Let $I$ be the incenter of triangle $ABC$. The tangent point of $\odot I$ on $AB,AC$ is $D,E$, respectively. Let $BI \cap AC = F$, $CI \cap AB = G$, $DE \cap BI = M$, $DE \cap CI = N$, $DE \cap FG = P$, $BC \cap IP = Q$. Prove that $BC = 2MN$ is equivalent to $IQ = 2IP$.
Problem
Source: 2018 CGMO Day 2 Problem 8
Tags: geometry, incenter
MarkBcc168
14.08.2018 12:33
Progress : If $BC = 2MN$. Let $T$ be the midpoint of $BC$. Recall that $\angle BMC = \angle BNC = 120^{\circ}$ so $TB=TC=TM=TN=MN$. Thus $\Delta TMN$ is equilateral $$\angle NCM = 30^{\circ} \implies \angle BIC = 120^{\circ}\implies \angle BAC = 60^{\circ}$$This implies $AIFG$ is cyclic or $IF=IG$. Thus $\overline{DPE}$ is Simson line of $I$ w.r.t. $\Delta AFG$. This implies $P$ is the foot from $I$ to $FG$ or $IP$ bisects $\angle FIG \equiv \angle BIC$. Hence $\Delta IMN\cup P\sim\Delta IBC\cup Q$, implying the conclusion.
TelvCohl
15.08.2018 17:43
Problem : Given a $ \triangle ABC $ with incenter $ I, $ intouch triangle $ \triangle DEF $ and incentral triangle $ \triangle XYZ. $ Let $ P $ be the intersection of $ EF, YZ $ and $ Q $ be the intersection of $ BC, IP. $ Prove that $ IP = 2IQ \Longleftrightarrow BC = 2UV $ where $ U,V $ is the intersection of $ EF $ with $ BI, CI, $ respectively. Proof : Clearly, $ B, C, U, Z $ lie on the circle with diameter $ BC, $ so $ UV = BC \cdot \sin \frac{1}{2} \angle A $ and hence $ BC = 2UV $ if and only if $ \angle A = 60^{\circ}. $ Let $ \tau \equiv \overline{A^*B^*C^*} $ be the perspectrix of $ \triangle ABC,\ \triangle XYZ $ and $ \overline{P}, \overline{Q}, \overline{R} $ be the intersection of the parallel from $ I $ to $ \tau $ with $ YZ, BC, AA^*, $ respectively. Since $ A^*(\overline{P},\tau; \overline{Q}, \overline{R}) = -1 = (\overline{P}, \overline{Q}; I, \overline{R}), $ so $ \overline{P}\overline{Q} = \overline{P}\overline{R} $ and $ I \overline{Q} = 2 I \overline{P}. $ Let $ T $ be the midpoint of $ EF, $ then note that $ AA^* \parallel EF $ we conclude that $$ IP = 2IQ\ \Longleftrightarrow\ \overline{P} \in EF\ \Longleftrightarrow\ \frac{AI}{TI} = 4\ \Longleftrightarrow\ \angle A = 60^{\circ}\ \Longleftrightarrow\ BC = 2UV. $$
RC.
16.08.2018 17:04
We have \(BN \perp CN\) and \(BM \perp CM \Rightarrow BCMN\) is cyclic \(\Rightarrow \Delta BIC \sim \Delta MIN \therefore \dfrac{BC}{MN}= \dfrac{IQ}{IP},\) Q.E.D
Rickyminer
16.08.2018 17:11
RC. wrote: We have \(BN \perp CN\) and \(BM \perp CM \Rightarrow BCMN\) is cyclic \(\Rightarrow \Delta BIC \sim \Delta MIN\), Q.E.D I don't think you have completed your proof.
Rickyminer
16.08.2018 18:02
RC. wrote: We have \(BN \perp CN\) and \(BM \perp CM \Rightarrow BCMN\) is cyclic \(\Rightarrow \Delta BIC \sim \Delta MIN \therefore \dfrac{BC}{MN}= \dfrac{IQ}{IP},\) Q.E.D Can you clarify that why $P,Q$ are corresponding points in the similar triangles?
trumpeter
29.03.2019 00:49
$IQ=2IP$ is equivalent to $\angle{BAC}=60^\circ$ or $120^\circ$ while $BC=2MN$ is equivalent to $\angle{BAC}=60^\circ$ only
. Assuming the problem is for an acute triangle instead, here is a solution.
I claim that both $BC=2MN$ and $IQ=2IP$ are equivalent to $\angle{BAC}=\frac{\pi}{3}$.
First, note that by the Iran Incenter Lemma, $M$ and $N$ are the feet of the $B$ and $C$ altitudes of $\triangle{BHC}$, where $H$ is the orthocenter of $\triangle{BIC}$. Thus \[MN=BC\left|\cos\angle{BHC}\right|=BC\left|\cos\left(\pi-\angle{BIC}\right)\right|=BC\left|\cos\left(\pi-\left(\frac{\pi}{2}+\angle{BAC}\right)\right)\right|=BC\sin\frac{\angle{BAC}}{2},\]so $BC=2MN$ is equivalent to $\sin\frac{\angle{BAC}}{2}=\frac{1}{2}$, which in turn is equivalent to $\angle{BAC}=\frac{\pi}{3}$.
Now, we apply barycentric coordinates on $\triangle{ABC}$ to prove that $IQ=2IP$ is equivalent to $\angle{BAC}=\frac{\pi}{3}$. Let $A=\left(1,0,0\right)$, $B=\left(0,1,0\right)$, $C=\left(0,0,1\right)$ such that $D=\left(s-b:s-a:0\right)$, $E=\left(s-c:0:s-a\right)$, $F=\left(a:0:c\right)$, $G=\left(a:b:0\right)$, and $I=\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)$ with $I$ in homogenized form. Let $P'=\left(\frac{a\left(s-a\right)}{bc},\frac{\left(a-c\right)\left(s-b\right)}{c\left(b-c\right)},\frac{\left(b-a\right)\left(s-c\right)}{b\left(b-c\right)}\right)$ in homogenized form, alternatively $P'=\left(a\left(b-c\right)\left(s-a\right):b\left(a-c\right)\left(s-b\right):c\left(b-a\right)\left(s-c\right)\right)$ in unhomogenized form. I claim that $P=P'$. It suffices to show that $D,E,P'$ collinear and $F,G,P'$ collinear.
To show that $D,E,P'$ collinear, we compute the determinant
\begin{align*}
\begin{vmatrix}
s-b & s-a & 0 \\
s-c & 0 & s-a \\
a\left(b-c\right)\left(s-a\right) & b\left(a-c\right)\left(s-b\right) & c\left(b-a\right)\left(s-c\right)
\end{vmatrix}
&=-b\left(a-c\right)\left(s-a\right)\left(s-b\right)^2-\left(s-a\right)\left[c\left(b-a\right)\left(s-c\right)^2-a\left(b-c\right)\left(s-a\right)^2\right]\\
&=-\left(s-a\right)\left[b\left(a-c\right)\left(s-b\right)^2+c\left(b-a\right)\left(s-c\right)^2-a\left(b-c\right)\left(s-a\right)^2\right]\\
&=-\left(s-a\right)\left[s^2\left(b\left(a-c\right)+c\left(b-a\right)-a\left(b-c\right)\right)-\left(2s-b\right)b^2\left(a-c\right)-\left(2s-c\right)c^2\left(b-a\right)+\left(2s-a\right)a^2\left(b-c\right)\right]\\
&=-\left(s-a\right)\left[-b^2\left(a^2-c^2\right)-c^2\left(b^2-a^2\right)+a^2\left(b^2-c^2\right)\right]\\
&=0
\end{align*}so $D,E,P'$ are collinear.
To show that $F,G,P'$ collinear, we compute the determinant
\begin{align*}
\begin{vmatrix}
a & 0 & c \\
a & b & 0 \\
a\left(b-c\right)\left(s-a\right) & b\left(a-c\right)\left(s-b\right) & c\left(b-a\right)\left(s-c\right)
\end{vmatrix}
&=abc\left(b-a\right)\left(s-c\right)+abc\left(a-c\right)\left(s-b\right)-abc\left(b-c\right)\left(s-a\right)\\
&=\frac{abc}{2}\left[\left(b-a\right)\left(b+a-c\right)+\left(a-c\right)\left(c+a-b\right)-\left(b-c\right)\left(b+c-a\right)\right]\\
&=\frac{abc}{2}\left[b^2-a^2-c\left(b-a\right)+a^2-c^2-b\left(a-c\right)-b^2+c^2+a\left(b-c\right)\right]\\
&=0
\end{align*}so $F,G,P'$ are collinear.
Thus $P=P'$ so we have the coordinates of $P$. Now let $Q=\left(0,y_Q,z_Q\right)$ in homogenized form. Then $\overrightarrow{IP}=\left(\frac{a\left(s-a\right)}{bc}-\frac{a}{2s},\ldots\right)$ and $\overrightarrow{IQ}=\left(-\frac{a}{2s},\ldots\right)$ in homogenized coordinate difference. Then $\frac{IP}{IQ}$ is the absolute value of the ratio between the $x$ components of these coordinate differences, so \[\frac{IP}{IQ}=\left|\frac{\frac{a\left(s-a\right)}{bc}-\frac{a}{2s}}{-\frac{a}{2s}}\right|=\left|\frac{2s\left(s-a\right)}{bc}-1\right|=\left|\frac{b^2+c^2-a^2}{2bc}\right|=\left|\cos\angle{BAC}\right|\]so $IQ=2IP$ is equivalent to $\cos\angle{BAC}=\frac{1}{2}$, which in turn is equivalent to $\angle{BAC}=\frac{\pi}{3}$.
Thus, both conditions are equivalent to $\angle{BAC}=\frac{\pi}{3}$, so $BC=2MN$ if and only if $IP=2IQ$.