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### Circle Inscribed in a Right Triangle

```Date: 09/09/97 at 21:44:25
From: Mary Ann
Subject: Circle inscribed in a right triangle

A circle is inscribed within a right triangle. What is the diameter
of the circle if the legs of the triangle are known to be A and B?

We tried dividing the triangle into several triangles. We have also
tried graphing.  I am stumped.
```

```Date: 09/13/97 at 20:01:51
From: Doctor Barney
Subject: Re: Circle inscribed in a right triangle

To explain how I solved this problem, I will need to define some
points on your triangle.  Let's call a the corner opposite side A,
and b the corner opposite side B. X will be the center of the circle.
Let r be the radius of the circle. Now let Theta be the angle at
point a.  tan(Theta) = A/B

To solve this problem, I used two subtle pieces of information.
First, X lies on the angle bisectors for all three angles. This is
easy to see if you just look at any two sides of the triangle at a
time, extend the legs out farther if you need to, and consider that
the circle is tangent to both sides of that angle.  Second, X is a
distance r away from both sides A and B.

Now if you will permit me to define one more point, q, on side B at
the point where the circle is tangent to B, I will ask you to consider
the right triangle axq. The internal angle of this triangle at a is
Theta/2, the length of the opposite leg is r, and the length of the
adjacent leg is B-r. Therefore, tan(Theta/2) = r/(B-r).

You can solve this algebraically for r to find the radius, and double
it to get the diameter.

Oddly enough, we could have solved this using an analogous triangle
originating from point b instead of point a, so A and B should be
interchangeable in the final answer.

-Doctor Barney,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```Date: 03/18/2003 at 22:58:54
From: Rhonda
Subject: Re: Circle Inscribed in a Right Triangle

I solved for the diameter, as Dr. Barney suggested, and I came up
with

2B * [tan( (tan^-1(A/B)) / 2 )]
diameter = -------------------------------
1 + [tan( (tan^-1(A/B)) / 2)]

It seems to be right, but I sometimes miss the obvious.  Can you tell
me if this is correct?
```

```Date: 03/19/2003 at 09:25:52
From: Doctor Rick
Subject: Re: Circle Inscribed in a Right Triangle (correction)

Hi, Rhonda.

Yes, that's correct.

However, I would not solve the problem this way at all; there is a
much simpler formula that you can reach by simple geometry and algebra.

Draw the figure as follows: right triangle ABC with the right angle
at C, and a circle of radius r, with center O, inscribed in ABC. Call
the length AC = b and BC = a. Then draw perpendiculars from O to the
three sides, meeting AB, BC, and AC at D, E, and F respectively; and
join O to vertices A and B.

Now, triangles AOD and AOF are congruent, as are triangles BOD and
BOE. Also, OF is parallel to BC, and OE is parallel to AC. Thus CF =
CE = r, so AF = b-r and BE = a-r.

By the congruent triangles, AD = b-r and BD = a-r, so AB = a+b-2r.
But by the Pythagorean theorem, AB^2 = a^2 + b^2. Combining these results,
you can write a quadratic equation in r, with two solutions of which
the correct one is

r = (a+b-c)/2

where c = AB, the length of the hypotenuse. That's a bit simpler,
isn't it? The formula is listed in the Dr. Math FAQ on Formulas:

http://mathforum.org/dr.math/faq/formulas/faq.triangle.html

In principle, you should be able to simplify your solution to get
it down to (a+b-c)/2. Try using one of the formulas for the tangent
of a half angle:

tan(x/2) = +-sqrt((1-cos(x))/(1+cos(x)))

= (1-cos(x))/sin(x)

= sin(x)/(1+cos(x))

You'll also need formulas for the sin and cos in terms of the tan:

sin(x) = +-tan(x)/sqrt(1+tan^2(x))

cos(x) = +-1/sqrt(1+tan^2(x))

-Doctor Rick,  The Math Forum
http://mathforum.org/dr.math/
```

```Date: 03/19/2003 at 14:33:55
From: Rhonda
Subject: Re: Circle Inscribed in a Right Triangle

Dr. Rick,

Thanks for clarifying this for me!  I definitely like the r=(a+b-c)/2
solution much better (especially since I'm using the formula in a
computer graphics application, so efficiency is important).

I happened to notice that in your response, you can get the result
directly from the "AD = b-r and BD = a-r, so AB = a+b-2r" equations
(substituting c for AB as you suggest, then solving for r).  In my
application, I don't know c directly, so I'm using the Pythagorean
theorem to compute it.

Thanks again for all your help! :~)

Sincerely,

Rhonda
```

```Date: 03/19/2003 at 14:40:31
From: Doctor Rick
Subject: Re: Circle Inscribed in a Right Triangle

Hi, Rhonda.

Thanks for noticing the simpler derivation! I didn't look at the
formula in the Dr. Math FAQ until I was done, so I wasn't expecting
it to simplify so much. I just plowed ahead with the quadratic
formula, which gave (a+b +- sqrt(a^2+b^2))/2, then I saw what the
square root term equaled. But you're right, c = a+b-2r, so
r = (a+b-c)/2, and that's all I needed to do.

-Doctor Rick,  The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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