A geometry problem by Kshetrapal Dashottar

Geometry Level 4

Let $ABC$ be a right triangle with $\angle B = 90^\circ$ . Let $E$ and $F$ be the mid-points of $AB$ and $AC$ , respectively . Suppose the incenter $I$ of triangle $ABC$ lies on the circumcircle of triangle $AEF$ . Find the ratio $\dfrac{BC}{AB}$ .

Give your answer to 2 decimal places.

The answer is 1.33.

2 solutions

Ahmad Saad
Jul 24, 2016

Kshetrapal Dashottar
Jul 22, 2016

$\angle AEF = \angle ABC = 90^\circ$ . <EF || BC> and EF = 2 BC So AF is diameter of the circumcircle of AEF  AIF = AEF = 90° CIF = AIC – AIF = 135° – 90° = 45° (AIC = 135 as I is the incentre of ABC) Now, AF = FC = 2 1 AC (F is the mid point of AC) Let FCI = BCI =  So, IAC = 180 – (AIC + ICA) = 180 – (135 + ) = 45 –  Apply sinrule is CIF IF

sin

CF sin45  IF = CF sin √2 ..(1) In AIF IF

sin(45  )

AF sin90  IF = sin(45 – ) AF ..(2) From Equation (1) and (2) √2 CF sin = sin(45 – ) AF √2 =    sin sin(45 ) √2 =     2 sin cos sin tan = 3 1 Now, tan 2 =    1 tan2

2tan

9 1 1 3 1 2 



3 2  8

9

4 3 In ABC tan ABC = tan 2 = 4

3

BC AB  AB

BC

3 4

This is a RMO problem. So u should try to solve it using Euclidean geometry only without using trigonometry.

Aditya Kumar - 4 years, 9 months ago

but in rmo trignometry is allowed

Kshetrapal Dashottar - 4 years, 9 months ago