Let $ABC$ be an acute triangle with ${AB\neq AC}$. The incircle ${\omega}$ of the triangle touches the sides ${BC, CA}$ and ${AB}$ at ${D, E}$ and ${F}$, respectively. The perpendicular line erected at ${C}$ onto ${BC}$ meets ${EF}$ at ${M}$, and similarly the perpendicular line erected at ${B}$ onto ${BC}$ meets ${EF}$ at ${N}$. The line ${DM}$ meets ${\omega}$ again in ${P}$, and the line ${DN}$ meets ${\omega}$ again at ${Q}$. Prove that ${DP=DQ}$. Ruben Dario & Leo Giugiuc (Romania)
Problem
Source: 2015 JBMO Shortlist G5
Tags: geometry, JBMO, similar triangles, incircle
navi_09220114
08.10.2017 13:41
Suffices to show PQ parallel BC. <=> (M,N;DI\cap EF,BC\cap EF)=-1 <=> (C,B;D,BC\cap EF)=-1, which is true
Devastator
20.06.2018 16:05
I have a proof with a little bit of trig
Let I be the incenter
Claim: triangles DBN and DCM are similar
Proof: It suffices to show that $\frac{NB}{BD}=\frac{MC}{CD}$, which is equivalent to $\frac{NB}{BF}=\frac{MC}{CE}$
By Sine Law, this is equivalent to
$\frac{sinBFN}{sinBNF}=\frac{sinCEM}{sinCME}$, but then BNF and CME sum up to 180 since BN and CM are parallel, and angles BFN and CEM are equal since AE=AF, so this is true
Thus, the Claim is proved
By Claim, angles BDN and CDM are equal, so angles IDN and IDM are equal a strong well, which immediately implies that DP=DQ
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sunken rock
20.06.2018 22:45
We need $\angle BDQ=\angle CDP$, or $\triangle NBD\sim\triangle MCD$, i.e. $\frac{BN}{BD}=\frac{CM}{CD}$. Let $R=EF\cap BC$. The parallel through $B$ to $AC$ intersects $EF$ at $S$, $\triangle BFS\sim\triangle AFE$, thus $BS=BF$, but$\frac{RB}{RC}=\frac{BS}{CE}=\frac{BF}{CE}=\frac{BD}{CD}$, but $\frac{RB}{RC}=\frac{BN}{CM}$, so $\frac{BN}{CM}=\frac{BD}{CD}$, done. Best regards, sunken rock
jayme
21.06.2018 10:29
Dear Mathlinkers, also at http://www.artofproblemsolving.com/community/c6t48f6h1276894_collinear_points Sincerely Jean-Louis
amar_04
23.11.2019 21:19
Here's probably a different solution... parmenides51 wrote: Let $ABC$ be an acute triangle with ${AB\neq AC}$. The incircle ${\omega}$ of the triangle κύκλος touches the sides ${BC, CA}$ and ${AB}$ at ${D, E}$ and ${F}$, respectively. The perpendicular line erected at ${C}$onto ${BC}$ meets ${EF}$at ${M}$, and similarly the perpendicular line erected at ${B}$onto ${BC}$ meets ${EF}$at${N}$. The line ${DM}$ meets ${\omega}$ again in ${P}$, and the line ${DN}$ meets ${\omega}$ again at ${Q}$. Prove that ${DP=DQ}$. Ruben Dario & Leo Giugiuc (Romania) Solution:- Let $DK\perp EF$ where $K\in EF$. Claim:- $\angle BKD=\angle CKD$. Let $EF\cap BC=T$. Note that $AD,BE,CF$ are concurrent at the Gregonne Point of $\triangle ABC$. Hence, $(T,D;B,C)=-1$ also $KD\perp EF$, hence, $\angle BKD=\angle DKC$. Next is just angle chase and applying Alternate Segment Theorem... $$\begin{cases}\angle BKD=\angle BND \\ \angle DKC=\angle DMC\end{cases}\implies \angle BND=\angle DMC\implies\angle NDB=\angle MDC\implies \angle DQP=\angle DPQ\implies\boxed{DP=DQ}\blacksquare$$
Mahdi_Mashayekhi
21.12.2021 21:57
we have to prove NB/BD = MC/CD NB/BD = NB/BF and MC/CD = MC/CE. NB/BF = sin NFB / sin BNF and MC/CE = sin MEC / sin EMC. ∠NFB = ∠MEC and ∠BNF = 180 - ∠EMC so sin NFB = sin MEC and sin BNF = sin EMC. we're Done.
parmenides51
27.03.2022 16:23
solved also here
ike.chen
15.08.2022 06:55
Let $T = EF \cap BC$. By Ceva-Menelaus on the Gergonne point, we have $-1 = (T, D; B, C)$, implying $$\frac{DB}{DC} = \frac{TB}{TC} = \frac{BN}{CM}.$$Now, $DBN \sim DCM$ follows from SAS, so $$\angle DPQ = \angle BDQ = \angle BDN = \angle CDM = \angle CDP = \angle DQP$$which finishes. $\blacksquare$ Remark: Cono Sur 2022/2 is actually the exact same problem.