Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need help with the following problem.

Given three circles $k, k_1, k_2$. $k_1$ and $k_2$ touch internally $k$ at points $M$ and $N$ respectively. $a$ is the common interior tangent to $k_1$and $k_2$ at points $R$ and $S$. $MR \cap k = A$ and $NS \cap k = B$. Prove that $a \perp AB$.

share|improve this question
    
What part of it do you need help with? –  Théophile May 8 '12 at 19:05
    
I don't have any idea how to prove it. Maybe you can give me a hint? –  Adam May 8 '12 at 19:10
    
I need clarification - aren't $M$ and $N$ points on $k$? If so, doesn't this imply $M=A$ and $N=B$? –  process91 May 8 '12 at 19:14
1  
HINT: First concentrate on one internal circle, $k_1$. Say $P$ and $P_1$ are the centers of $k$ and $k_1$, respectively. For point $U_1 \neq M$ any (other) point on $k_1$, let $V_1$ be the "second intersection point" of $MU_1$ with $k$. What can you say about $PV_1$ and $P_1U_1$? ... ... When $U_1$ coincides with your point $R$ (so $V_1$ coincides with $A$), what can you say about $PA$? ... ... How does a similar understanding of $k_2$ and $PB$ help? (Pay attention to how $P_1R$ and $P_2S$ relate.) –  Blue May 8 '12 at 21:40
1  
@Adam: Great. You've reduced the problem to the core issue, which isn't too tricky. Focus on $\triangle MP_1U_1$ and $\triangle MPV_1$. –  Blue May 8 '12 at 22:40
show 2 more comments

1 Answer

up vote 4 down vote accepted

enter image description here

Here is the diagram of the construction described, with the addition of the centers of $k$, $k_1$, and $k_2$; $T$, $P$, and $Q$.

Note that $\triangle PMR\simeq\triangle TMA$ and $\overline{PR}||\overline{TA}$.

Note that $\triangle QNS\simeq\triangle TNB$ and $\overline{QS}||\overline{TB}$.

Since $\overline{PR}\perp\overline{RS}$, we have $\overline{TA}\perp\overline{RS}$.

Since $\overline{QS}\perp\overline{RS}$, we have $\overline{TB}\perp\overline{RS}$.

Therefore, $\overline{TA}||\overline{TB}||\overline{AB}$.

Thus, $\overline{AB}\perp\overline{RS}=a$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.