Difference between revisions of "1997 AHSME Problems/Problem 26"
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==Solution== | ==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 == | == 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 14:15, 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
The product of two lengths with a common point brings to mind the Power of a Point Theorem.
Since , we can make a circle with radius that is centered on , and both and will be on that circle. Since , we can see that point will also lie on the circle, since the measure of arc is twice the masure of inscribed angle , which is true for all inscribed angles.
Since is a line, we have , which gives , or .
We now extend radius to diameter . Since is a line, we have , which gives , or .
Finally, we apply the Power of a Point theorem to point . This states that . Since and , the desired product is , which is .
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 |