1966 AHSME Problems/Problem 31


[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:

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


We will prove that $\triangle DOB$ and $\triangle COD$ is isosceles, meaning that $CD=OD=BD$ and hence $\fbox{D}$.

Let $\angle A=2\alpha$ and $\angle B=2\beta$. Since the incentre of a triangle is the intersection of its angle bisectors, $\angle OAB=\alpha$ and $\angle ABO=\beta$. Hence $\angle DOB=\alpha +\beta$. Since quadrilateral $ABCD$ is cyclic, $\angle CAD=\alpha=\angle CBD$. So $\angle OBD=\angle OBC+\angle CBD=\alpha + \beta = \angle DOB$. This means that $\triangle DOB$ is isosceles, and hence $BD=OD$.

Now let $\angle C=2\gamma$ which means $\angle ACO=COD=\gamma$. Since $ABCD$ is cyclic, $\angle DAB=\alpha=\angle DCB.$ Also, $\angle DAC\alpha$ so $\angle DOC=\alpha + \gamma$. Thus, $\angle OCD=\angle OCB+\angle BCD=\gamma + \alpha=\angle DOC$ which means $\triangle COD$ is isosceles, and hence $CD=OD=BD$.

Thus our answer is $\fbox{D.}$

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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