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/
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