2016 AIME I Problems/Problem 15

Problem

Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ , $Y$ , $D$ are collinear, $XC = 67$ , $XY = 47$ , and $XD = 37$ . Find $AB^2$ .

Solution

By radical axis theorem $AD, XY, BC$ concur at point $E$ .

Let $AB$ and $EY$ intersect at $S$ . Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel theorem $AXBE$ are cyclic as well. Thus $\angle AEX = \angle ABX = \angle XCB = \angle XYB$ and $\angle XEB = \angle XAB = \angle XDA = \angle XYA.$ Thus $AY // EB$ and $YB // EA$ so $AEBY$ is a parallelogram. Hence $AS = SB$ and $SE = SY$ . But notice that $DXE$ and $EXC$ are similar by $AA$ Similarity, so $XE^2 = XD \cdot XC = 37 \cdot 67$ . But $XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.$ Hence $AB^2 = 47^2 - 37 \cdot 67 = 270.$

2016 AIME I (Problems • Answer Key • Resources)
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2016 AIME I Problems/Problem 15

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