Answer to Question #147540 in Geometry for Reem

Question

Let AK, BL, CN be angle bisectors of triangle ABC, and they intersect at point I. It is known that the ratios of areas of triangles ANL , BKN and CLK to the area of triangle ABC are equal to (7/39), (9/65) and (7/15) respectively. Find the ratio of IK:AI

Answer 1

Lets denote a=BC, b=AC, c=AB.

Incenter I divides the angle bisector in a defined ratio of the length of sides of the triangle

$\frac{AI}{IK}=\frac{b+c}{a}, \ \frac{BI}{IL}=\frac{a+c}{b}$

Let x=AN, y=BN, n=BK, m=KC, p=AL, q=CL.

An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

$\frac{x}{y}=\frac{b}{a}, \ x+y=c$

$\frac{q}{p}=\frac{a}{c}, \ p+q=b$

$\frac{n}{m}=\frac{c}{b}, \ n+m=a$

Then

$x=\frac{bc}{a+b}, y=\frac{ac}{a+b}$

$n=\frac{ac}{b+c}, m=\frac{ab}{b+c}$

$p=\frac{bc}{a+c}, q=\frac{ab}{a+c}$

The ratios of areas of triangles ANL, BKN and CLK to the area of triangle ABC are equal to 7/39 , 9/65 and 7/15 respectively, therefore:

$\frac{px}{bc}=\frac{7}{39},\ \frac{ny}{ac}=\frac{9}{65}, \ \frac{mq}{ab}=\frac{7}{15}$

Then

$\frac{bc}{a+c}\cdot\frac{bc}{a+b}=\frac{7}{39}bc$

$\frac{ac}{b+c}\cdot\frac{ac}{a+b}=\frac{9}{65}ac$

$\frac{ab}{b+c}\cdot\frac{ab}{b+c}=\frac{7}{15}ab$

Let b/a=u and c/a=v, then:

$uv=\frac{7}{39}(1+u)(1+v)$

$v=\frac{9}{65}(u+v)(1+u)$

$u=\frac{7}{15}(u+v)(1+v)$

$\frac{9}{65}(u+v)(1+u)\cdot\frac{7}{15}(u+v)(1+v)=\frac{7}{39}(1+u)(1+v)$

$(u+v)^2=\frac{325}{117}=\frac{25}{9}$

$\frac{AI}{IK}=\frac{b+c}{a}=\frac{b}{a}+\frac{c}{a}=u+v=\frac{5}{3}$

Answer: $\frac{5}{3}$