Let $\triangle ABC$ be a nondegenerate isosceles triangle with $AB=AC$, and let $D, E, F$ be the midpoints of $BC, CA, AB$ respectively. $BE$ intersects the circumcircle of $\triangle ABC$ again at $G$, and $H$ is the midpoint of minor arc $BC$. $CF\cap DG=I, BI\cap AC=J$. Prove that $\angle BJH=\angle ADG$ if and only if $\angle BID=\angle GBC$. Proposed by David Stoner
Problem
Source: ELMO Shortlist 2013: Problem G11, by David Stoner
Tags: geometry, circumcircle, geometry unsolved
v_Enhance
28.07.2013 05:49
Here's the not-very-satisfying solution I found for this problem: v_Enhance wrote: I have no idea why I did this, but for some reason it made my day. Anyway, hello. By barycentric coordinates on $\triangle ABC$ it is easy to obtain $G=(a^2+c^2 : -b^2 : a^2+c^2)$. Then, one can compute $I=(a^2+c^2 : a^2+c^2 : b^2+2(a^2+c^2)$, from which it follows that $J = (a^2+c^2 : 0 : b^2+2(a^2+c^2))$. Now we use complex numbers. Set $D=0$, $C=1$, $B=-1$, $A=ri$ for $r \in \mathbb R^+$, $K=\frac{r}{3}$, and $H = -\frac{i}{r}$. Now, upon using the vector definition for barycentric coordinates, we obtain $I= \textstyle\frac{(r^2+5)(ri) + (r^2+5)(-1) + (3r^2+11)(1)}{5r^2+21}$, or \[ I = \frac{2r^2+6}{5r^2+21} + \frac{r(r^2+5)}{5r^2+21} i. \] Similarly, we can get \[ J = \frac{3r^2+11}{4r^2+16} + \frac{r(r^2+5)}{4r^2+16}i. \] Claim 1: $\angle BID = \angle GBC \iff r^6+9r^4-17r^2-153 = 0.$
Let $V(a+bi) = \textstyle\frac ba$ for $a,b \in \mathbb R$, and note $V(nz) = V(z)$ for all $n \in \mathbb R$. Then, \[ \angle BID = \angle GBC \iff V\left( \frac{D-I}{B-I} \right) = V\left( \frac{G-B}{C-B} \right) \] Obviously the right-hand side is $\textstyle\frac{r}{3}$. Meanwhile,
\begin{align*}
\frac{-I}{1-I} &= \frac{I}{I+1} \\
&= \frac{ \frac{2r^2+6}{5r^2+21} + \frac{r(r^2+5)}{5r^2+21} i }{ \frac{7r^2+27}{5r^2+21} + \frac{r(r^2+5)}{5r^2+21} i } \\
&= \frac{1}{\text{real}} \left[ \left( (2r^2+6)+r(r^2+5)i \right)\left( (7r^2+26)-r(r^2+5)i \right) \right] \\
&= \frac{1}{\text{real}} \left[ \left( r^6 + 24r^4+121r^2+162 \right) + (5r^2+21)(r)(r^2+5)i \right] \\
\end{align*}
Hence, $V\left( \textstyle\frac{I}{I+1} \right) = \textstyle\frac{(5r^2+21)(r)(r^2+5)}{r^6+24r^4+121r^2+161}$. This is equal to $r/3$ if and only if \[ r^6+24r^4+121r^2+162-3(5r^2+21)(r^2+5) = 0. \] Expanding gives the conclusion. $\blacksquare$
Claim 2: $\angle BJH = \angle ADG \iff 2r^8+8r^6-28r^r-136r^2-102=0$.
We proceed in the same spirit. It's evident that $V\left( \frac{K-D}{G-D} \right) = V(I)^{-1} = \frac{2r^2+6}{r(r^2+5)}$. On the other hand, we can compute
\begin{align*}
\frac{-\frac{1}{r} \cdot i - J}{-1-J} &= \frac{rJ+i}{r(1+J)} \\
&= \frac{1}{r} \cdot \frac{\frac{r(3r^2+11)}{4r^2+16} + \frac{r^2(r^2+5) + (4r^2+16)}{4r^2+16} i}{\frac{7r^2+27}{4r^2+16} + \frac{r(r^2+5)}{4r^2+16} i} \\
&= \frac{1}{\text{real}} \left[ \left( r(3r^2+11) + (r^4+9r^2+16 \right)i \right]\left[ (7r^2+27)-r(r^2+5)i \right] \\
&= \frac{1}{\text{real}} \left[ r(r^6+35r^4+219r^2+377) + i(4r^6+64r^4+300r^2+432) \right]
\end{align*}
Hence, $V\left( \textstyle\frac{H-J}{B-J} \right) = \textstyle\frac{4r^6+64r^4+300r^2+432}{r(r^6+35r^4+219r^2+377)}$. So, the equality occurs when \[ (r^2+5)(4r^6+64r^4+300r^2+432) - (2r^2+6)(r^6+35r^4+219r^2+377) = 0. \] Expand again. $\blacksquare$
Now all that's left to do is factor these polynomials! The former one is $(r^4-17)(r^2+9)$, and the latter is $2(r^2+1)(r^2+3)(r^4-17)$. Restricted to positive $r$ we see that both are zero if and only if $r = \sqrt[4]{17}$. Therefore the conditions are equivalent, occuring if and only if $AD = \sqrt[4]{17}$.
oneplusone
10.08.2013 06:14
I can show $\angle BID=\angle GBC\implies \angle BJH=\angle ADG$ without mentioning $\sqrt[4]{17}$. $\angle BID=\angle GBC\implies DI\cdot DG=DB^2=DC^2$, so $\angle IDC=180-\angle DIC-\angle ICD=180-\angle DCG-\angle GBD=\angle BGC=\angle BAC$. So if $DG$ intersects $AC$ at $K$, then $AKDB$ is cyclic so $BK\perp AC$ so $BK\parallel HC$. Let $BK$ intersect $AH$ at $L$, then $DL=DH$. Let $M$ be on $DC$ such that $CM=2DM$. Let $AD$ intersect $BE$ at $N$. Now we length chase. \[\frac{KI}{ID}=\frac{KC}{CA}\frac{AN}{ND}=\frac{2KC}{CA} \]\[\frac{KJ}{JC}=\frac{KI}{ID}\frac{DB}{BC}=\frac{KC}{CA} \]\[\frac{KC}{CA}=\frac{KC-KJ}{CA-JC}=\frac{CJ}{AJ} \]\[\frac{CJ}{JA}\frac{AH}{HD}\frac{DM}{MC}=\frac{KC}{CA}\frac{AH}{HL}=1 \]Thus $H,M,J$ are collinear. Note that $\triangle HDC\sim\triangle BDA$ so $\angle DHM=\angle DBN=\angle BID\implies \angle BJH=\angle ADG$.