2019 Japan Junior Mathematical Olympiad Finals 5

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Source: 2019 Japan Junior Mathematical Olympiad (JJMO) Finals 5

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#1 h Feb 12, 2019, 6:09 AM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle with $AB\neq AC$ . Let $M$ be the midpoint of $BC$ , and $N$ be the midpoint of the arc $BC$ of the circumcircle of the triangle $ABC$ containing $A$ . Let $H$ be the foot of the perpendicular from $N$ to $AC$ . The circumcircle of the triangle $AMC$ and the line $CN$ meet at $K(\neq C)$ . Prove that $\angle AKH = \angle CAM$ .

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#2 h Feb 12, 2019, 9:09 AM • 2 Y

Y by AlastorMoody, Adventure10

WLOG $AB<AC$ . We need to show that $\angle{NKA}-\angle{NKH}=\angle{NAM}-\angle{NAC}$ .
Invert the whole figure around circle $(N,NB)$ yields the following problem:

Quote:

Given an isosceles triangle $NBC$ . A point $A'$ lies on $BC$ such that $A',B,C$ collinear in this order. Let $H'$ and $M'$ be the antipode of $N$ with respect to $(A'NC)$ and $(BNC)$ respectively. Let $K'$ be the antipode of $M'$ with respect to $(A'M'C)$ . Show that $\angle{NA'K'}-\angle{NH'K'}=\angle{NM'A'}-\angle{NCA'}$ .

First off, note that $\angle{NM'A'}-\angle{NCA'}=\angle{NM'A'}-\angle{NM'B}=\angle{BM'A'}$ .

Let $O_1$ and $O_2$ be the circumcenter of $\triangle{A'M'C}$ and $\triangle{A'NC}$ .
From $O_1O_2\perp A'C$ , we get that $O_1O_2\parallel NM'$ . Note that $K',O_1,M'$ and $H',O_2,N$ collinear.
Easy side chasing gives $K'H'\parallel NM'$ . Therefore, $\angle{NH'K'}=\angle{H'NM'}$ .

Also, by spiral similarity, we get $\triangle{A'NH'}\sim \triangle{A'K'M'}$ .
This gives $\triangle{A'NK'} \sim \triangle{A'H'M'}\implies \angle{NA'K'}=\angle{H'A'M'}$ .

Hence, we have to show that $\angle{H'A'M'}-\angle{H'NM'} =\angle{BM'A'}$ . Indeed, we have
$\begin{align*} \angle{H'A'M'}-\angle{H'NM'} & = (\angle{H'A'C}-\angle{M'A'C})-(\angle{H'NC}-\angle{M'NC}) \\ & = \angle{M'NC}-\angle{M'A'C}\\ & = \angle{M'BC} -\angle{M'A'C} \\ & = \angle{BM'A'}. \qquad \blacksquare \end{align*}$

This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Feb 12, 2019, 9:10 AM

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#3 h Feb 12, 2019, 9:42 AM • 5 Y

Y by AlastorMoody, ABCCBA, sttsmet, Adventure10, Mango247

Nice problem! I felt this a bit harder than the MO P4.
$[asy] size(9cm); defaultpen(fontsize(10pt)); pair A = dir(120); pair B = dir(220); pair C = dir(320); pair N = dir(90); pair M = (B+C)/2; pair H = foot(N,A,C); pair T = C+N-M; pair P = extension(M,T,A,N); pair K = 2*foot(circumcenter(M,A,C),C,N)-C; draw(A--B--C--cycle,linewidth(1.5)); draw(circumcircle(A,B,C)); draw(circumcircle(A,M,C)); draw(circumcircle(K,N,T),dashed); draw(M--N--T--C); draw(N--H); draw(A--P--M); draw(C--K); draw(B--N,dashed); dot(A);dot(B);dot(C);dot(N);dot(M);dot(H);dot(T);dot(P);dot(K); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$N$",N,1.5*dir(70)); label("$M$",M,S); label("$H$",H,1.5*dir(255)); label("$P$",P,NE); label("$T$",T,dir(345)); label("$K$",K,NW); [/asy]$

Let $T$ be the point such that $CMNT$ is rectangle and let $P=AN\cap MT$ . Notice that $MT\parallel BN$ thus Reim theorem on $\overline{BMC}$ and $\overline{ANP}$ (with $\odot(ACBN)$ ) gives $A,C,M,P$ are concyclic. Moreover, since $NT\parallel MC$ , thus Reim theorem on $\overline{CNK}$ and $\overline{MTP}$ (with $\odot(AKMC)$ ) gives $K,N,T,P$ are concyclic.

Now notice that $K$ is Miquel point of $\{AN, MT, AM, NT\}$ thus $\triangle KAN\sim\triangle KMT$ . But $\triangle AHN\sim\triangle MCT$ thus quadrilateral $KAHN$ and $KMCT$ are similar. Hence we can conclude that $\angle AKH = \angle MKC = \angle MAC$ .

This post has been edited 2 times. Last edited by MarkBcc168, Feb 12, 2019, 10:00 AM

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#4 h Feb 12, 2019, 11:27 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Great problem. Here's my solution: Note that the given condition is equivalent to showing that $AM$ is tangent to $\odot (AHK)$ . Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the angle bisector of $\angle BAC$ to get the following equivalent problem-

Inverted Problem wrote:

Let $\omega$ be the circumcircle of $\triangle ABC$ , and $F$ be the point where the $A$ -symmedian meets $\omega$ . Also, let $N$ be the foot of the internal angle bisector of $\angle BAC$ . Suppose $CF \cap \odot (ANC)=K$ and $AC \cap \odot (FNC)=T$ . Show that $KT \parallel AF$ .

Note that $N$ is the center of spiral similarity mapping $KF$ to $AT$ , i.e. $\triangle NKF \sim \triangle NAT$ . Also $F$ is the center of spiral similarity which takes $NT$ to $BA$ , i.e. $\triangle FNT \sim \triangle FBA$ . Thus, combining these two facts, we have that $\frac{KF}{AT}=\frac{NF}{NT}=\frac{BF}{AB}=\frac{CF}{AC} \Rightarrow \frac{CF}{KF}=\frac{CA}{TA} \Rightarrow KT \parallel AF \quad \blacksquare$

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jayme

#5 h Feb 13, 2019, 1:03 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,

the key of this problem is to show that the circle (AKH) is tangent to AM at A....

Sincerely
Jean-Louis

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jayme

#6 h Feb 16, 2019, 11:43 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
the proof #3 is very nice and ingenious...

Sincerely
Jean-Louis

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jayme

#7 h Feb 16, 2019, 12:47 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
who knows the author of this problem?

Sincerely
Jean-Louis

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#8 h Feb 16, 2019, 2:35 PM • 1 Y

Y by Adventure10

Here is a short proof.
By simple angle chasing $H$ and $M$ are isogonal conjugates in triangle $AKN$ .
The conclusion follows.

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9774 posts

jayme

#9 h Mar 30, 2019, 10:41 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Problemes%20divers%20aimants%20et%20aimables%201.pdf p. 7...

Sincerely
Jean-Louis

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#10 h Feb 12, 2020, 7:12 PM • 1 Y

Y by Adventure10

We need to move the angle $\angle AKH$ somewhere.. Draw circumcircle of triangle $AKH$ and let it intersect $CN$ at $Q$ . We claim $AQ \parallel BC$ . By setting a phantom point $Q'$ such that $AQ' \parallel BC$ we prove $\triangle AHQ' \sim \triangle CMA$ . And then we get $\angle AMC = \angle AHQ' = 180^{\circ} - \angle AKC$ which leads to $Q=Q'$ . So we are done. (the bashy part is by los to prove $\frac{AH}{AQ'}=\frac{CM}{AC}$ )

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