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Two circles intersect at A and B. A common external tangent is tangent to the circles at T and U as shown. Let M be the intersection of line AB and TU. If AB = 9 and BM = 3 find TU.

Picture: https://latex.artofproblemsolving.com/c/5/0/c50a548001dc09c6eb63031398f6a77787276a50.png

 Mar 12, 2019
 #1
avatar+234 
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Power of a point on M, you get TU = 12. AM * BM = MU2 = 36. So we get MU = 6. Do this on the other side, and you find that TM = 6 as well. So TU = 12

 Mar 12, 2019
 #2
avatar+22896 
+3

Two circles intersect at A and B. A common external tangent is tangent to the circles at T and U as shown.

Let M be the intersection of line AB and TU.

If AB = 9 and BM = 3 find TU.

https://latex.artofproblemsolving.com/c/5/0/c50a548001dc09c6eb63031398f6a77787276a50.png

 

Since we've only given AM and BM, we can simplify:


\(\text{Let r $ = \dfrac{AB}{2}+BM$}=4.5+3=7.5\)


Pythagoras:

\(\begin{array}{|rcll|} \hline MU^2 + \left(\dfrac{AB}{2}\right)^2&=&r^2 \\ MU^2 &=&r^2 - \left(\dfrac{AB}{2}\right)^2 \quad | \quad r = \dfrac{AB}{2} + BM \\ MU^2 &=& \left(\dfrac{AB}{2} + BM \right)^2 - \left(\dfrac{AB}{2}\right)^2 \\ MU^2 &=& \left(\dfrac{AB}{2}\right)^2 + AB\cdot BM + BM^2 - \left(\dfrac{AB}{2}\right)^2 \\ MU^2 &=& AB\cdot BM + BM^2 \\ MU^2 &=& BM\cdot(AB + BM) \\ MU^2 &=& BM\cdot AM \\ MU^2 &=& 3\cdot 12 \\ MU^2 &=& 36 \\ MU &=& 6 \\\\ TM &=& MU \\ &=& 6 \\\\ TU &=& TM+MU \\ &=& 6+6 \\ \mathbf{TU} &\mathbf{=}& \mathbf{12} \\ \hline \end{array}\)

 

laugh

 Mar 12, 2019

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