Angle bisectors and areas

Geometry Level 2

In a triangle $\triangle{ABC}$ , $\overline{AB}$ is the larger side. The angle bisector of $\widehat{A}$ divides the triangle in two triangles of areas $231$ and $189$ . The angle bisector of $\widehat{B}$ divides the triangle in two triangles of areas of $200$ e $220$ . Find $AB^2$ . The answer is in the form $n\sqrt{2}$ , find $n$ .

The answer is 847.

1 solution

A Former Brilliant Member
Apr 9, 2019

Let BC=a, CA=b, AB=c, s=(a+b+c)/2. Then b/c=189/231, a/c=200/220=10/11, s=(c/2)(6930/2541), s-a=(c/2)(2310/2541), s-b=(c/2)(2772/2541), s-c=(c/2)(1848/2541). The area of the triangle ABC is 200+220=420. Therefore (c^2)/4=420(2541)^2/√((6930)(2310)(2772)(1848)). This yields AB^2=c^2=847√2

Angle bisectors and areas

The answer is 847.

1 solution

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