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PA is the radius of a circle with center P, and QB is the radius of a circle with center Q, so that AB is a common internal tangent of the two circles, Let M be the midbout of AB and N be the point of line PQ so that line MN is perpendicular to PQ. Z is the point where AB and PQ intersects. If PA=5, QB=10, and PQ=17. compute PN.

So I tried to compute the problem above and I found the ratio between triangle ZMN:PAN:BQZ is 1:2:4. After finding that I discovered that the distance from both circles is 2, so after some work I found MN to be 2.5 and MZ to be 17/6 but when I used the pythogerom therom to find ZN thus getting a weird answer (8/6). Ultimately my answer for PN was incorrect and I don't know how to solve this problem. Please help me.

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Yes midbout is the same as midpoint. M is the the midpoint of AB. – Kelly Rocks Dec 1 '11 at 4:08
You mention triangles $ZMN$, $PAN$, and $BQZ$. Don't you mean $ZMN$, $PAZ$ and $BQZ$? – Joel Reyes Noche Dec 1 '11 at 4:36
The actual answer is 107/17. I did compute an answer of 7. – Kelly Rocks Dec 1 '11 at 4:53
up vote 1 down vote accepted

Let $\theta=\angle APZ$. Since $PA=5$ and $PZ=17/3$, we have $\cos\theta=15/17$, and therefore $\sin\theta=8/17$. It follows that $AZ=8/3$.

Since $M$ is the midpoint of $AB$, we have $ZM=(1/2)(AZ)$ (I got fooled here in my first calculation, and thought that $ZM=AZ$). So $ZM=4/3$, and therefore $ZN=(4/3)\sin\theta=32/51$. Now $PN=17/3+32/51=107/17$.

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Since $BQ=10$, $AP=5$ and triangles $BQZ$ and $APZ$ are similar, we get $QZ=2PZ$. Because $PQ=17$, we get $PZ=17/3$ and $QZ=34/3$. Using the Pythagorean theorem, we get $BZ=16/3$ and $AZ=8/3$, and thus $AB=8$. Since $MZ=AB/6$, we get $MZ=8/6$ (and not 17/6 as you computed). Could you do the rest of the computation?

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