Let us consider the incircle touching the sides BC, CA, AB at Ai , Bi , Ci respectively

Problem 11881

(American Mathematical Monthly, Vol.123, January 2016) Proposed by R. Bagby (USA).

Given a triangle T , let Tⁱ be the triangle formed by the points of tangency of the incircle and let T^e be the triangle whose vertices are the points where the three excircles of T are tangent to the sides of T . Prove that

Area(Tⁱ) = Area(T^e) = r

2R· Area(T ).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let A, B, C the vertices of T , and let a = |BC|, b = |CA|, c = |AB|.

Let us consider the incircle touching the sides BC, CA, AB at Ai , Bi , Ci respectively.

Then it is known that

2|ABi| = 2|ACi| = b + c − a, 2|BCi| = 2|BAi| = c + a − b, 2|CAi| = 2|CBi| = a + b − c.

Therefore,

Area(Tⁱ)

Area(T ) = Area(T ) − Area(ABⁱCi) − Area(BCⁱAi) − Area(CAⁱBi) Area(T )

= 1 − |ABⁱ||ACⁱ|

|AB||AC| −|BCⁱ||BAⁱ|

|BC||BA| −|CAⁱ||CBⁱ|

|CA||CB|

= 1 − (b + c − a)²

4cb −(c + a − b)²

4ac −(a + b − c)² 4ba

= (a + b − c)(b + c − a)(a + c − b)

4abc = (4Area(T ))²

4abc(a + b + c) = r 2R where we used Heron’s formula and the following known facts,

Area(T ) = abc

4R = r(a + b + c)

2 .

Now, let us consider the excircles touching the sides BC, CA, AB at Ae, Be, Cerespectively.

Then it is known that

2|ABe| = 2|BAe| = a + b − c, 2|BCe| = 2|CBe| = b + c − a, 2|CAe| = 2|ACe| = c + a − b.

Therefore, Area(T^e)

Area(T ) = Area(T ) − Area(AB^eCe) − Area(BC^eAe) − Area(CA^eBe) Area(T )

= 1 − |AB^e||AC^e|

|AB||AC| −|BC^e||BA^e|

|BC||BA| −|CA^e||CB^e|

|CA||CB|

= 1 − (a + b − c)(c + a − b)

4cb −(b + c − a)(a + b − c)

4ac −(c + a − b)(b + c − a) 4ba

= (a + b − c)(b + c − a)(a + c − b)

4abc = r

2R.