CHKMO 2018 P2

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YanYau

#1 h Dec 2, 2017, 3:52 PM • 3 Y

Y by integrated_JRC, Adventure10, Mango247

Suppose $ABCD$ is a cyclic quadrilateral. Extend $DA$ and $DC$ to $P$ and $Q$ respectively such that $AP=BC$ and $CQ=AB$ . Let $M$ be the midpoint of $PQ$ . Show that $MA\perp MC$ .

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#2 h Dec 2, 2017, 6:01 PM • 3 Y

Y by Tawan, Vietjung, Adventure10

Let $B'$ be a point on $(ABCD)$ such that $BB'CA$ is an isosceles trapezoid, we have $B'C=AB=QC$ and $B'A=CB=PA$ .
Let $N$ be the second intersection of $(PAB')$ with $PQ$ .
We've $\angle{B'NP}=\angle{B'AD}=180^{\circ}-\angle{B'CD}$ . So, $\angle{B'CD}=180^{\circ}-\angle{B'NP}=\angle{B'NQ}$ .
This means $N'$ also lies on $(B'CQ)$ .
Note that $CB'=CQ$ and $AB'=AP$ . So $\angle{ANC}=\angle{B'NC}+\angle{B'NA}=\frac{\angle{B'NQ}+\angle{B'NP}}{2}=90^{\circ}$ .
Let $(ANC)$ intersects $PQ$ again at $M'$ , note that $\angle{AM'C}=90^{\circ}$ .
It's enough to prove that $M'=M$ , i.e. $M'$ is the midpoint of $PQ$ .
Note that we've $\angle{M'AC}=\angle{CNQ}=\angle{CB'Q}$ .
Reflect $A$ and $C$ over $M'$ to get $A'$ and $C'$ , respectively. Also, reflect $B'$ over $AC$ to get $B_2$ .
We've $CB'=CQ,CA=CA'$ and $\angle{ACA'}=\angle{B'CQ}\Rightarrow \angle{B'CA}=\angle{QCA'}$ .
This gives $\triangle{B'CA}\equiv \triangle{QCA'}$ . It follows that $\triangle{AB_2C}$ is the reflection of $\triangle{A'QC}$ over $CM'$ .
Similarly, $\triangle{CB_2A}$ is the reflection of $\triangle{C'PA}$ over $AM'$ .
Hence, $\angle{PAM'}=\angle{B_2AM'}=\angle{B_2AC}+\angle{CAM'}=\angle{B'AC}+\angle{CA'M'}=\angle{CA'Q}+\angle{CA'M'}=\angle{M'A'Q}$ .
So, $AP\parallel A'Q$ . Since $M'$ is the midpoint of $AA'$ , $M'$ is also the midpoint of $PQ$ , done.

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Dec 2, 2017, 6:01 PM

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643 posts

#3 h Dec 3, 2017, 2:47 AM • 1 Y

Y by Adventure10

Solution: We flip $\Delta ABC$ . Note that this preserve all the properties of the configuration.
We use complex numbers to prove $\angle AMC=\pi/2$ . Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral $ABCD$ lie on the unit circle.
Then since $d,c,q$ are collinear we have,
$\frac{d-c}{q-c}=\frac{\bar{d}-\bar{c}}{\bar{q}-\bar{c}}\implies \bar{q}=\frac{c-q+d}{cd}$ Again since $BC=CQ$ ,
$|b-c|=|c-q|\implies (c-b)(\frac{1}{c}-\frac{1}{b})=(\frac{1}{c}-\frac{c-q+d}{cd})(c-q)$ $\implies q=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+c$ Similarly,
$p=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+a$ We can rotate the quadrilateral, so WLOG we can assume $b=1$ .
It can be seen that exactly one of them is $+$ and the other has $-$ sign. WLOG assume $p$ has $+$ sign.
Therefore,
$m=\frac{p+q}{2}=\frac{\sqrt{d}(a-c)+c+a}{2}$ Let $\delta=\frac{2a-2m}{2c-2m}$ , we wish to show $\delta+\bar{\delta}=0$ . Now,
$\delta=\frac{2a+\sqrt{d}(c-a)-c-a}{2c+\sqrt{d}(c-a)-c-a}=\frac{(c-a)(\sqrt{d}-1)}{(c-a)(\sqrt{d}+1)}=\frac{\sqrt{d}-1}{\sqrt{d}+1}$ Hence,
$\bar{\delta}=\frac{\frac{1}{\sqrt{d}}-1}{\frac{1}{\sqrt{d}}+1}=\frac{1-\sqrt{d}}{1+\sqrt{d}}$ Thus, $\delta+\bar{\delta}=0$ . And we are done. $\blacksquare$

Remark: I am a beginner in complex bashing, so...

This post has been edited 2 times. Last edited by ayan.nmath, Dec 3, 2017, 2:53 AM

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#4 h Dec 3, 2017, 4:52 AM • 2 Y

Y by Adventure10, Mango247

YanYau wrote:

Suppose $ABCD$ is a cyclic quadrilateral. Extend $DA$ and $DC$ to $P$ and $Q$ respectively such that $AP=BC$ and $CQ=AB$ . Let $M$ be the midpoint of $PQ$ . Show that $MA\perp MC$ .

From medians formula we need to prove
$MA^2+MC^2=AC^2$ .

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44 posts

#5 h Dec 23, 2017, 6:16 PM • 1 Y

Y by Adventure10

ayan.nmath wrote:

Solution: We flip $\Delta ABC$ . Note that this preserve all the properties of the configuration.
We use complex numbers to prove $\angle AMC=\pi/2$ . Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral $ABCD$ lie on the unit circle.
Then since $d,c,q$ are collinear we have,
$\frac{d-c}{q-c}=\frac{\bar{d}-\bar{c}}{\bar{q}-\bar{c}}\implies \bar{q}=\frac{c-q+d}{cd}$ Again since $BC=CQ$ ,
$|b-c|=|c-q|\implies (c-b)(\frac{1}{c}-\frac{1}{b})=(\frac{1}{c}-\frac{c-q+d}{cd})(c-q)$ $\implies q=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+c$ Similarly,
$p=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+a$ We can rotate the quadrilateral, so WLOG we can assume $b=1$ .
It can be seen that exactly one of them is $+$ and the other has $-$ sign. WLOG assume $p$ has $+$ sign.
Therefore,
$m=\frac{p+q}{2}=\frac{\sqrt{d}(a-c)+c+a}{2}$ Let $\delta=\frac{2a-2m}{2c-2m}$ , we wish to show $\delta+\bar{\delta}=0$ . Now,
$\delta=\frac{2a+\sqrt{d}(c-a)-c-a}{2c+\sqrt{d}(c-a)-c-a}=\frac{(c-a)(\sqrt{d}-1)}{(c-a)(\sqrt{d}+1)}=\frac{\sqrt{d}-1}{\sqrt{d}+1}$ Hence,
$\bar{\delta}=\frac{\frac{1}{\sqrt{d}}-1}{\frac{1}{\sqrt{d}}+1}=\frac{1-\sqrt{d}}{1+\sqrt{d}}$ Thus, $\delta+\bar{\delta}=0$ . And we are done. $\blacksquare$

Remark: I am a beginner in complex bashing, so...

Good evening! We have BC = AP , not BC = CQ ! Although it is not a dammage for the whole idea!

This post has been edited 2 times. Last edited by DimPlak, Dec 23, 2017, 6:26 PM
Reason: change opinion

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#6 h Dec 23, 2017, 6:38 PM • 2 Y

Y by Adventure10, Mango247

@above Umm... Actually it is "Good night" for me

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I wrote "We flip $\triangle ABC$ " in the very first line of my proof, so I think it should be fine.

This post has been edited 2 times. Last edited by ayan.nmath, Dec 23, 2017, 6:42 PM

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#7 h Dec 24, 2017, 8:40 AM • 1 Y

Y by Adventure10

ayan.nmath wrote:

@above Umm... Actually it is "Good night" for me

.
I wrote "We flip $\triangle ABC$ " in the very first line of my proof, so I think it should be fine.

Sorry, but I don't understand the term "flip" ! Im a beginner too, more than you!

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#8 h Mar 12, 2018, 5:54 AM • 2 Y

Y by Adventure10, Mango247

Make a trapezoids $ABCP_{1}$ and $ABCQ_{1}$ such that $AP=BC=AP_{1}$ and $CQ=AB=CQ_{1}$ .Let $CP_{1}\cap AQ_{1}=N$ then $180-\angle ADC=\angle ABC=\angle ANC=\angle P_{1}NQ_{1}$ wich gives $\angle AP_{1}N=\angle AP_{1}=\angle ADCC=\angle ADC=\angle P_{1}NA$ so $P_{1}A=AN=AP$ and similarly $NQ_{1}=NC=CQ$ .Now $\angle PDQ+\angle DPN+\angle DQN=\angle PNQ...(1)$ but $\angle PDQ=180-\angle ABC=180-\angle ANC=180-(\angle ANP+\angle PNQ+\angle CNQ)=180-(\angle DPN+\angle PNQ+\angle DQN )$ and replacing in $(1)$ gives $180 =2\angle PNQ$ so $\angle PNQ=90$ and so $MN=MP=MQ$ but we proved $PA=AN$ and $CN=CQ$ thus $MA$ bisects $\angle NMP$ and $MC$ bisects $\angle NMQ$ so $\angle AMC=\frac{\angle NMP+\angle NMQ}{2}=90$ as desired. $\blacksquare$

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#9 h Mar 12, 2018, 11:51 AM • 7 Y

Y by tenplusten, MarkBcc168, jchang0313, BALAMATDA, Lsyaaaaaaaa, Adventure10, Mango247

Let $K,N$ be the midpoints of $CA,CP$ , then $\triangle ABC\sim \triangle MNK$ with ratio $2$ , so $KM=\frac{AC}{2}=KA=KC$ $\implies MA\perp MC$ .

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IstekOlympiadTeam

542 posts

IstekOlympiadTeam

#10 h Mar 12, 2018, 5:05 PM • 7 Y

Y by tenplusten, luizp, Godfather2043, BALAMATDA, Adventure10, Mango247, JJeom

Let $A'$ be the reflection of point $A$ through point $M$ $\implies$ $APA'Q$ is a parallelogram.
Since $\angle APQ + \angle CQP=\angle ABC$ , it follows that that $\angle A'QC=\angle CQP+\angle A'QP=\angle CQP+\angle APQ=\angle ABC$ . $\implies$ $\boxed{\angle A'QC=\angle ABC}$ .
And we also have that $A'Q=AP$ and $CQ=AB$ $\implies$ $\boxed{\triangle ABC= \triangle A'QC}$
$\implies$ $AC=A'C$ $\implies$ $CM\perp AM$

This post has been edited 2 times. Last edited by IstekOlympiadTeam, Mar 12, 2018, 5:12 PM

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#11 h Sep 3, 2020, 1:34 PM

Y by

Here is a routine spiral similarity solution.

Let $T = \odot(DAC)\cap \odot(DPQ)$ be the spiral center. Then, since $\triangle TAP\sim\triangle TCQ$ , $\tfrac{TA}{TC} = \tfrac{AP}{CQ} = \tfrac{BC}{AB}$ , which means that $BT$ passes through the midpoint $K$ of $AC$ . Seeing this, we observe that $\triangle TAP\sim\triangle TKM$ or
$\frac{KM}{TK} = \frac{AP}{AT} = \frac{BC}{AT} = \frac{CK}{KT}$ or $KM=KC=KA$ as desired.

This post has been edited 1 time. Last edited by MarkBcc168, Sep 3, 2020, 1:34 PM

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#12 h Sep 10, 2020, 2:01 AM • 4 Y

Y by math31415926535, Mango247, Mango247, Mango247

See my solution to this problem on my Youtube channel here:

https://www.youtube.com/watch?v=5aTTapplHQQ

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Krits

#13 h Jan 13, 2021, 3:55 PM • 1 Y

Y by Mango247

Let $B'$ be the reflection of $B$ about perpendicular bisector of $AC$ . Let $A,C$ be the midpoints of $PS,QT$ respectively. So $\triangle B'PS\overset{+}{\sim}\triangle B'TQ$ hence $B',P,S,PT\cap QS$ are cyclic. The fact that $AM//QS$ and $MC//PT$ finishes.

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