Thread: Incircle of a Triangle

Let the incircle of triangle ABC have radius 2 and let it be tangent to BC at D. Suppose BD = 3 and DC = 4. What is the length of the longest side of ABC?

Can't comment on YOUR work on this: can't see it!

Anyway, start with a diagram; like E = tangent point on AB, F = tangent point on AC.
You now have BD=BE, CD=CF and AE=AF.
Use pythagorean theorem and Law of Sines.

Originally Posted by Wilmer

Can't comment on YOUR work on this: can't see it!

Anyway, start with a diagram; like E = tangent point on AB, F = tangent point on AC.
You now have BD=BE, CD=CF and AE=AF.
Use pythagorean theorem and Law of Sines.

I used the Pythagorean Theorem to get the lengths of BE and CF, but I'm stuck on getting EA and AF. I know the Law of Sines, but I don't know where to apply it.

Originally Posted by UNLVRich

I used the Pythagorean Theorem to get the lengths of BE and CF, but I'm stuck on getting EA and AF. I know the Law of Sines, but I don't know where to apply it.

You seem unaware of the basics; like, no need to use pythagorean theorem to get BE and CF;
BE=3 and CF=4. And similarly, AE = AF.
Were you aware that the 2 tangent lines from a point outside circle to the tangent points are equal?

You would use Law of Sines to calculate angles ABC and ACB, leaving angle BAC = 180 - ABC - ACB.
Let M = inner circle center. Now work with the 3 inner triangles AMB, AMC and BMC.

Were you aware of this formula:
radius-of-inner-circle = 2[(area-of-triangle) / (perimeter of triangle)] ?
If we let x = AE = AF, then perimeter becomes 2x + 14 and area (using Heron's formula)
becomes sqrt[12(x^2 + 7x)].
This would be another way of solving, NOT needing the Law of Sines.

Hope this helps...

Originally Posted by Wilmer

Were you aware that the 2 tangent lines from a point outside circle to the tangent points are equal?

I had a feeling this was true, but I didn't use it because I hadn't proved it yet. I just proved it so I'll use it. Thank you!!

I knew those formulas you mentioned, but the key was the fact that the two tangent lines were congruent. Thanks for your help!!

Originally Posted by UNLVRich

I had a feeling this was true, but I didn't use it because I hadn't proved it yet. I just proved it so I'll use it. Thank you!!

Good stuff Rich! Btw, don't think you needed to prove it: it's a theorem, right?

Originally Posted by Wilmer

Good stuff Rich! Btw, don't think you needed to prove it: it's a theorem, right?

Yeah...it's a theorem, but I don't like to use things that I can't prove by myself.