UNIQUE TRIANGLE PUZZLE.

Question:
In a triangle ABC, AB = 3, BC = 5, and CA = 6. The line segment DE is tangent to the incircle (inscribed circle) of the triangle. What is the perimeter of triangle CDE?

The principle of equal tangents is key to solving this problem: two tangents from an external point to a circle have equal length.

We can prove this as follows. For an external point A tangent at point B to a circle with center O, ABO will be a right triangle (since AB is tangent to BO). Then BO equals the radius of the circle r, and AO is some length w.

For another tangent point C, we end up with another right triangle ACO with hypotenuse AO = w and the leg CO = r.

Two right triangles that have congruent legs and hypotenuses are congruent, so the corresponding legs are also congruent (AB = AC).

We will use this principle of equal tangents repeatedly to solve the original problem.

Re-shaping the perimeter of CDE

First let’s simplify the perimeter of CDE. Let’s label tangent points from the incircle, with F for AC, J for DE, and G for BC.

 Now we come to 

Next part,

Suppose EJ = x and DJ = y. By the principle of equal tangents, EG = xand DF = y.

We can re-write the perimeter of CDE as follows:

CD + CE + DE
= CD + CE + x + y
= CD + y + CE + x

But since:

CF = CD + y
CG = CE + x

We then have:

CD + CE + DE
= CD + y + CE + x
= CF + CG

Now we use the principle of equal tangents one more time: CF = CG. So we have:

CD + CE + DE
= 2(CF)

As show in figure:👇

There’s another way to derive this expression too using a “shape-shifting” triangle CDE. Since DE only has to be tangent to the incircle of ABC, we can imagine pushing point E towards point C. In the limiting case, we end up with E and C being coincident, which will happen when D is a tangent point to the incircle.

In this limiting case, CE = 0, and CD = DE = CF, so we have the perimeter is equal to:

CD + CE + DE
= CF + 0 + CF
= 2(CF)

Remarkable!

Now let’s use this simplified diagram to solve the original problem.

Solving for the perimeter of CDE

We will let CF = p/2, and we want to solve for p.

We will “walk around the triangle” using equal tangents and known side lengths.

First we have CG = CF by equal tangents, so CG = p/2. Then we have BC = 5, which makes BG = 5 – p/2.

Then we have BG = BH by equal tangents, so BH = 5 – p/2. Then we have AB = 3, which makes AH = 3 – (5 – p/2) = p/2 – 2.

Then we have AH = AF by equal tangents, so AF = p/2 – 2. Then we have CA = 6, but we also have derived it is equal to AF + FC = p/2 – 2 + p/2. So we have:

p/2 – 2 + p/2 = 6
p = 8

And by going around the triangle we have solved for the perimeter of CDE, pretty neat!
That's the solution of unique triangle question.

Thanks to Mr. Presh Talwalkar
Credit:👉Presh Talwalkar.
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