Triangle $ABC$ has a right angle at $C$, and $AC=BC= 1$. Let $I$ be the incenter of triangle $ABC$. Let $D$, $E$, and $F$ be the midpoints of $AI$, $BI$, and $CI$, respectively. Furthermore, let $J$ be the intersection of $AE$ and $BD$, $K$ be the intersection of $BF$ and $CE$, and $L$ be the intersection of $CD$ and $AF$.
Find the area of hexagon $DJEKFL$.
Write your answer in the form $a/b$, where $a$ and $b$ are positive integers and $\gcd(a, b) = 1$.
Answer of this problem is $1/6$.
But my answer is 3/8.
Here's My approach
Let $M$ be the midpoint of $AB$. Since triangle $ABC$ is isosceles, $M$ is also the midpoint of $CI$. Hence, $DM$ is a median of triangle $ADI$, and $EF$ is a median of triangle $BDI$. Therefore, by the triangle median property, $DM = 1/2 AD$ and $EF = 1/2 BD$.
Since $I$ is the incenter of triangle $ABC$, we have angle $AIC = \angle BIC = 90 + \angle C/2$. Thus, $\angle AID = 180 - \angle AIC = 90 - \angle C/2$. Similarly, $\angle BID = 90 - \angle C/2$.
Therefore, triangles $AID$ and $BID$ are similar, and we have $AD/BD = AI/BI = \sin(B/2)/\sin(A/2) = \cos(C/2)/\sin(A/2)$. Since triangle $ABC$ is isosceles, we have $\angle BAC = 90 - \angle C/2$. Thus, $\sin(A/2) = \cos(BAC/2) = \cos((90 - \angle C/2)/2) = \sin(C/4)/\sqrt2$.
Therefore, $AD/BD = \cos(C/2)/(\sin(C/4) \sqrt2)$. It follows that $DM/EF = (1/2 AD)/(1/2 BD) = \cos(C/2)/\sin(C/4) \sqrt2$.
By the angle bisector theorem, we have $AE/BE = AC/BC = 1$. Thus, $J$ is the midpoint of $AE$, and $K$ is the midpoint of $BF$. Therefore, the line segments $DJ$ and $EK$ divide triangle $DEF$ into three triangles, each of which is similar to triangle $BDI$.
The ratio of the area of triangle $DJK$ to the area of triangle $DEF$ is $(DJ/EJ)^2 = (EK/BK)^2 = (BD/2EF)^2 = (2 \cos(C/2)/\sin(C/4))^2 = 4 \cos^2(C/2)/\sin^2(C/4)$.
The ratio of the area of triangle $DJL$ to the area of triangle $DEF$ is $(DJ/DL)^2 = (1/2 DM/DF)^2 = (\cos(C/2)/\sin(C/4) \sqrt2/EF)^2 = 2 \cos^2(C/2)/\sin^2(C/4)$.
The ratio of the area of triangle $EKL$ to the area of triangle $DEF$ is $(EK/EL)^2 = (1/2 EF/EM)^2 = (1/2 EF/CI)^2 = (1/2 EF/(2EF/\cos(C/2)))^2 = 4 \cos^2(C/2)$.
Therefore, the area of hexagon $DJEKFL$ is the difference between the area of triangle $DEF$ and the sum of the areas of triangles $DJK$, $DJL$, and $EKL$. Hence,
$Area(DJEKFL) = (1/2) EF^2 (1 - 4 \cos^2(C/2)/\sin^2(C/4) - 2 \cos^2(C/2)/\sin^2(C/4) - 4 \cos^2(C/2))$.
Substituting $\cos(C/2) = \sqrt{1 - \sin^2(C/2)}$ and $\sin(C/2) = 1/\sqrt{1 + \cot^2(C/2)}$, we obtain
$Area(DJEKFL) = (1/4) (1 - \sin^2(C/4)) (1 + \cos(C/2) - \cos^2(C/2) - 2 \cos^4(C/2))$.
Simplifying, we obtain $Area(DJEKFL) = (1/4) (1 - (1 - \cos(C/2))/(2\sin(C/4))) (2 \cos^4(C/2) - \cos^3(C/2) - \cos^2(C/2) - \cos(C/2) + 1)$.
Simplifying, we obtain
$Area(DJEKFL) = (1/16) (2\sin(C/4) + \cos(C/2) - 1) (2 \cos^4(C/2) - \cos^3(C/2) - \cos^2(C/2) - \cos(C/2) + 1)$.
Now, we can use the identity $\cos(2x) = 2\cos^2(x) - 1$ to express $\cos^4(C/2)$ in terms of $\cos(2C)$:
$\cos^4(C/2) = (1/4) (2\cos^2(C/2) - 1)^2 = (1/4) (2\cos^2(C) - 3)^2 = (1/4) (\cos^2(C) - 4\cos^2(C) + 4) = (3/4) - (1/4) \cos^2(C)$.
Similarly, we can use the identity $\cos(3x) = 4\cos^3(x) - 3\cos(x)$ to express $\cos^3(C/2)$ and $\cos^2(C/2)$ in terms of $\cos(C)$:
$\cos^3(C/2) = (1/2) \cos(C/2) (1 + \cos(C)) = (1/2) \sqrt{1 - \sin^2(C/2)} (1 + \cos(C))$,
$\cos^2(C/2) = 1 - \sin^2(C/2) = \cos^2(C/2) - \sin^2(C/2) = \cos(C) - \sin^2(C/2)$. Substituting these expressions into the formula for the area of hexagon $DJEKFL$ and simplifying, we obtain
$Area(DJEKFL) = (3/16) (2\sin(C/4) + \cos(C/2) - 1) (1 - \cos(C) - \cos^2(C/2) - \cos(C/2) + 1)$.
Simplifying further, we obtain
$Area(DJEKFL) = (3/16) (2\sin(C/4) + \cos(C/2) - 1) (\sin^2(C/2) - \cos(C/2) - \cos(C) + 1)$.
Finally, we can use the half-angle identity again to express $\sin^2(C/2)$ in terms of $\cos(C/2)$:
$\sin^2(C/2) = (1 - \cos(C))/2$.
Substituting this into the formula for the area of hexagon $DJEKFL$ and simplifying, we obtain
$Area(DJEKFL) = (3/32) (\cos(C/2) - 2(8\cos^4(C/4) - 8\cos^2(C/4) + 1) + 2(2\cos^2(C/4) - 1) + 1)$.
Simplifying further, we obtain
$Area(DJEKFL) = (3/8) \cos^4(C/4) - (9/8) \cos^2(C/4) + (1/4)$.
Therefore, the area of hexagon $DJEKFL$ is $(3 \cos^4(C/4) - 9 \cos^2(C/4) + 2)/8$. To write this in the form $a/b$, we can multiply the numerator and denominator by $8$ to obtain
$Area(DJEKFL) = (3 \cos^4(C/4) - 9 \cos^2(C/4) + 2)/8 = (3 \cos^4(C/4) - 9 \cos^2(C/4) + 2)/(2^3) = 3/8 * (\cos^4(C/4) - 3 \cos^2(C/4) + 2/3)$.
Since $\gcd(3, 8) = 1$, the answer is $3/8 * (\cos^4(C/4) - 3 \cos^2(C/4) + 2/3)$.
