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Difference between revisions of "1966 AHSME Problems/Problem 31"

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== Problem ==
 
== Problem ==
 +
<asy>
 +
draw(circle((0,0),10),black+linewidth(1));
 +
draw(circle((-1.25,2.5),4.5),black+linewidth(1));
 +
dot((0,0));
 +
dot((-1.25,2.5));
 +
draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5));
 +
draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5));
 +
draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5));
 +
draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5));
 +
MP("O'", (0,0), W);
 +
MP("O", (-2,2), W);
 +
MP("A", (-10,-2), W);
 +
MP("B", (10,-2), E);
 +
MP("C", (-2,sqrt(96)), N);
 +
MP("D", (sqrt(96)-2.5,7), NE);
 +
</asy>
 
Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have:
 
Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have:
  

Revision as of 03:50, 23 September 2014

Problem

[asy] draw(circle((0,0),10),black+linewidth(1)); draw(circle((-1.25,2.5),4.5),black+linewidth(1)); dot((0,0)); dot((-1.25,2.5)); draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5)); draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5)); MP("O'", (0,0), W); MP("O", (-2,2), W); MP("A", (-10,-2), W); MP("B", (10,-2), E); MP("C", (-2,sqrt(96)), N); MP("D", (sqrt(96)-2.5,7), NE); [/asy] Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then we must have:

$\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD$

Solution

$\fbox{D}$

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
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

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