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Difference between revisions of "1997 AHSME Problems/Problem 26"

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Finally, we apply the power of a point theorem to point <math>D</math>.  This states that <math>AD \cdot DC = DB \cdot DE</math>.  Since <math>DB = 1</math> and <math>DE = 5</math>, the desired product is <math>5</math>, which is <math>\boxed{A}</math>.
 
Finally, we apply the power of a point theorem to point <math>D</math>.  This states that <math>AD \cdot DC = DB \cdot DE</math>.  Since <math>DB = 1</math> and <math>DE = 5</math>, the desired product is <math>5</math>, which is <math>\boxed{A}</math>.
 +
 +
==Solution 2==
 +
Construct the angle bisector of <math>\angle APD,</math> and let it intersect <math>AD</math> at <math>E.</math> From the angle bisector theorem, we have <math>AE=3a</math> and <math>DE=2a</math> for some <math>a.</math> Then, note that <math>\angle EPD = \angle BCD = x,</math> so <math>EPCB</math> is cyclic. Then, <math>\frac{PD}{ED} = \frac{CD}{BD}</math> or <math>\frac{2}{2x} = \frac{CD}{1}.</math> Thus, <math>AD \cdot DC = 5x \cdot DC = 5,</math> or <math>\boxed{A}.</math>
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=25|num-a=27}}
 
{{AHSME box|year=1997|num-b=25|num-a=27}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:01, 9 March 2019

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 by
Problem 25 		Followed by
Problem 27
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All AHSME Problems and Solutions

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