+0  
 
0
1161
2
avatar+14 

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

 Mar 9, 2019

Best Answer 

 #1
avatar+26388 
+1

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
 #1
avatar+26388 
+1
Best Answer

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

heureka Mar 12, 2019
 #2
avatar+14 
+1

Thanks so much!

 Mar 14, 2019

0 Online Users