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/