Given an acute triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
with
![\angle B > \angle C](/media/m/f/b/c/fbcf95fc3cf2504af0093e150fe7de36.png)
. Point
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
is the incenter, and
![R](/media/m/4/d/7/4d76ce566584cfe8ff88e5f3e8b8e823.png)
the circumradius. Point
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
is the foot of the altitude from vertex
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
. Point
![K](/media/m/e/1/e/e1ed1943d69f4d6a840e99c7bd199930.png)
lies on line
![AD](/media/m/6/9/6/69672822808d046d0e94ab2fa7f2dc80.png)
such that
![AK = 2R](/media/m/a/e/0/ae0c7e025ea4dd63d238d325a462672c.png)
, and
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
separates
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
and
![K](/media/m/e/1/e/e1ed1943d69f4d6a840e99c7bd199930.png)
. Lines
![DI](/media/m/c/4/4/c4428a036d22a395c2bf542fe0a39a5d.png)
and
![KI](/media/m/d/1/d/d1d86e523bf3ff8514ca00f76265e400.png)
meet sides
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
at
![E,F](/media/m/e/d/9/ed9b3965cc5e0b074cdb261c465e4fe2.png)
respectively. Let
![IE = IF](/media/m/a/c/5/ac54a7ac44ecc3201c37af8778b721be.png)
.
Prove that
![\angle B\leq 3\angle C](/media/m/a/6/0/a603e3a1075d1a1b3be1a6ba18ab43c5.png)
.
Author: Davoud Vakili, Iran
%V0
Given an acute triangle $ABC$ with $\angle B > \angle C$. Point $I$ is the incenter, and $R$ the circumradius. Point $D$ is the foot of the altitude from vertex $A$. Point $K$ lies on line $AD$ such that $AK = 2R$, and $D$ separates $A$ and $K$. Lines $DI$ and $KI$ meet sides $AC$ and $BC$ at $E,F$ respectively. Let $IE = IF$.
Prove that $\angle B\leq 3\angle C$.
Author: Davoud Vakili, Iran