Two parallel lines

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Source: Iranian third round 2015 geometry problem 2

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824 posts

andria

#1 h Sep 10, 2015, 11:41 AM • 2 Y

Y by mathmaths, Adventure10

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Let $K$ be the midpoint of $AH$ . point $P$ lies on $AC$ such that $\angle BKP=90^{\circ}$ . Prove that $OP\parallel BC$ .

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2312 posts

#2 h Sep 10, 2015, 11:56 AM • 3 Y

Y by enhanced, Adventure10, Mango247

This problem is a particular case of the problem Perpendicular line

.

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9775 posts

jayme

#3 h Sep 11, 2015, 9:58 AM • 2 Y

Y by Infinityfun, Adventure10

Dear Mathlinkers,

1. E the foot of the B-altitude of ABC
B' the circumtrace of BE,
A'', B'' the antipoles of A, B wrt (O),
(1) the circle with diameter BP
A' the second point of intersection of (1) with (O).
2. by considering two time a converse of the Reim theorem, B''P goes through A', then AH goes through A'
3. By the Pascal theorem, we are done...

Sincerely
Jean-Louis

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228 posts

ATimo

#4 h Sep 13, 2015, 7:17 PM • 7 Y

Y by Mehdijahani1998, K.N, Dilshodbek, gemcl, hakN, Adventure10, Mango247

Let $P$ be the point on $AC$ such that $OP\parallel BC$ , we will prove that $\angle PKB=90$ . Let $M$ be the foot of the perpendicular line from $P$ to $BC$ . Suppose that $N$ is the midpoint of $BC$ . Then we have $PM=ON=AK=KH$ . $AH\parallel PM$ so $APMK$ and $KPMH$ are parallelograms. So $MK\parallel AC$ and $MH\parallel PK$ . So we have to say that $MH$ is perpendicular to $BK$ . $MK\parallel AC$ , so $BH$ is perpendicular to $MK$ . $KH$ is also perpendicular to $BM$ , So $H$ is the orthocenter of the triangle $\triangle BKM$ . And we are done.

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42 posts

#5 h Sep 14, 2015, 10:45 AM • 3 Y

Y by Dilshodbek, Adventure10, Mango247

My solution:

Let $AD,BE$ are the altitudes of $\triangle ABC.$
Easy to see that $BKEP$ is cyclic.
$\Longrightarrow \angle KBP =\angle AEK =\angle KAE =\angle EBC \Longrightarrow \angle KBE =\angle PBC .$
We have $\angle OBP =\angle OBC -\angle PBC = 90^{o} - \angle BAC -\angle KBE =\angle ABE - \angle KBE =\angle ABK (1).$
In the other hand, $\angle BAK = \angle PAO.(2)$ (well-known).
From $(1),(2)$ we deduce $O, K$ is isogonal conjugate WRT $\triangle ABP.$
$\Longrightarrow \angle BPO = \angle APK = \angle KBE = \angle PBC \Longrightarrow OP \parallel BC.$ Done

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670 posts

#6 h Sep 14, 2015, 5:54 PM • 2 Y

Y by Adventure10, Mango247

Solution:Let $B'$ be the symmetric point of $B$ wrt point $K$ .We shall start with the following lemma:

Lemma: $m(\measuredangle{KBP})=90^0-m(\measuredangle{ACB})$

Proof of the lemma:Quadrilateral $ABHB'$ is parallelogram,so $m(\measuredangle{HAB'})=m(\measuredangle{BHA})=180^0-m(\measuredangle{ACB})$ and since $m(\measuredangle{HAC})=90^0-m(\measuredangle{ACB})$ ,it results that $B'A\perp AC$ .Hence, $m(\measuredangle{B'AP})=90^0=m(\measuredangle{B'KP})$ ,resulting that quadrilateral $AKPB'$ is concyclic.
Therefore $m(\measuredangle{KBP})=m(\measuredangle{KB'P})=m(\measuredangle{KAP})=90^0-m(\measuredangle{ACB})$ ,which ends the proof. $\blacksquare$

Back to the main problem,using the above lemma we obtain $m(\measuredangle{KBP})=90^0-m(\measuredangle{ACB})=m(\measuredangle{ABO})$ ,which implies $\measuredangle{ABB'}\equiv\measuredangle{OBP}(\bigstar)$ .
But,let's observe that $\triangle{AOB}\sim\triangle{B'PB}(\text{case A.A.})$ .This helps us see that $\frac{AB}{BO}=\frac{BB'}{BP}(\bigstar\bigstar)$ .
Finally,from $(\bigstar)$ and $(\bigstar\bigstar)$ we get $\triangle{BAB'}\sim\triangle{BOP}$ .So, $m(\measuredangle{BOP})=m(\measuredangle{BAB'})=90^0+m(\measuredangle{BAC})=180^0-m(\measuredangle{OBC})$ ,following that $OP\parallel BC$ ,which is what we wanted to prove.

This post has been edited 1 time. Last edited by huricane, Sep 14, 2015, 6:14 PM
Reason: no need for point L...

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243 posts

#7 h Mar 27, 2016, 5:24 PM • 3 Y

Y by Adventure10, Mango247, shafikbara48593762

This is nice problem but not too hard to complex bash: $k=a+\frac {b+c}{2}$ ,because $P$ is on chord $AB$ , $\overline{p}=\frac {a+c-p}{ac}$ , now because $BK$ perpendicular to $KP$ so $\frac{b-k}{\overline{b}-\overline{k}}= - \frac{p-k}{\overline{p}-\overline{k}}$ so after some calculations (about 15 minutes) we get $p=\frac{b^2c+abc+b^2a-c^2b}{b^2+bc-ac-ab}$ and finaly aply $p$ in formula for $OP || BC$ : $\frac{p}{\overline{p}}=\frac{b-c}{\overline{b}-\overline{c}}$ and result follows (I only needed 30 minutes for this bash ).

This post has been edited 1 time. Last edited by Garfield, Mar 27, 2016, 5:24 PM

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111 posts

#8 h Jul 31, 2018, 7:14 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

Let BH cuts (O) at E, midpoint of AD is F. We observe that BDK+KDP=HAC+BCA=90 so BDP=90. Because HGP=HFP=90 then FHPG is cylic therefore AG.AP=AF.AH=AE.AK. Note that BDP=BEP=90 so BEPD is cyclic too and K lies on (DPE) then K lies on (BEP). So BKP=BEP=90, q.e.d

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3042 posts

#9 h Dec 25, 2018, 5:35 AM • 1 Y

Y by Adventure10

andria wrote:

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Let $K$ be the midpoint of $AH$ . point $P$ lies on $AC$ such that $\angle BKP=90^{\circ}$ . Prove that $OP\parallel BC$ .

Solution: Let $M$ be the midpoint of $BC$ , $X$ be the feet from $P$ on $BC$ and $E$ be the feet from $B$ on $AC$ . We have $B,K,P,E,X$ are concyclic with $BP$ as diameter. Therefore $\angle PXK = \angle PBK = \angle AEK = \angle KAE.$ Now as $AK||PX$ , we have $AKXP$ is parallelogram or $PX = AK = OM$ and therefore $POMX$ is a rectangle and we are done. $\square$

This post has been edited 1 time. Last edited by TheDarkPrince, Dec 25, 2018, 5:35 AM

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816 posts

#10 h Mar 11, 2019, 9:29 AM • 1 Y

Y by Adventure10

We make use of complex numbers.Assume without loss of generality that $ABC$ is the unit circle and let the coordinates of $X$ be $x$ respectivly.It is well-known that $h=a+b+c$ and so $k=\frac{2a+b+c}{2}$ .Assume that $P$ is the intersection of line passing throught $O$ and parallel to $BC$ with $AC$ then we have:

$P$ lies on a line parallel to $BS$ passing throught $O$ so $\frac{p}{\overline{p}}=-bc$ .

$P$ lies on $AC$ so $\frac{a-p}{\overline{a}-\overline{p}}=-ac$

Then we have $P=\frac{b(a+c)}{a+b}$ Then it is straightforward calculation to show $PK$ is perpendicular to $BK$ .

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2125 posts

#11 h Sep 29, 2019, 7:24 PM • 2 Y

Y by Adventure10, Mango247

Iran Round 3 2015 G2 wrote:

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Let $K$ be the midpoint of $AH$ . point $P$ lies on $AC$ such that $\angle BKP=90^{\circ}$ . Prove that $OP\parallel BC$ .

Solution: Let $H',H''$ be reflection of $H$ over $\overline{BC}$ , $\overline{AC}$ . Hence, $KE||AH''$ . By Converse of Reim's Theorem, $BH'PEK$ is cyclic,
$\angle PAH'=\angle KEA=\angle AH'P \implies AP=PH' \implies OP ||BC \qquad \blacksquare$

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214 posts

#12 h Oct 9, 2020, 6:38 AM

Y by

a cute problem
let $P$ the point on $AC$ such that $OP \parallel BC$ we will prove that $PK \perp KB$
let $M,N$ be the projections from $O,P$ onto $BC$
since $\vec{PN}=\vec{OM}=\frac{1}{2}\vec{AH}=\vec{AK}$ we have $AKNP$ is parallelogram
so $KN \parallel AC \implies BH \perp KN$ so $H$ is the orthocenter of $\triangle BKN$
so $NH \perp BK$
and we win

This post has been edited 3 times. Last edited by Ali3085, Oct 9, 2020, 6:40 AM

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23531 posts

#13 h Apr 8, 2021, 5:23 PM

Y by

Rename $K$ as $P$ and let a line through $O$ parallel to $BC$ meet $AC$ at $F$ . We will show that $F=P$ by proving $\angle CMF=90^\circ$ .

We use complex numbers with $(ABC)$ as the unit circle and $OF$ as the real line. Observe that $B$ and $C$ are reflections across the imaginary line so $bc=-1$ .

Clearly we have $M=\frac{A+H}2=\frac{a+a+b+c}2=\frac{2a+b+c}2$ . By Lemma 8, we have $F=\frac{a+b}{ab+1}$ and clearly $C=c$ . Their conjugates are $\overline{M}=\frac{\frac2a+\frac1b+\frac1b}{2},\overline{F}=F=\frac{a+b}{ab+1},\overline{C}=\frac1c$ , respectively.

It remains to check that $\frac{m-f}{m-c}+\overline{\left(\frac{m-c}{m-f}\right)}=0$ which can be rewritten as $\frac{m-f}{\overline{m}-\overline{f}}+\frac{m-c}{\overline{m}-\overline{c}}=0$ . Upon plugging in the values and substituting $c=-\frac1b$ , we now wish to show $\frac{\frac{2a+b-1/b}2-\frac{a+b}{ab+1}}{\frac{2/a+1/b-b}2-\frac{a+b}{ab+1}}+\frac{\frac{2a+b-1/b}2-\left(-\frac1b\right)}{\frac{2/a+1/b-b}2-(-b)}=0$ which happens to be true upon expansion. $\blacksquare$

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#14 h Feb 12, 2022, 7:24 AM

Y by

Let $PS$ be perpendicular to $BC$ . It's well known that $PS = AM/2$ and $PS || AH$ so $PAKS$ and $SPKH$ are parallelogram.
we know $BH$ is perpendicular to $AC$ so $BH$ is also perpendicular to $KS$ . $KH$ is perpendicular to $BS$ so $H$ is orthocenter of $BKS$ so $SH$ is perpendicular to $BK$ so $PK$ is also perpendicular to $BK$ as wanted.
we're Done.

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#15 h Aug 27, 2022, 6:41 AM

Y by

let circumcircle of $ABC$ is the unit circle and $B,C$ are relative to the width axis so :
$A=a,B=b,C=-\bar{b},H=a+b-\frac{1}{b},K=\frac{1}{2}(2a+b+\frac{1}{b})$
let $X$ be a point on $AB$ such that $OX||BC \Rightarrow x=\bar{x}$ and $\frac{x-a}{x-b}=\frac{x-\frac{1}{a}}{x-\frac{1}{b}}$ so $x=\frac{a+b}{ab+1}$
now we show that $XK \perp KC \Longleftrightarrow \frac{k-x}{k+\frac{1}{b}} + \frac{\bar{k}-x}{\bar{k}+b} = 0$

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