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Can you please help me find the solution for this problem. I have a right $\triangle ABC$. Now I draw its altitude $CD$. Obviously we have two triangles formed after drawing the altitude . The radii of these two triangles are respectively $r_1$ for $\triangle ACD$ and $r_2$ for $\triangle CDB$. The question is: find the radius of $\triangle ABC$.

I've tried to use the area of a triangle using the radius and its semiperimeter, but as I noticed, it was useless.

I hope you'll help me. Thank you!

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Draw a picture. All the triangles in the picture are similar. Let $r$ be the "radius" of the big triangle.

By similarity, one of the small triangles is the big triangle scaled by the factor $\frac{r_1}{r}$, and the other small triangle is the big triangle scaled by the factor $\frac{r_2}{r}$.

So the area of one small triangle is the area of the big one, scaled by the factor $\left(\frac{r_1}{r}\right)^2$. The area of the other is the area of the big triangle, scaled by the factor $\left(\frac{r_2}{r}\right)^2$.

But the sum of the areas of the small triangles is the area of the big one. So $\frac{r_1^2+r_2^2}{r^2}=1$, and now we know $r$.

Remark: This can be written up in more conventional style. But I like scaling arguments. Note we did not even have to know what "radius" meant. Inradius? Probably. Circumradius? Probably not, but the argument works for that also. We can also work with more exotic radii, such as the radius of the smallest circle that contains the triangle. Same result.

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  • $\begingroup$ Thank you very much. You helped me a lot! and finally I solved it. $\endgroup$ – Ivan Gandacov Apr 3 '14 at 14:09
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    $\begingroup$ You are welcome. The semiperimeter stuff will work. By similarity, if the semiperimeter of one of the little circles is $cr_1$, the semiperimeter of the other is $cr_2$, and the semiperimeter of the big one is $cr$. Then by area we have $(cr_1)(r_1)+(cr_2)(r_2)=(cr)(r)$. $\endgroup$ – André Nicolas Apr 3 '14 at 14:16
  • $\begingroup$ Incidentally, we can also see from the final equation that the three radii can be placed together to form another right triangle. $\endgroup$ – Théophile Apr 3 '14 at 14:20

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