# 1997 AHSME Problems/Problem 26

## Problem

Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$, angle $APB$ is twice angle $ACB$, and $\overline{AC}$ intersects $\overline{BP}$ at point $D$. If $PB = 3$ and $PD= 2$, then $AD\cdot CD =$ $[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("A",A,SW);label("B",B,SE);label("C",C,N);label("D",D,NE + N);label("P",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]$ $\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

## Solution

The product of two lengths with a common point brings to mind the Power of a Point Theorem.

Since $PA = PB$, we can make a circle with radius $PA$ that is centered on $P$, and both $A$ and $B$ will be on that circle. Since $\angle APB = \widehat {AB} = 2 \angle ACB$, we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the masure of inscribed angle $\angle ACB$, which is true for all inscribed angles. $[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3.06,0.9); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); pair E = (0,4.5); dot(A);dot(B);dot(C);dot(P);dot(D);dot(E); label("A",A,SW);label("B",B,SE);label("C",C,NW);label("D",D,NE + N);label("P",P,N);label("E", E, NW); draw(A--B--P--cycle); draw(A--C--B--cycle); draw(circle(P, 2.46)); draw(P--E);[/asy]$

Since $PDB$ is a line, we have $PD + DB = PB$, which gives $3 = DB + 2$, or $DB = 1$.

We now extend radius $PB = 3$ to diameter $EB = 6$. Since $EDB$ is a line, we have $ED + DB = EB$, which gives $ED + 1 = 6$, or $ED = 5$.

Finally, we apply the power of a point theorem to point $D$. This states that $AD \cdot DC = DB \cdot DE$. Since $DB = 1$ and $DE = 5$, the desired product is $5$, which is $\boxed{A}$.

## Solution 2

Construct the angle bisector of $\angle APD,$ and let it intersect $AD$ at $E.$ From the angle bisector theorem, we have $AE=3a$ and $DE=2a$ for some $a.$ Then, note that $\angle EPD = \angle BCD = x,$ so $EPCB$ is cyclic. Then, $\frac{PD}{ED} = \frac{CD}{BD}$ or $\frac{2}{2x} = \frac{CD}{1}.$ Thus, $AD \cdot DC = 5x \cdot DC = 5,$ or $\boxed{A}.$

## See also

 1997 AHSME (Problems • Answer Key • Resources) Preceded byProblem 25 Followed byProblem 27 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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