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

Revision as of 14:15, 21 August 2011

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 $AD \cdot DC = DB \cdot BE$ . Since $DB = 1$ and $BE = 5$ , the desired product is $5$ , which is $\boxed{A}$ .

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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Difference between revisions of "1997 AHSME Problems/Problem 26"

Revision as of 14:15, 21 August 2011

Problem

Solution

See also

@@ Line 20: / Line 20: @@
 ==Solution==
+The product of two lengths with a common point brings to mind the [[Power of a Point Theorem]].
+Since <math>PA = PB</math>, we can make a circle with radius <math>PA</math> that is centered on <math>P</math>, and both <math>A</math> and <math>B</math> will be on that circle.  Since <math>\angle APB = \widehat {AB} = 2 \angle ACB</math>, we can see that point <math>C</math> will also lie on the circle, since the measure of arc <math>\widehat {AB}</math> is twice the masure of inscribed angle <math>\angle ACB</math>, 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 <math>PDB</math> is a line, we have <math>PD + DB = PB</math>, which gives <math>3 = DB + 2</math>, or <math>DB = 1</math>.
+We now extend radius <math>PB = 3</math> to diameter <math>EB = 6</math>.  Since <math>EDB</math> is a line, we have <math>ED + DB = EB</math>, which gives <math>ED + 1 = 6</math>, or <math>ED = 5</math>.
+Finally, we apply the [[Power of a Point]] theorem to point <math>D</math>.  This states that <math>AD \cdot DC = DB \cdot BE</math>.  Since <math>DB = 1</math> and <math>BE = 5</math>, the desired product is <math>5</math>, which is <math>\boxed{A}</math>.
 == See also ==
 {{AHSME box|year=1997|num-b=25|num-a=27}}