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# Tangents to a Circle

06/11/2010 1:16 PM

When tangents are drawn to a circle of radius R at three points, a triangle is formed. The sides of the triangle are 3R, 4R, 5R. Find sum of squares of sides

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#1

### Re: tangents to a circle

06/11/2010 1:27 PM

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

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#2

### Re: Tangents to a Circle

06/11/2010 3:24 PM

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

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#3

### Re: Tangents to a Circle

06/11/2010 3:38 PM

Yes Sir, Thank You

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#4

### Re: Tangents to a Circle

06/11/2010 8:15 PM

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?

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#5

### Re: Tangents to a Circle

06/11/2010 8:42 PM

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

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#6

### Re: Tangents to a Circle

06/11/2010 9:33 PM

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.

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#9

### Re: Tangents to a Circle

06/12/2010 11:43 AM

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]

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#10

### Re: Tangents to a Circle

06/12/2010 4:10 PM

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).

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#11

### Re: Tangents to a Circle

06/12/2010 8:13 PM

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.

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#12

### Re: Tangents to a Circle

06/12/2010 8:47 PM

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.

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#7

### Re: Tangents to a Circle

06/12/2010 5:35 AM

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

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#8

### Re: Tangents to a Circle

06/12/2010 8:13 AM

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

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#13

### Re: Tangents to a Circle

09/20/2019 11:14 AM

Yes, it geometrically must = 50R2

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