HK bisect QS

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lssl

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lssl

#1 h Jan 5, 2012, 8:39 AM • 2 Y

Y by jhu08, Adventure10

In a concyclic quadrilateral $PQRS$ , $\angle PSR=\frac{\pi}{2}$ , $H,K$ are perpendicular foot from $Q$ to sides $PR,RS$ , prove that $HK$ bisect segment $SQ$ .

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SCP

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SCP

#2 h Jan 5, 2012, 9:19 AM • 4 Y

Y by vsathiam, jhu08, Adventure10, Mango247

This question is trivial because $Q$ is on the circumscribed circle, $HK$ the simsonline and $S$ the orthocenter of $PRS$ , so the Eulerline goes through this point and the wallace line through the midpoint.

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Kenny_cz

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Kenny_cz

#3 h Jan 6, 2012, 5:04 PM • 4 Y

Y by vsathiam, jhu08, Adventure10, Mango247

SCP wrote:

This question is trivial because $Q$ is on the circumscribed circle, $HK$ the simsonline and $S$ the orthocenter of $PRS$ , so the Eulerline goes through this point and the wallace line through the midpoint.

That's a bit confusing...

Let $K'$ be the projection of $Q$ to $SP$ , then $K$ , $K'$ , and $H$ are collinear (Simson line of $Q$ wrt $\triangle PRS$ ). Now, we are done because $K'QKS$ is a rectangle and thus it's diagonals bisect each other.

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sayantanchakraborty

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sayantanchakraborty

#4 h Sep 13, 2013, 7:43 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Let <PSQ=x.Then <PRQ=<HQP=x.Now points H,Q,P,K are concyclic (Why?) Thus <HKS=x.Thus HK=HS=HQ

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Ferid.---.

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Ferid.---.

#5 h Mar 22, 2017, 8:05 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Let $\triangle PSR$ -- $HK$ be a Simson line.
$\text{Lemma:}$ Let $P\in(ABC),$ and $H$ be the orthocenter of this triangle,then Simson line bisect $HP.$
Then $\triangle PSR$ --- $S$ be the orthocenter, and $Q\in(ABC).$ Then $HK$ bisect $QS.$

This post has been edited 4 times. Last edited by Ferid.---., Mar 22, 2017, 8:06 AM

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Tumon2001

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Tumon2001

#6 h Mar 22, 2017, 8:14 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Let $KH\cap RS = L$ . Since, $KH$ is a Simson line, so $QL\perp SR$ . Thus, $LSKQ$ is a rectangle and the result follows.

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Delray

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Delray

#7 h Apr 17, 2017, 4:57 PM • 3 Y

Y by jhu08, Adventure10, Mango247

On phone, so can't type might be kind of rushed. Let the foot of the altitude from $Q$ to $SR$ be $L$ . Recognizing that $K$ , $H$ , and $L$ are collinear (Simpson Line) it suffices to show that $KL$ bisects $SQ$ , which is trivial, since they are both diagonals of rectangle $KQLS$ . $\square$

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Delray

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Delray

#8 h Apr 17, 2017, 5:56 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Kenny_cz wrote:

SCP wrote:

This question is trivial because $Q$ is on the circumscribed circle, $HK$ the simsonline and $S$ the orthocenter of $PRS$ , so the Eulerline goes through this point and the wallace line through the midpoint.

That's a bit confusing...

Let $K'$ be the projection of $Q$ to $SP$ , then $K$ , $K'$ , and $H$ are collinear (Simson line of $Q$ wrt $\triangle PRS$ ). Now, we are done because $K'QKS$ is a rectangle and thus it's diagonals bisect each other.

What's the Wallace Line?

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Dr_Vex

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Dr_Vex

#9 h Jun 2, 2020, 7:09 PM • 1 Y

Y by jhu08

@above it's other name of Simson's line.

Small contribution to problem:
Let $KH\cap QS =X$
By angle chasing we find that $PKHQ$ is cyclic. Then again by angle chasing prove that $\Delta XSK$ and $\Delta XKQ$ are isosceles and then we're done.

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PCChess

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PCChess

#10 h Sep 7, 2020, 7:50 PM • 2 Y

Y by jhu08, Mango247

Since $QK$ and $QH$ are perpendiculars, dropping the perpendicular from $Q$ to $SR$ and calling the foot point $L$ , we see that $K, H, L$ are collinear by the Simson Line Theorem. Since $\angle QKS=\angle KSL =\angle SLQ=90$ , $QKSL$ is a rectangle, and since diagonals in a rectangle bisect each other, $HK$ bisects $SQ$ .

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nayersharar

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nayersharar

#11 h Sep 21, 2020, 8:11 AM • 4 Y

Y by jhu08, Mango247, Mango247, Mango247

Let $T\in SR$ such that $QT\perp SQ$ $\Longrightarrow$ by simpson line lemma $H,K,T$ are collinear.

Let $Z$ be a point one the same side of $TQ$ as $PR$ such that $TZ\parallel QS.$

Now, $\angle STZ=\angle TSQ=\angle QPR=\dfrac{\pi}{2}-\angle QRH=\angle HQR$

$\because QHTR$ is cyclic $\Longrightarrow \angle HQR=\angle HTR \Longrightarrow \angle STZ=\angle HQR =\angle HTR$

Let $P\infty$ be the point at infinity along $TZ$ and $QS$ and let $TH\cap QS=M$

Now, $\angle QTS=\dfrac{\pi}{2}$ and $\angle MTS=\angle ZTS$

$\therefore$ By a well known lemma $(Q,S;M,TZ\cap QS)=-1$ $\Longrightarrow (Q,S;M,P\infty)=-1$

$\therefore M$ is the midpoint of $QS$

$(A.W.D)$

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IAmTheHazard

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IAmTheHazard

#12 h Apr 26, 2021, 2:43 PM • 4 Y

Y by centslordm, jhu08, Mango247, Mango247

EGMO lol
Let $\overline{HK}$ intersect $\overline{RS}$ at $T$ . It is known that $\overline{QT} \perp \overline{RS}$ . Thus $\angle QKS=\angle KST=\angle STQ=90^\circ$ , hence $\angle TQK=90^\circ$ and $QKST$ is a rectangle. This implies that $\overline{QS}$ is bisected by $\overline{KT}$ , which implies the desired result as $K,H,T$ are collinear.

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Mogmog8

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Mogmog8

#13 h Sep 9, 2021, 1:43 AM • 2 Y

Y by jhu08, centslordm

Solution

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554183

484 posts

554183

#14 h Sep 10, 2021, 12:12 PM

Y by

Lol wut
Consider $\triangle{RSP}$ and $Q$ lying on its circumcircle. It’s orthocenter is $S$ . The result now follows directly as it is well known that $QS$ is bisected by the Simpson line ( $HK$ )

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BVKRB-

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BVKRB-

#15 h Sep 10, 2021, 12:48 PM

Y by

Since $\triangle PSR$ is a right triangle we conclude that $S$ is the orthocenter of $\triangle PSR$
But the simson line $HK$ bisects the segment joining the orthocenter $S$ and the point $Q$ which finishes the proof $\blacksquare$

Prabh2005 wrote:

Lol wut
Consider $\triangle{RSP}$ and $Q$ lying on its circumcircle. It’s orthocenter is $S$ . The result now follows directly as it is well known that $QS$ is bisected by the Simpson line ( $HK$ )

The typo was made on purpose right? :rotfl:

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Taco12

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Taco12

#16 h Dec 26, 2021, 6:49 PM

Y by

oops I solved this while on vacation (also this is really easy if you have even a small knowledge of simson lines).

Note that $KH$ is the simson line of $Q$ wrt $\triangle PSR$ . Since $S$ is the orthocenter of $\triangle PSR$ , by Simson Line Bisection (well-known result), we are done.

This post has been edited 1 time. Last edited by Taco12, Dec 26, 2021, 6:58 PM

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Karencando

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Karencando

#17 h May 13, 2022, 9:59 PM

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Do I need to prove the Simson Line Bisection result or is it fine to state without proof in contest?

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th1nq3r

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th1nq3r

#18 h Jul 18, 2022, 5:14 PM

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Solution to Hong Kong MO 1998/1

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franzliszt

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franzliszt

#19 h Jul 25, 2022, 3:33 PM

Y by

lol easy problem

Observations:

-By Thales, $PR$ is the diameter of $(SPQR)$ .
- $HK$ is the Simson line of $Q$ wrt $\triangle PSR$ .
- $S$ is the orthocenter of $\triangle PSR$ .

Now we are done by EGMO Lemma 4.4 (which states "For a point $Q$ on the circumcircle of triangle $ABC$ , the Simson line of $Q$ bisects $QH$ .").

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YaoAOPS

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YaoAOPS

#20 h Jun 12, 2023, 8:33 PM

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Note that $HK$ is the simson line of $Q$ with respect to $\triangle PRS$ . Since $S$ is the orthocenter of the triangle, the desired result follows.

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peace09

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peace09

#21 h Jan 3, 2024, 7:00 PM

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Lol

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BestAOPS

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BestAOPS

#22 h Mar 12, 2024, 7:51 PM

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Let $F$ be the foot of the perpendicular from $Q$ to $SR$ . By Simson line, $K,H,F$ are collinear. Notice that $KQFS$ is a rectangle. Thus, the diagonals $FK$ and $QS$ bisect each other, which is the desired result.

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bodannikolov

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bodannikolov

#23 h Jun 19, 2024, 4:58 PM

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we can just use complex numbers (unit circle around pqrs obvi) let $p=1$ and $q=-1$ and let $m$ be defined as $(q+s)/2$ and $h$ is given by $\frac{1}{2}*(p+q+r-\frac{pr}{q})$ and $k$ as $\frac{1}{2}(s+p+q-\frac{sp}{q})$ or $h=\frac{1}{2}*(q+\frac{1}{q})$ and $k=\frac{1}{2}*(s+q+1-\frac{s}{q})$ plugging in collinearity formula we get $\frac{1-\frac{s}{q}}{1-\frac{q}{s}}=\frac{\frac{1}{q}-s}{q-\frac{1}{s}}$ or $\frac{s(q-s)}{q(s-q)}=\frac{1-\frac{s}{q}}{1-\frac{q}{s}}=\frac{\frac{1}{q}-s}{q-\frac{1}{s}}=\frac{(1-sq)*s}{(sq-1)*q}$

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iMqther_

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iMqther_

#25 h Sep 10, 2024, 6:06 PM

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Let $M=KH\cap SQ$ , $L=KM\cap PS$ and $O$ the circumcenter of $\triangle SPQ$ . Notice that $\angle QHR+\angle RKQ=180^{\circ} \implies HRKQ$ is cyclic therefore $\angle HRQ=\angle QKH$ but $\angle LSQ=\angle HKQ \implies \angle LSQ=LKQ$ so $LSQK$ is cyclic and consequently it's a rectangle so $QL$ is an altitude therefore $QL$ and $QO$ are isogonal conjugates which means $\angle PQL=\angle OQS$ . Note that $\angle LQR=90^{\circ}-\angle LQP$ and since $\angle LQK=90^{\circ} \implies \angle KQR=\angle LQP$ then $\angle RQK=\angle KHR$ but $\angle RHK=\angle MHO \implies \angle MQO=\angle MHO \implies MHQO$ is cyclic therefore $\angle OMQ=\angle QHO=90^{\circ}$ . Finally since $\triangle OSQ$ is isosceles we have that $OM$ is the perpendicular bisector of $SQ$ , finishing the proof.

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This post has been edited 2 times. Last edited by iMqther_, Sep 11, 2024, 5:14 AM
Reason: Typo

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LeYohan

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LeYohan

#26 h May 5, 2025, 2:34 AM

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We notice that $HK$ is the $Q-$ simson line of $\triangle SPR$ , so if we drop the $Z$ perpendicular foot from $Q$ to $SR$ , we know that $H - K - Z$ .
Since $\angle PSR = \angle QZS = 90 \implies QZ \parallel PS$ . Additionally, $\angle QPS = \angle ZSP = 90 \implies PQ \parallel SR$ .
This means that $PQZS$ is a rectangle, which is also a parallelogram, and the diagonals of a parallelogram bisect each other, implying the result. $\square$

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