Difference between revisions of "2024 AMC 12A Problems/Problem 23"

Revision as of 21:36, 20 March 2025

The following problem is from both the 2024 AMC 12A #23 and 2024 AMC 10A #25, so both problems redirect to this page.

Problem

In parallelogram $ABCD$ , let $\omega$ be the circle with diameter $\overline{AD}$ and suppose $P$ and $Q$ are points on $\omega$ such that both lines $BP$ and $BQ$ are tangent to $\omega$ . If $BC = 8$ , $BP = 3$ , and line $PQ$ bisects $\overline{CD}$ , what is $AC^{2}$ ?

$\textbf{(A)}~180\qquad\textbf{(B)}~181\qquad\textbf{(C)}~182\qquad\textbf{(D)}~183\qquad\textbf{(E)}~184$

Solution 1

Let $M$ , $O$ , and $N$ be the midpoints of $\overline{CD}$ , $\overline{AD}$ , and $\overline{AB}$ , respectively. Let $\overline{PQ}$ intersect $\overline{BO}$ at $R$ and note that the radius of $\omega$ is $\tfrac{AD}{2} = \tfrac{8}{2} = 4$ . The Pythagorean theorem applied to $\triangle BPO$ gives $BO = 5$ , and the similarity $\triangle BPR \sim \triangle BOP$ implies $BR = \tfrac{BP^{2}}{BO} = \tfrac{9}{5}$ .

Let $\overline{MN}$ intersect $\overline{BO}$ at point $E$ . Since $ABCD$ is a parallelogram, $E$ is the midpoint of $\overline{BO}$ and $EN = \tfrac{AO}{2} = \tfrac{4}{2} = 2$ . Because $MN = BC = 8$ , we have $ME = 6$ and $RE = BE - BR = \tfrac{5}{2} - \tfrac{9}{5} = \tfrac{7}{10}$ . It follows that $\cos(\angle MEO) = \cos(\pi - \angle MER) = -\cos(\angle MER) = -\tfrac{7}{60}.$

$[asy] size(7cm); defaultpen(fontsize(10pt)+linewidth(0.4)); pair A = (0, 0), B = (55/12, sqrt(3551)/12), C = (151/12, sqrt(3551)/12), D = (8, 0), M = (C + D)/2, N = (A + B)/2, O = (A + D)/2, E = (B + O)/2, P = intersectionpoints(circle(O,abs(A-O)),circle(E,abs(B-O)/2))[1], Q = intersectionpoints(circle(O,abs(A-O)),circle(E,abs(B-O)/2))[0], R = (P + Q)/2; draw(A--B--C--D--cycle); draw(arc(O,abs(A-O),0,180)); draw(B--P, gray); draw(B--Q, gray); draw(B--O, dashed); draw(P--M); draw(rightanglemark(M, R, E)); fill(M--E--O--cycle, pink); dot("$A$", A, dir(263)); dot("$B$", B, dir(83)); dot("$C$", C, dir(83)); dot("$D$", D, dir(263)); dot("$E$", E, dir(353)); dot("$M$", M, dir(353)); dot("$O$", O, dir(263)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(45)); dot("$R$", R, dir(45)); dot("$N$", N, dir(187)); [/asy]$

Applying the Law of Cosines in $\triangle MEO$ , $MO^{2} = 6^{2} + \left(\frac{5}{2}\right)^{2} + 2 \cdot \frac{5}{2} \cdot 6 \cdot \frac{7}{60} = 36 + \frac{25}{4} + \frac{7}{2} = \frac{183}{4}.$ A double homothety at $D$ gives $AC^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}$ .

Solution 2

Let $M$ , $O$ , and $N$ be the midpoints of $\overline{CD}$ , $\overline{AD}$ , and $\overline{AB}$ respectively. Let $E$ be the midpoint of $\overline{BO}$ and let $\gamma$ be the circumcircle of $\triangle{BPQ}$ .

First, $OP = \tfrac{AD}{2} = \tfrac{BC}{2} = 4$ , so the radius of $\omega$ is $4$ . Since $\angle{OPB} = \angle{OQB} = 90^{\circ}$ , $\overline{BO}$ is a diameter of $\gamma$ , and $\gamma$ has center $E$ . The Pythagorean theorem applied to either $\triangle{BPO}$ or $\triangle{BQO}$ gives that $BO = 5$ , so the radius of $\gamma$ is $\tfrac{5}{2}$ . Since $E$ is the midpoint of $\overline{BO}$ , and $ABCD$ is a parallelogram, we must have that $E$ lies on $\overline{MN}$ , and lies $\tfrac{3}{4}$ of the way from $M$ to $N$ , giving $ME = \tfrac{3MN}{4} = 6$ .

Since $M$ lies on line $PQ$ , the radical axis of $\gamma$ and $\omega$ , it has equal power with respect to both circles. Thus $\text{Pow}_{\gamma}{(M)} = ME^{2} - EB^{2} = 6^{2} - \left(\frac{5}{2}\right)^{2} = \frac{119}{4} = \text{Pow}_{\omega}{(M)} = MO^{2} - OA^{2} = MO^{2} - 16,$ and $MO^{2} = \tfrac{183}{4}$ . A double homothety at $D$ finishes, and $AC^{2} = (2MO)^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}$ .

Solution 3

Let $M$ and $O$ be the midpoints of $\overline{CD}$ and $\overline{AD}$ , respectively. Consider taking segment $\overline{BO}$ , translating it $8$ units in the direction of vector $\vec{AD}$ . Then perform a homothety with ratio $\tfrac{1}{2}$ at $D$ , mapping $B$ to $M$ and $O$ to $O^{\prime}$ . Since $CO^{\prime} = 2$ and $CO = 4$ , we have that $OO^{\prime} = 6$ . Also, $MO^{\prime} = \tfrac{OB}{2} = \tfrac{5}{2}$ .

Let $\overline{BO}$ and $\overline{PQ}$ intersect at $R$ , and note that $OR = \tfrac{16}{5}$ . Since $\overline{OR} \perp \overline{PQ}$ , we also have $\overline{O^{\prime}M} \perp \overline{PQ}$ . Applying the Pythagorean theorem on right trapezoid $ORMO^{\prime}$ , we have $RM^{2} = 6^{2} - \left(\tfrac{16}{5} - \tfrac{5}{2}\right)^{2} = \tfrac{3551}{100}$ . The Pythagorean theorem in $\triangle ORM$ can again be used to calculate $MO^{2} = \left(\tfrac{16}{5}\right)^{2} + \tfrac{3551}{100} = \tfrac{183}{4}$ . As in other solutions, $AC^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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