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avatar+1230 

In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $3$, and $AB = 1$, then find $\frac{AC}{BC}.$

 Jun 14, 2024
 #1
avatar+129733 
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This is impossible

Leg AC  would be >  6  while  AB (the hypotenuse)  would only  = 1

 

cool cool cool

 Jun 15, 2024

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