# Triangle with incircle [closed]

This question is from the German mathematics competition Känguru der Mathematik. In this competition students have to solve 30 mathematical tasks like this in 90 minutes without calculator. Actually they are given 5 possible answers, but for this community a bit of additional challenge does not hurt.

You are given a triangle $ABC$ with $\overline{AB} = 3$, $\overline{BC} = 5$ and $\overline{CA} = 6$ (image not to scale). The line segment $DE$ is tangential to the incircle of the triangle $ABC$. What is the perimeter of the triangle $CDE$?

• Usually when talking about triangles, we call the sum of the edges the perimeter, and reserve the word circumference for the perimeter of a circle. – micsthepick Oct 17 '17 at 21:32
• @micsthepick Thank you, that was new to me. I changed it in the question. – A. P. Oct 17 '17 at 21:37
• It is not necessarily incorrect, I would think that perimeter is used in this case more often. – micsthepick Oct 17 '17 at 21:38
• @micsthepick I read it as the circumference of the circumcircle, which completely changes the question. – boboquack Oct 17 '17 at 22:01

This is not a puzzle but I want to share the answer:

First, we know that

The $|AB|$ is equal to $3$.

So

$|AF|+|BG|$ has to be equal to $3$ as well since the tangents to the circle from a point has to be equal to each other.

Moreover, E and D point location does not matter but the sum

$|EF|+|DG|$ is equal to $|DE|$ with the same rule. which is a part of the circumference of the big triangle except the grey part $|EC|+|DC|$.

As a result, whatever E and D point you choose, the circumference of $CDE$ will be

Circumference of CDE = Circumference of ABC - 6 = $6+5+3-6=8$

• Isn't it incorrect to say "circumference of CDE/ACD," since they are triangles? Isn't it "perimeter," since circumference is a term used specifically for circles and similar shapes? – Aryaman Feb 16 '19 at 7:02

If I were to simply solve this like a multiple choice question, I would consider the extreme case where the tangent is co-linear with AC or BC, the length will be twice the distance from C to either intersection between the circle and the triangle that is on a side that ends at C. the length of these such sides are always equal. using this fact you can set up some simultaneous equations and solve for c:

a+b=3 (x)
a+c=5 (y)
b+c=6 (z)
2*(a+b+c) = 14 (x+y+z)
a+b+c = 7 (i)
c = 4 (i-x)


c = 4 and 2*c = 8, therefore the answer is 8. As for a proof that It is the same in between the extremes, you can refer to other answers.

• You still need to prove that the circumference is constant. Otherwise theoretically the multiple choice could come into play. – boboquack Oct 17 '17 at 21:23
• Wow, you are really fast. But as you have more than the average 3 minutes per question, could you also add your calculations that made you find $4$ as length between $C$ and the intersection between the triangle side and the incircle? It would be even better with a proof that shows that this is true not only for this extreme position of the tangent. – A. P. Oct 17 '17 at 21:41