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Source: CJMO 2022 P1

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890 posts

#1 h Mar 12, 2022, 4:53 AM

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Let $\triangle{ABC}$ has circumcircle $\Gamma$ , drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$ , similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$ . Prove that if $AB=DE, \angle{ACB}=60^{\circ}$
(sorry it is from my memory I can't remember the exact problem, but it means the same)

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awesomeming327.

1687 posts

awesomeming327.

#2 h Mar 12, 2022, 4:58 AM

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lucky imagine getting a doable geo

our geo was so hard

anyway, AB=DE implies AE || BD which implies $\angle EAD =\angle EBD$ which by orthocenter reflection stuff blah blah implies BHD and AHE both equilateral so we're done

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890 posts

#3 h Mar 12, 2022, 5:06 AM • 1 Y

Y by Vladimir_Djurica

This is my solution provided during the exam
solution

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30629 posts

#4 h Mar 19, 2022, 3:57 AM

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Let $ABC$ be an acute angled triangle with circumcircle $\Gamma$ . The perpendicular from $A$ to $BC$ intersects $\Gamma$ at $D$ , and the perpendicular from $B$ to $AC$ intersects $\Gamma$ at $E$ . Prove that if $|AB| = |DE|$ , then $\angle ACB = 60^o$ .

This post has been edited 4 times. Last edited by parmenides51, May 6, 2024, 11:14 AM

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8791 posts

#5 h May 3, 2022, 11:57 PM • 4 Y

Y by Vladimir_Djurica, Mango247, Mango247, Mango247

jamboard solve!!!

pretty easy??, nothing used beyond the fact that angles intercepting the same arc are equal + angles in a triangle sum to 180.

Attachments:

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1080 posts

#6 h Jul 6, 2022, 1:30 PM • 2 Y

Y by centslordm, Vladimir_Djurica

Notice $\triangle ABE\cong\triangle BED$ by SAS so $H=\overline{AD}\cap\overline{BE}$ is the center of $\Gamma.$ Hence, $2\angle C=\angle AHB=180-\angle C$ and $\angle C=60.$ $\square$

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50 posts

#7 h Jul 6, 2022, 2:00 PM

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Let $AD \cap BE = H$ , $AC \cap BE = X$ , $BC \cap AD = Y$ . Using the fact that angles subtended by arcs of equal lengths are equal, $\angle DBE = \angle ADB$ so $HB = HD$ . Since $AXYB$ is cyclic, $\angle HBY = \angle HAX$ . By construction $ABDC$ is cyclic so $\angle HAX = \angle DBC$ so $BY$ is the perpendicular bisector of $\triangle HBD$ thus $BD = HB$ . Hence $\triangle HBD$ is equilateral and $\angle ACB = \angle ADB = \angle HDB = 60^\circ$ .

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3146 posts

yofro

#8 h Aug 8, 2022, 5:51 PM • 2 Y

Y by Mango247, Mango247

Let $X$ be the foot from $A$ to $BC$ and $Y$ be the foot from $B$ to $AC$ . By reflecting the orthocenter, $XY=\frac{1}{2}ED=\frac{1}{2}AB$ . Since $AYXB$ is cyclic, we have $\triangle CYX\sim \triangle CBA$ so $CX=\frac{1}{2}AC$ and hence $\cos C=\frac{1}{2}\implies C=60^{\circ}$ .

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1162 posts

#9 h Aug 25, 2022, 6:22 AM

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Let $AD \cap BC = X$ and $BE \cap CA = Y$ . The Orthocenter Reflection Lemma yields $XY = \frac{DE}{2} = \frac{AB}{2}.$ Now, since $CXY \overset{-}{\sim} CAB$ , we know $| \cos ACB | = \frac{CX}{CA} = \frac{XY}{AB} = \frac{1}{2}$ so $\angle ACB = 60^{\circ}$ or $\angle ACB = 120^{\circ}$ must hold. $\blacksquare$

Remarks: I'm pretty sure $\angle ACB = 120^{\circ}$ is possible too.

Also, what is the CJMO? Personally, I only know of one CJMO, and this problem isn't from there.

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8791 posts

#10 h Aug 25, 2022, 4:32 PM • 2 Y

Y by oolite, ike.chen

Canada Junior Math Olympiad?

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#11 h Sep 4, 2022, 5:24 AM

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ike.chen wrote:

Also, what is the CJMO? Personally, I only know of one CJMO, and this problem isn't from there.

Canadian Junior Math Olympiad

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30629 posts

#12 h May 6, 2024, 11:14 AM

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Notation : $(XY)$ stands for small arc $XY$

$(DE)= (EC)+(CD)=2\angle EAC+2<DBC=2(180^o-\angle AEB) + 2(180^o-\angle DBC) = 180^o +180^o-4(AB)-4(AB)=360^o-8(AB)=360^o- 4\angle ACB=360^o-4 \cdot 60^o=360^o -240^o=120^o =\angle (DCE) \Rightarrow AB= DE$

my solution might be modified to prove that the converse is also true,
and so the problem could have been asked with iff condition

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DensSv

#13 h Sep 14, 2024, 4:04 PM

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Sol.

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958 posts

#14 h Mar 30, 2025, 6:23 PM

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Angle chase using $\angle AEB= \angle EBD$ and the orthocentre reflection lemma.

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