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2024 AMC 12A Problems/Problem 23

Revision as of 21:36, 20 March 2025 by Maa is stupid (talk | contribs)
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, $AC^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}$.

See also

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