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: $\frac{5}{3}$