Difference between revisions of "1997 AHSME Problems/Problem 26"
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+ | ==Problem== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | |||
+ | |||
== 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 and point in the same plane are given. Point is equidistant from and , angle is twice angle , and intersects at point . If and , then
Solution
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 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 |