Geometry with a lot of orthogonality

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Geometry with a lot of orthogonality

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Source: European Mathematical Cup 2017 Junior,P3

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1008 posts

#1 h Dec 28, 2017, 6:26 PM • 4 Y

Y by buratinogigle, nguyendangkhoa17112003, Adventure10, Mango247

Let $ABC$ be an acute triangle. Denote by $H$ and $M$ the orthocenter of $ABC$ and the midpoint
of side $BC,$ respectively. Let $Y$ be a point on $AC$ such that $YH$ is perpendicular to $MH$ and let $Q$ be a point
on $BH$ such that $QA$ is perpendicular to $AM.$ Let $J$ be the second point of intersection of $MQ$ and the circle
with diameter $MY.$ Prove that $HJ$ is perpendicular to $AM.$

(Steve Dinh)

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3454 posts

#2 h Dec 29, 2017, 12:03 AM • 2 Y

Y by Adventure10, Mango247

Solution with skipiano and wu2481632

Let $D,E,F$ be the feet of the altitudes to $BC, AC, AB$ respectively. Let $EF$ and $BC$ intersect at $T$ .

Lemma 1. $H$ is the orthocenter of $ATM.$

Proof. This is the result of a USA TST problem; clearly, $M$ is the circumcenter of $(BFEC),$ so by Brokard, $M$ is the orthocenter of $ATH$ and $MH\perp AT$ . Since also $AD\perp TM,$ $H$ is the orthocenter of $ATM$ as desired.

Lemma 2. Triangles $HYM$ and $AQM$ are similar.

Proof. This is bashable with a variety of techniques (coordinate, trig, bary). We spent a very long time looking for a synthetic solution but we couldn't find it

Main problem. By Lemma 1, we know that $TH\perp AM.$ Thus, using phantom points, the condition we're asked to prove is equivalent to showing that $J,$ the intersection of $TH$ and $QM,$ is the projection of $Y$ onto $QM.$ In other words, it suffices to show that $HYJM$ is cyclic.

Since $AQ\perp AM,$ we have $AQ\parallel TH,$ from which we deduce that $\angle AQM=\angle HJM.$ But by Lemma 2, $\angle HYM=\angle AQM,$ so $\angle HYM=\angle HJM,$ implying the desired cyclicy.

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4233 posts

#3 h Dec 29, 2017, 1:56 AM • 1 Y

Y by Adventure10

Any synthetic proof for Lemma 2 above?

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abeker

#4 h Dec 29, 2017, 8:53 AM • 3 Y

Y by ILOVEMYFAMILY, Adventure10, Mango247

Here is a synthetic proof of Lemma 2 from post #2 . By simple angle chasing we can get $\angle HAY = \angle MBH$ and $\angle AHY = \angle BMH$ , hence $\triangle AHY \sim \triangle BMH$ . Similarly, we have $\angle AHQ = \angle MCA$ and $\angle AQH = \angle MAC$ , so $\triangle AQH \sim \triangle MAC$ . From these similarities we obtain $\frac{HY}{HM} = \frac{AH}{BM}$ and $\frac{AQ}{AM} = \frac{AH}{MC}$ . Since $BM = MC$ , it follows that $\frac{HY}{HM} = \frac{AQ}{AM}$ , which combined with the fact that $\angle MAQ = \angle MHY = 90^\circ$ implies $\triangle AQM \sim \triangle HYM$ .

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#5 h Dec 29, 2017, 10:20 AM • 2 Y

Y by Adventure10, Mango247

Let $E$ be the foot from $B$ to $AC$ . The key claim is

Claim : $\Delta AQM\sim\Delta HYM$ .

Proof : Let $U, V$ be midpoints of $CE, BH$ . Since $\angle QAM=90^{\circ}$ , $AM$ is tangent to $\odot(QAE)$ which implies $\measuredangle AQE=\measuredangle UAM$ therefore $\Delta AUM\sim \Delta QEA$ .

Proceeding similarly on other side, we get that $\measuredangle HYE=\measuredangle VHM$ . Therefore $\Delta HVM\sim\Delta YEH$ .

Since $\Delta MUV$ is the midpoint triangle of $\Delta EBC$ and $\angle EBC=\angle EAH$ , therefore $\Delta MUV\sim\Delta EBC\sim\Delta EAH$ . Therefore
$\frac{AQ}{AM}=\frac{AE}{MU} =\frac{HE}{MV}=\frac{HY}{HM}$ implying the desired similarity.

Back to the main problem, note that $\measuredangle HJM=\measuredangle HYM = \measuredangle AQM$ . Therefore $JH\parallel AQ\perp AM$ as desired.

This post has been edited 2 times. Last edited by MarkBcc168, Dec 29, 2017, 10:28 AM
Reason: Add more detail

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#6 h Dec 29, 2017, 12:00 PM • 7 Y

Y by anantmudgal09, Einstein314, microsoft_office_word, zuss77, TurtleKing123, Adventure10, Mango247

This is a very beautiful problem! I'll provide one more synthetic solution of the fact $\triangle AQM\sim\triangle HYM$ :
Let $HY\cap AQ=L$ . Note that it is enough to prove that $\angle{HMY}=\angle{AMQ}$ . Let $HY$ intersect $AB$ at $X$ , by butterfly we have that $\triangle{MHX} \equiv \triangle{MHY}$ , thus it is enough to show that $\angle{XMH}=\angle{AMQ}$ . Now applying Isogonality Lemma on $\{Q,X\}$ wrt $\angle{AMH}$ , we have that it is enough to show that $\angle{AML}=\angle{BMH}$ , but because $ALMH$ is cyclic, we have the following relations:
$\angle{AML}=\angle{AHL}=90^{\circ}-\angle{MAH}-\angle{ALH}=90^{\circ}-\angle{MAH}-\angle{AMH}=\angle{BMH} \ \ \ \blacksquare$

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#7 h Dec 29, 2017, 8:46 PM • 2 Y

Y by Adventure10, Mango247

It's pretty tricky with synthetic techniques. I'll post the trig solution which many posters rightly claim to be feasible. Also, @above, nice solution!

Ferid.---. wrote:

Let $ABC$ be an acute triangle. Denote by $H$ and $M$ the orthocenter of $ABC$ and the midpoint
of side $BC,$ respectively. Let $Y$ be a point on $AC$ such that $YH$ is perpendicular to $MH$ and let $Q$ be a point
on $BH$ such that $QA$ is perpendicular to $AM.$ Let $J$ be the second point of intersection of $MQ$ and the circle
with diameter $MY.$ Prove that $HJ$ is perpendicular to $AM.$

(Steve Dinh)

Clearly, we want $HJ \parallel AQ \iff \angle MYH=\angle MJH=\angle MQA \iff \triangle MHY \sim \triangle MAQ$ . Equivalently, we'll show that $\tan \angle MYH=\tan \angle MQA$ . Observe that $\frac{AQ}{\cos A}=\frac{AB}{\sin CAM}\iff AQ=(2R\cos A) \left(\frac{\sin ACM}{\sin CAM}\right)=\frac{2AM \cdot AH}{BC}.$ Hence $\tfrac{AQ}{AM}=2\cot A$ . Let $BB_1$ be the altitude in $\triangle ABC$ and $N$ be the midpoint of $CB_1$ . Then $HMNY$ is cyclic, so $\frac{HY}{HM}=\tan \angle B_1NH=\frac{2HB_1}{CB_1}=2\cot A$ proving $\tan \angle MYH=\tfrac{1}{2}\tan A=\tan \angle MQA$ as desired. $\blacksquare$

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2487 posts

#8 h Jan 1, 2018, 5:02 PM • 2 Y

Y by Adventure10, Mango247

Possibly insightful solution that looks at the bigger side of what's going on.

Let $E,F$ be the feet of $B,C$ on the opposite sides. $X=YH \cap AB$ , $P=AQ\cap CH$ . From butterfly theorem on $BCEF$ twice, once with $A$ and then with $H$ , we have $HXM \cong HYM$ and $APM \cong AQM$ . Letting the midpoints of $BF,CE,BE,CF$ to be $R,S,T,U$ we have that since $ABE \cup \{ T \} \sim HBF \cup \{ R \}$ so $\angle MQA = 90^\circ - \angle AMQ 90^\circ - \angle ATQ = 90^\circ - \angle ATE = 90^\circ -\angle HRF = \angle HRM = \angle HXM = \angle HYM$ , so $HJ \parallel AQ$ so $HJ \perp AM$ . $\blacksquare$

NB: The angle chase setup we used can also be used to give another proof of the Butterfly theorem.

This post has been edited 1 time. Last edited by WizardMath, Jan 1, 2018, 5:02 PM

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RC.

#9 h Jan 8, 2018, 3:01 AM • 3 Y

Y by zuss77, Adventure10, Mango247

Clearly, $\Delta AQH \sim \Delta MAC \Rightarrow \dfrac{AH}{MC} = \dfrac{AQ}{AM} -(i)$

Let $YH \cap AB = Z$ , by butterfly theorem, $HY = HZ$ , also we have: $\Delta AHZ \sim \Delta CMH \Rightarrow \dfrac{AH}{CM} = \dfrac{HZ}{HM} =\dfrac{HY}{HM} - (ii)$

From $(i),(ii): \dfrac{AQ}{AM} =\dfrac{HY}{CM}$ also, $\angle MAQ = \angle MHY =90^{\circ}$
$\Rightarrow \Delta MAQ \sim \Delta MHY \Rightarrow \angle AQM = \angle HYM = \angle HJM \Rightarrow HJ || AM \Longrightarrow HJ \perp AM.$

I see @above now, essentially the same.

This post has been edited 3 times. Last edited by RC., Jul 16, 2020, 10:49 AM

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#10 h Jan 13, 2019, 5:42 PM • 2 Y

Y by Adventure10, Mango247

Here’s my solution (read sketch) which I guess is not yet posted

Let $X,Z$ be the reflection of $A,H$ over $M$ .
Angle chasing yields $HYZC$ and $ABQX$ to be cyclic.
Further angle chasing yields $\angle{YZH} = \angle{QAX}$ .
Also, we know that $\angle{YHZ} = \angle{PAX} = 90^{\circ}$
So, $\Delta AXQ \sim \Delta HRY$
It is not difficult to see that $\Delta AQM \sim \Delta HYM$ .
Note that the above condition is equivalent to $HJ \parallel AQ$
Thus, $AM \perp HJ$ . $\blacksquare$

This post has been edited 1 time. Last edited by MathInfinite, Jan 13, 2019, 5:51 PM

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#11 h May 15, 2020, 9:01 AM • 3 Y

Y by Mango247, Mango247, Mango247

RC. wrote:

Clearly, $\Delta AQH \sim \Delta MAC \Rightarrow \dfrac{AH}{MC} = \dfrac{AQ}{AM} -(i)$

Let $YH \cap AB = Z$ , by butterfly theorem, $HY = HZ$ , also we have: $\Delta AHZ \sim \Delta CMH \Rightarrow \dfrac{AH}{CM} = \dfrac{HZ}{HM} =\dfrac{HY}{CM} - (ii)$

From $(i),(ii): \dfrac{AQ}{AM} =\dfrac{HY}{CM}$ also, $\angle MAQ = \angle MHY =90^{\circ}$
$\Rightarrow \Delta MAQ \sim \Delta MHY \Rightarrow \angle AQM = \angle HYM = \angle HJM \Rightarrow HJ || AM \Longrightarrow HJ \perp AM.$

I see @above now, essentially the same.

excuse me, may I ask to which cyclic quadrilateral do you apply Butterfly theorem?

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zuss77

#12 h May 15, 2020, 4:56 PM

Y by

@above. To a circle with diameter $BC$ .
Also there's a typo: $\dfrac{HY}{CM}$ needs to be $\dfrac{HY}{HM}$

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#13 h Nov 30, 2021, 3:08 PM

Y by

First of all let's $D$ and $E$ be foot of perpendicular from $A$ and $B$ ,respectively.
$HJ\perp AM$ $\iff$ $HJ||AQ$ $\iff$ $\angle AQM=\angle HJM$ $\iff$ $\angle AQM=\angle HYM$ $\iff$ $\triangle AQM\sim \triangle HYM$ $\iff$ $\frac{AQ}{AM}=\frac{HY}{HM}$ . Now from angle chasing you can easily obsserve that
$\angle HAY=\angle HBM$ and $\angle AYH=\angle BHM$ . So $\triangle AYH\sim \triangle BHM$ . So $\frac{HY}{HM}=\frac{AH}{BM}$ .

Since $\angle AHQ=\angle ACM$ and $\angle AQH=\angle MAC$ w eget $\triangle AHQ\sim \triangle MCA$ . So $\frac{AQ}{AM}=\frac{AH}{MC}=\frac{AH}{BM}$ $\implies$ $\frac{AQ}{AM}=\frac{AH}{BM}=\frac{HY}{HM}$ . So we are done!

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#14 h Dec 25, 2021, 4:16 PM

Y by

A problem filled with similarities ...

Let MH meet AQ at S. we have to prove HJ || AQ or in fact ∠MHJ = ∠MSQ.
Let's go and prove some similarities first'.
step1 : HAY and MBH are similar.
∠HAY = ∠MBH and ∠AYH = 90 - ∠YHQ = ∠QHS = ∠BHM.

step2 : AQH and MAC are similar.
∠AHQ = ∠ACM and ∠AMC = 180 - ∠AMB = 180 - ∠HAS = ∠HAQ.

step3 : MHY and MAQ are similar.
∠MHY = 90 = ∠MAQ and MH/HY = MB/AH = MC/AH = AM/AQ.

Now that we have MHY and MAQ are similar we have AHM and QYM are similar as well.
now let's get back to our angle chasing.
∠MHJ = 90 - ∠JHY = 90 - ∠JMY = 90 - ∠AMH = ∠MSA so HJ || AQ.
we're Done.

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