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Source: European Mathematical Cup 2022, Junior Division, Problem 3

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1280 posts

#1 h Dec 19, 2022, 11:48 PM • 1 Y

Y by Rounak_iitr

Let $ABC$ be an acute-angled triangle with $AC > BC$ , with incircle $\tau$ centered at $I$ which touches $BC$ and $AC$ at points $D$ and $E$ , respectively. The point $M$ on $\tau$ is such that $BM \parallel DE$ and $M$ and $B$ lie on the same halfplane with respect to the angle bisector of $\angle ACB$ . Let $F$ and $H$ be the intersections of $\tau$ with $BM$ and $CM$ different from $M$ , respectively. Let $J$ be a point on the line $AC$ such that $JM \parallel EH$ . Let $K$ be the intersection of $JF$ and $\tau$ different from $F$ . Prove that $ME \parallel KH$ .

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#2 h Dec 20, 2022, 12:37 AM • 2 Y

Y by ike.chen, Mango247

I initially thought in Pascal's theorem because I felt that I needed something to deal with a circle and collinear points or concurrent lines.

But then:
Let $\alpha = \angle{EMH}$ . Since $EC$ is tangent to the incircle, then $\alpha = \angle{CEH}$ .
Let $G = MF\cap EC$ .

$ME \| KH \iff \angle{EMH} = \angle{KEM} = \angle{KFM} = \angle{JFG} \iff GJ^2 = GF \cdot GM$
So we want to prove that $G$ is the midpoint of $JE$ because we know that $GE^2 = GF \cdot GM$

Then I just wanted to play with the cross ratio:
$(MJ, ME; MG, p(M, IE)) = -1 \iff (p(I, HE), p(I, ME); p(I, DE), IE) = -1 \iff (EH, EM; ED, EC) = -1$

And the latter is true since $CD, CE$ are tangents to the incircle and $M, H, C$ are collinear. We're done!
(The first equality comes from lemma applied around point $I$ and the second one comes from lemma around point $E$ )

Remarks

This post has been edited 5 times. Last edited by on_gale, Dec 20, 2022, 12:41 AM
Reason: Minor typo errors

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#3 h Dec 20, 2022, 12:45 AM

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Trig bash is love, Trig bash is life! <3

Important Remark

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1389 posts

#4 h Dec 20, 2022, 7:31 AM

Y by

Here is a simple synthetic solution using only angle chasing and similar triangles.

We want $ME \parallel HK$ , so we need $\angle MFK = \angle CEH$ . This motivates to let $MF \cap EH=G$ and we want $GEFJ$ being cyclic. We have that $\angle EGF = \angle HED= \angle HMD =\angle EMN$ (where $N$ is midpoint of $ED$ and it lies on $FH$ by the harmonic quadrilateral $MEHD$ ). Thus, we want $\triangle EMN \sim \triangle EJF \iff \frac{EF}{EJ}=\frac{EN} {EM} \iff \frac {EJ} {EM}=\frac {DM} {DN}$ . The last one follows from the similarity of the triangles $DNM$ and $EMJ$ ( $\angle NDM =\angle JEM$ and $\angle EMJ = \angle MEH = \angle MND$ ), done.

This post has been edited 4 times. Last edited by VicKmath7, Dec 20, 2022, 7:33 AM

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1280 posts

#5 h Dec 20, 2022, 7:45 AM • 1 Y

Y by VicKmath7

@above well the need of harmonic quadrilateral is a bit out of the ordinary similarity stuff, especially for juniors

Anyway, nice solution!

This post has been edited 1 time. Last edited by Assassino9931, Dec 20, 2022, 7:45 AM

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256 posts

#6 h Dec 20, 2022, 11:16 AM

Y by

A bit boring problem.

Step 1: $\triangle JEF \sim \triangle MHD$
Proof: Observe that $EDFM$ is isosceles trapezium, therefore by symmetry we have $\triangle CEF = \triangle CDM$ . Moreover $\angle HDC = \angle CMD = \angle CFE$ , which implies that $\triangle CHD \sim \triangle CEF$ . This means that we have:
$\frac{CE}{EH}=\frac{EF}{HD}$ Since $EH \parallel JM$ we have $\frac{CE}{CH}=\frac{JE}{MH}$ . We conclude that:
$\frac{CE}{EH}=\frac{EF}{HD} =\frac{JE}{MH}$ Moreover, $\angle JEF = \angle EMF = \angle DFM = \angle DHM$ . This implies that $\triangle JEF \sim \triangle MHD$ as desired.

We define point $T \neq M$ to be the intersection of $JM$ and $\tau$ .

Step 2: $CJ \parallel KT$
Proof: Observe that since $EDMF$ and $EHMT$ are isosceles trapezium, then $\widehat{EF}=\widehat{DM}$ and $\widehat{ET}=\widehat{MH}$ . Therefore $\widehat{FT}=\widehat{ET}-\widehat{EF}= \widehat{MH}-\widehat{DM}=\widehat{DH}$ . Thus $\angle EJF = \angle DMH = \angle FKT$ , which implies $CJ \parallel KT$ as desired.

Step 3: Finish
Finish Observe that:
$\angle KTM = \angle CJM = \angle CEH = \angle HME$ This means that $\widehat{EH} = \widehat{KM}$ , therefore $EHKM$ is isosceles trapezium, implying that $ME \parallel KH$ as desired.

Attachments:

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#7 h Dec 12, 2023, 11:36 PM

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Consider a homothety at $C$ sending $E$ to $J$ . By $EH || JM$ we get that $H$ gets sent to $M$ , let's say that $D$ gets sent to $D'$ . We have that $MF,D'J$ are both perpendicular to $CI$ , so by symmetry $JD'MF$ is a trapezium. Finally we have
$\sphericalangle{HDE} =\sphericalangle{MD'J} = \sphericalangle{D'JF}=\sphericalangle{MFK}\implies EH=MK \implies HK || EM.$

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