively. By De Moivre's theorem, $\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$ . When you multiply by $i$ , we can ... | 834 |
5,351 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | 1 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$ . Melanie is swimming against the current, so she must aim upstream from point $A$ , to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must... | 550 |
5,352 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | 2 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | Claim
Median $AM$ and altitude $AH$ are drawn in triangle $ABC$ $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$
Prove that \begin{align*}2ax = c^{2} - b^{2}\end{align*}
Proof \[BH + CH = a,\] \begin{align*} BH^{2} - CH^{2} = c^{2} - b^{2}\implies BH - CH &= \frac{c^{2} - b^{2}} {a},\end{align*} \[BH = \f... | 550 |
5,353 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | 3 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot("Finish",A,1.75*N,linewidth(5)); dot("Sherry",B,1.75*S,linewidth(5)); dot("Melanie",C,1.75*S,linewidth(5)); Label ... | 550 |
5,354 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | 4 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | We can break down movement into two components: the $x$ -component and the $y$ -component. Suppose that Melanie travels a distance of $a$ in the $x$ -direction and a distance of $c$ in the $y$ -direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$ -directi... | 550 |
5,355 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | 1 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$ . Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$ . Since $a < b$ , there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in m... | 228 |
5,356 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | 2 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} a... | 228 |
5,357 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | 3 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | Denote $S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}$
Denote by $A$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$
Denote by $B$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$
Hence, $C$ i... | 228 |
5,358 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | 4 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | divide cases into $7\leq a<20; 21\leq a\leq28$ .(Notice that $a$ can't be equal to $6,20$ , that's why I divide them into two parts.
There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$ .(NOTICE that $5,20,35,50$ IS NOT A VALID SEQUENCE!)
So when $7\leq a<20$ , there are $10+11+12+...+2... | 228 |
5,359 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | 1 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \... | 289 |
5,360 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | 2 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.
To bash efficiently, we can start out with $7\cdot8\cdot9$ as our denominator. This, however, leaves us with the numbers $1, 2, 3, 4, 5,$ and $6$ left. The smallest we can make out of this is $1\cdot5\cdot6 - 2\cd... | 289 |
5,361 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | 3 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | To minimize the numerator, we must have $abc - def = 1$ . Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value $(d, e, f) = (1, 3, 5)$ cannot result in a difference of $1$ , and the next smallest product, $(d, e, f) = (1, 3, 7)$ ca... | 289 |
5,362 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | 1 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$ [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W... | 378 |
5,363 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | 2 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | [asy] /* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = inter... | 378 |
5,364 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | 3 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\omega_A, \omega_B,\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \[CC' = 2R = C'K + KC = r + \frac{r}{\sin 30^\circ} = 3r.\] \[r = \frac{2R}{3} = 12.\]... | 378 |
5,365 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | 4 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | Let $O$ be the center of $\omega$ $X$ be the intersection of $\omega_B,\omega_C$ further from $A$ , and $O_A$ be the center of $\omega_A$ . Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$ -mixtilinear inradius $R_A$ is $\tfrac{r}{\cos^2\left(\frac{\angle A}{2}\right)} = \tfrac{9}{\cos^2\left(30^{\circ}... | 378 |
5,366 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9 | 1 | Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For ... | Consider this position chart: \[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\] Since there has to be an even number of spaces between each pair of the same color, spots $1$ $3$ $5$ $7$ $9$ , and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \c... | 247 |
5,367 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9 | 2 | Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For ... | We can simply use constructive counting. First, let us place the red blocks; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \cdot 6$ ways to place the red blocks. Similarly, there are $10 \cdot 5$ ways to place the bl... | 247 |
5,368 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | 1 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$ . We take a cross-section that contains $A$ and $B$ , which contains these two spheres but not the third, as shown below: [asy] size(400); pa... | 756 |
5,369 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | 2 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately.
According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$
The desired ... | 756 |
5,370 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | 3 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | Denote by $r$ the radius of three congruent circles formed by the cutting plane.
Denote by $O_A$ $O_B$ $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$ $B$ $C$ , respectively.
Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$ $O_A O_C = 11 + 19 = 30$
We have ... | 756 |
5,371 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | 1 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over t... | Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$ . The problem states that for $n=2, S_n$ is $4$ . Let's try it out for $n=3$
Let's perform casework on the number of elements in $A, B$
$\textbf{Case 1:} |A| = |B| = 1$
In this case, the only possible equivalencies wil... | 245 |
5,372 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | 2 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over t... | We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \cdot \frac{k}{n} \cdot \frac{k}{n}$ by linearity of expectation because for... | 245 |
5,373 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | 3 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over t... | For each element $i$ , denote $x_i = \left( x_{i, A}, x_{i, B} \right) \in \left\{ 0 , 1 \right\}^2$ , where $x_{i, A} = \Bbb I \left\{ i \in A \right\}$ (resp. $x_{i, B} = \Bbb I \left\{ i \in B \right\}$ ).
Denote $\Omega = \left\{ (x_1, \cdots , x_n): \sum_{i = 1}^n x_{i, A} = \sum_{i = 1}^n x_{i, B} \right\}$
Denot... | 245 |
5,374 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 | 2 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive intege... | Denote $BC = a$ $CA = b$ $AB = c$
Let the splitting line of $\triangle ABC$ through $M$ (resp. $N$ ) crosses $\triangle ABC$ at another point $X$ (resp. $Y$ ).
WLOG, we assume $c \leq b$
$\textbf{Case 1}$ $a \leq c \leq b$
We extend segment $AB$ to $D$ , such that $BD = a$ .
We extend segment $AC$ to $E$ , such that $C... | 459 |
5,375 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 | 3 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive intege... | We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$ . It can be shown that $3|a$ and $3|b$ , so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.
Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$ , where $\omega=e^{i\frac{2\pi}{3}}$ . Thus, \[x^2+xy+... | 459 |
5,376 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 | 1 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$ , as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$ , as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x +... | 154 |
5,377 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 | 2 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12.
Call the amount of people in the beginning $x$ .Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12.
Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the c... | 154 |
5,378 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 | 3 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Let $a$ be the number of adults before the bus arrived and $x$ be the total number of people at the concert. So, $\frac{a}{x}=\frac{5}{12}$ . Solving for $x$ in terms of $a$ $x = \frac{12}{5}a$ . After the bus arrives, let's say there are an additional $y$ adults out of the 50 more people who enter the concert. From th... | 154 |
5,379 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_2 | 1 | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probabili... | Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.
2022AIMEIIP2.png
This problem can be solved by using $2$ cases.
$\textbf{Case 1:}$ $C$ 's opponent for the semifinal is $A$
The... | 125 |
5,380 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | 1 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, \[\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}\] Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$ , so by the definition of logs, $a = \log_... | 112 |
5,381 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | 2 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We could assume a variable $v$ which equals to both $\log_{20x} (22x)$ and $\log_{2x} (202x)$
So that $(20x)^v=22x \textcircled{1}$ and $(2x)^v=202x \textcircled{2}$
Express $\textcircled{1}$ as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}$
Substitute $\textcircled{{2}}$ to $\textcircled... | 112 |
5,382 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | 3 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We have \begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}
We have \begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}
Because $\log_{20x} (22x)=\... | 112 |
5,383 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | 4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$ . Dividing, we get \begin{align*} \dfrac{(20x)^a}{(2x)^a} &= \dfrac{22x}{202x}. \\ \left(\dfrac{20x}{2x}\right)^a &= \dfrac{22x}{202x}. \\ 10^a &= \dfrac{11}{101}. \\ \end{align*} Thus, we see that $\log_{10} \left(\dfrac{11}{101}\right) = a = \log_{... | 112 |
5,384 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | 5 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$ , or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\frac{a-c}{b-d}$ . We use this identity and find that $k=\frac{\log 202 -\log 22}{\... | 112 |
5,385 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | 6 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By change of base formula, \[\frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1\] \[\log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1\] \[\frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} = \log_{2x} 1... | 112 |
5,386 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6 | 1 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. F... | To find the greatest value of $x_{76} - x_{16}$ $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$ . If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$ . The other number... | 841 |
5,387 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6 | 2 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. F... | Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is $1$ $|s_N|+|s_P|=1$
Since the sum of all the numbers is $0$ $s_N=-s_P\implies |s_N|=|s_P|$
Therefore, $|s_N|=|s_P|=\frac 12$ , so $s_N=-\frac 12$ and $s_P=\frac 12$ s... | 841 |
5,388 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 | 1 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | [asy] //Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)... | 192 |
5,389 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 | 2 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$ . Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$ , respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$... | 192 |
5,390 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 | 3 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | [asy] //Created by isabelchen and edited by afly size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0))... | 192 |
5,391 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | 1 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn fo... | We can use recursion to solve this problem:
1. Fix 7 points on $\ell_A$ , then put one point $B_1$ on $\ell_B$ . Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$ . For example, $f(1) = 6$ because there are 6 regions.
2. Now, put the second point ... | 244 |
5,392 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | 2 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn fo... | We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$ . Now let's focus on $f(m+1, n)-f(m, n)$ , which is the difference caused by adding one point to the existing $m$ points of line $\ell_A$ . This new poin... | 244 |
5,393 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | 3 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn fo... | Let some number of segments be constructed. We construct a new segment. We start from the straight line $l_B.$ WLOG from point $B_3.$ Segment will cross several existing segments (points $A,B,C,...$ ) and enter one of the points of the line $l_A (A_1).$
Each of these points adds exactly 1 new bounded region (yellow bou... | 244 |
5,394 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | 4 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn fo... | When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points.
WLOG, let the poi... | 244 |
5,395 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | 5 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn fo... | We know the by Euler's Formula for planar graphs that $F-E+V=2$ , where $F$ is the number of bounded faces, plus the outer region, $E$ is the number of edges, and $V$ is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that:
$V_{i}=7+5=12$
$E_{i}=6+4+7\cdot5=... | 244 |
5,396 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | 1 | 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 $ABCD.$ | [asy] defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M... | 180 |
5,397 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | 2 | 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 $ABCD.$ | Denote by $M$ the midpoint of segment $BC$ .
Let points $P$ and $Q$ be on segment $AD$ , such that $AP = AB$ and $DQ = DC$
Denote $\angle DAM = \alpha$ $\angle BAD = \beta$ $\angle BMA = \theta$ $\angle CMD = \phi$
Denote $BM = x$ . Because $M$ is the midpoint of $BC$ $CM = x$
Because $AM$ is the angle bisector of $\an... | 180 |
5,398 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | 3 | 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 $ABCD.$ | Claim
In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$
Proof
Let $A = 2\alpha.$ Then $\angle DBC = \angle DCB = \alpha.$
Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the c... | 180 |
5,399 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | 4 | 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 $ABCD.$ | Extend $AB$ and $CD$ so they intersect at a point $X$ . Then note that $M$ is the incenter of $\triangle{XAD}$ , implying that $M$ is on the angle bisector of $X$ . Now because $XM$ is both an angle bisector and a median of $\triangle{XBC}$ $\triangle{XBC}$ is isosceles. Then we can start angle chasing:
Let $\angle{BAM... | 180 |
5,400 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | 5 | 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 $ABCD.$ | As shown in paragraph one of solution 4, extending $AB$ and $CD$ to $X$ , we realize that $\triangle{XBC}$ is isosceles, thus $XM \perp BC$ . Let $XB = XC = x$ . And, midpoint $M$ is the incenter of $\triangle{XAD}$ . Construct perpendiculars $ME, MF, MG$ to sides $AD, AX, DX$ respectively (constructing the radii of th... | 180 |
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