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

(Created page with "== See also == {{AHSME box|year=1997|num-b=25|num-a=27}}")
 
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==Problem==
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Triangle <math>ABC</math> and point <math>P</math> in the same plane are given. Point <math>P</math> is equidistant from <math>A</math> and <math>B</math>, angle <math>APB</math> is twice angle <math>ACB</math>, and <math>\overline{AC}</math> intersects <math>\overline{BP}</math> at point <math>D</math>. If <math>PB = 3</math> and <math>PD= 2</math>, then <math>AD\cdot CD =</math>
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<asy>
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defaultpen(linewidth(.8pt));
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dotfactor=4;
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pair A = origin;
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pair B = (2,0);
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pair C = (3,1);
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pair P = (1,2.25);
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pair D = intersectionpoint(P--B,C--A);
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dot(A);dot(B);dot(C);dot(P);dot(D);
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label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N);
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draw(A--B--P--cycle);
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draw(A--C--B--cycle);</asy>
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<math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9 </math>
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==Solution==
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== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=25|num-a=27}}
 
{{AHSME box|year=1997|num-b=25|num-a=27}}

Revision as of 13:12, 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

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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