Tangents to a Circle

You need to clarify this puzzle a little better with a diagram. Unless you think that 9R² + 14R² + 25R² = 48R² is tricky puzzle.

Typographic correction: 9R²+16R²+25R² = 50R².

As 3,4,5 is a right triangle,

wouldn't that be

9*R²+16*R² = 25*R²

sum of squares of the two sides is the square of the hypotenuse?

Maybe, that's why I asked for a clarification.

That is true, of course, but the wording of the OP seemed to ask for the sum of squares for all 3 sides.

For a bit more of a challenge, prove that the configuration is possible; i.e., that the circle is indeed tangent to the hypotenuse as well as the legs.

Tornado,

How about this:

The radius, R of an inscribed circle would be Area/k where k is (a + b + c)/2, a, b and c being the length of each side.

In this case, k = (3 + 4 +5)R/2 = 6R

Area = 3R*4R/2 = 6R²

Radius = Area/k = R [QED]

That's one way, but it relies on:

"The radius, R of an inscribed circle would be Area/k where k is (a + b + c)/2, a, b and c being the length of each side."

Which may be a bit obscure, and would itself require proof.

There are no doubt several ways to show this; mine was to use the triangle (0,0), (3,0), and (0,4) and show that the circle was tangent to the hypotenuse at (1.8, 1.6).

Tornado, you are a hard taskmaster.

Inscribe a circle in any triangle with sides a, b and c. Assume the radius is r. Draw a line from the center of the circle to each vertex, dividing the triangle into three smaller triangles. The area of each smaller triangle is a*r/2, b*r/2 and c*r/2 and the area of the original triangle is r(a + b + c)/2.

In our case, the area is r(3R + 4R +5R)/2 = 6rR.

Since we have a right angle, the area is 3R*4R/2 = 6R², so r = R.

Quite right. Shortly after suggesting that (a+b+c)/2 might be "obscure," I realized it was simple, just as in your description. (Too many years since I studied geometry; I forget or don't see the most elegant solutions.) Nicely explained.

This is just so pathetic. The OP doesn't know how to do even simple algebra like this!!!!

Uhhh... Wouldn't that just be 50R²? 9R² + 16R² + 25R² = 50R².

Yes, it geometrically must = 50R²