Question

Triangle $ABC$ is inscribed in circle $\omega.$ Points $P$ and $Q$ are on side $\overline{AB}$ with $AP \lt AQ.$ Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP = 4,$ $PQ = 3,$ $QB = 6,$ $BT = 5,$ and $AS = 7,$ then $ST = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

By Power of a Point at $Q$ in $\omega,$ $QC \cdot QT = QA \cdot QB = 7 \cdot 6 = 42.$ Extend $\overline{AB}$ beyond $B$ to the point $R$ with $BR = 8,$ so that $QR = QB + BR = 14$ and $QP \cdot QR = 3 \cdot 14 = 42 = QC \cdot QT.$ By the converse of Power of a Point, $C,$ $P,$ $T,$ and $R$ are concyclic.

In circle $CPTR,$ $\angle BRT = \angle PRT = \angle PCT,$ and in $\omega,$ $\angle PCT = \angle SCT = \angle SAT$ (both subtend arc $ST$ ). Also $ASTB$ is cyclic, so the exterior angle of the quadrilateral at $B$ equals the opposite interior angle: $\angle RBT = \angle AST.$ Hence $\triangle AST \sim \triangle RBT.$

Therefore $\frac{ST}{BT} = \frac{AS}{RB},$ so $ST = 5 \cdot \frac{7}{8} = \frac{35}{8},$ and $m + n = 35 + 8 = 43.$

2016 AIME II Problem 10

Solution:

Problem 10 in Other Years