Reflection about Euler line

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Reflection about Euler line

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Source: INMO 2010 Problem 5

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#1 h Jan 18, 2010, 10:56 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle with altitude $AK$ . Let $H$ be its ortho-centre and $O$ be its circum-centre. Suppose $KOH$ is an acute-angled triangle and $P$ its circum-centre. Let $Q$ be the reflection of $P$ in the line $HO$ . Show that $Q$ lies on the line joining the mid-points of $AB$ and $AC$ .

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gilcu3

#2 h Jan 18, 2010, 3:24 PM • 3 Y

Y by Sx763_, Adventure10, and 1 other user

Let $R$ the center of the nine points circle. Let $Q_1$ , $P_1$ and $R_1$ the projections of $Q$ , $P$ and $R$ on the altitude $AK$ . Then for do the problem we have to prove that $Q_1$ is the midpoint of $AK$ . But is easy to prove that $Q_1P_1=2P_1R_1$ .

So:

$Q_1K=Q_1P_1+P_1K=2P_1R_1+P_1K=2(R_1K-P_1K)+P_1K=2R_1K-P_1K=\frac{1}{2}AH+HK-P_1K=\frac{1}{2}AK$ as claimed.

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#3 h Jan 19, 2010, 6:01 PM • 1 Y

Y by Adventure10

By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?

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#4 h Jan 21, 2010, 4:25 AM • 6 Y

Y by Wizard_32, SHREYAS333, Adventure10, Mango247, Math_DM, and 1 other user

Soution 1 (own)

Solution 2 (Due to Akashnil)

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madhusudan kale

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madhusudan kale

#5 h Jan 24, 2010, 3:47 PM • 2 Y

Y by Adventure10, Mango247

i have a very easy solution using pure geometry

Let ABC be an acute-angled triangle with altitude AK. Let H be its ortho-centre and O be its circum-centre. Suppose KOH is an acute-angled triangle and P its circum-centre. Let Q be the reflection of P in the line HO.
it is very easy to prove that
quadrilateral HPOQ is a rhombus.

Thus now consider the \triangle HKO
in this triangle
$\angle HPO = 2\angle HKO$ or $1/2\angle HPO = \angle HKO = \angle PQO ------------------(1)$
since P is the circumcenter and if we draw the circle we will get that \angle HPO is subtended at the center and \angle HKO is subtended at any other point by the same arc .

now we know that AK\perp BC since it is the altitude and $UN\perp BC$ thus $UN \parallel AK$ .
hence $\angle HKO = \angle KOT$

but $1/2\angle HKO = \angle HPO$
hence $1/2\angle HPO = \angle KON$
since HPOQ is a rhombus ,
$\angle HQP = \angle PQO$

we get $\angle PQO = \angle KON$ OR $\angle MQO = \angle KOT$
consider $\triangle KOT$ and $\triangle OQM$
$\angle MQO = \angle KON$
and $\angle QMO = \angle OTK$
thus $\triangle KOT$ and $\triangle OQM$ are similar so we get
$\angle OKT= \angle QOM$

now since UT is a straight line
$\angle QOX + \angle QOM + \angle HOK + \angle KOT=180$
$\angle QOX + \angle OKT + \angle HOK + \angle KOT=180$ since $\angle OKT= \angle QOM$
$\angle QOX + \angle HOK +90=180$
$\angle QOX + \angle HOK=90$
but $\angle QOX=90-\angle OQX$
thus $\angle OQX= \angle HOK$ $--------------(2)$
now consider $\triangle HLN$ and $\triangle NMQ$
$\angle HLN= \angle NMQ$
and $\angle HNL= \angle NQM$
thus $\triangle HLN$ and $\triangle NMQ$ are similar.
hence $\angle LHN=\angle NQM$
but $\angle LHN and \angle KOH$ are same.
hence we get $\angle KOH= \angle NQM=\angle RQP$
we know that in a triangle sum of all angles is 180.
Thus in $\triangle KOH$ ,

$\angle HKO + \angle KOH + \angle KHO=180.$
or $\angle PQO + \angle RQP + \angle OQX =180$ from (1) and (2),
this implies that RQX is a straight line or point Q lies on RS.

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#6 h Jan 24, 2010, 5:29 PM • 5 Y

Y by div5252, thewitness, mathetillica, Understandingmathematics, Adventure10

keyree10 wrote:

Let $ABC$ be an acute-angled triangle with altitude $AK$ . Let $H$ be its ortho-centre and $O$ be its circum-centre. Suppose $KOH$ is an acute-angled triangle and $P$ its circum-centre. Let $Q$ be the reflection of $P$ in the line $HO$ . Show that $Q$ lies on the line joining the mid-points of $AB$ and $AC$ .

Let $a,b,c,h,p,q,z_O,k$ be the complex numbers corresponding to the points $A,B,C,H,P,Q,O,K$ in that order.

Wlog assume that $z_O = 0, |a| = |b| = |c| = 1$ . Then $h = 3(g - z_O) + z_O = a + b + c$ . $k = \frac {1}{2} \left ( a + b + c - \overline{a}bc \right )$ is easy to verify.

A point $Z(z)$ lies on the line joining the midpoints of $AB$ and $AC$ iff $\frac {z - \frac {a + b}{2}}{\overline{z - \frac {a + b}{2}}} = \frac {\frac {a + c}{2} - \frac {a + b}{2}}{\overline{\frac {a + c}{2} - \frac {a + b}{2}}} \iff$
$2z + 2bc\overline{z} = a + b + c + \overline{a}bc$ . $bc = \frac {h - k}{\overline{h - k}}$ and $a + b + c + \overline{a}bc = 2h - 2k$ , so:
$\iff z + \frac {h - k}{\overline{h - k}}\overline{z} = h - k$

$P$ is the orthocenter of $\triangle OKH$ so:
$p = \frac {hk(\overline{h - k})}{\overline{h}k - h\overline{k}}$

$Q$ is obtained by the reflection of $P$ in $OH$ , so:
$q = \overline{p} \frac {h}{\overline{h}} = \frac { - h\overline{k}(h - k)}{\overline{h}k - h\overline{k}}$

Inserting it is obvious that:
$q + \frac {h - k}{\overline{h - k}}\overline{q} = (h - k) \frac {\overline{h}k - h\overline{k}}{\overline{h}k - h\overline{k}} = h - k$ , and therefore $Q$ lies on the line joining the midpoints of $AB$ and $AC$ , and we are done

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#7 h Jan 25, 2010, 7:50 AM • 2 Y

Y by Adventure10, Mango247

keyree10 wrote:

Let $ABC$ be an acute-angled triangle with altitude $AK$ . Let $H$ be its ortho-centre and $O$ be its circum-centre. Suppose $KOH$ is an acute-angled triangle and $P$ its circum-centre. Let $Q$ be the reflection of $P$ in the line $HO$ . Show that $Q$ lies on the line joining the mid-points of $AB$ and $AC$ .

Nice Problem!
Let the midpoints of $AB,CD$ be $Q,R$ respectively, and let the reflection of $P$ wrt $OH$ be $P'$ .
We consider a half turn wrt $T$ , the midpoint of $OH$ (and the center of the nine point circle of $\triangle ABC$ ). This maps $P' \to P$ , $QR \to Q'R'$ .
It is enough to prove that $P$ lies on $Q'R'$ .
We know that $Q'R' \parallel CB$ . So $Q'R'$ intersect $HK$ at the midpoint, and also $HK \perp Q'R'$ . So $P$ lies on $Q'R'$ .

groshanlal wrote:

By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?

Though I have not ever took part in InMO (I am from Bangladesh), I believe that it depends on many facts, such as: the quality of the contestants, difficulty level of the problems etc. In general, in Olympiads the contestants who solve half of the problem gets some sort of awards. I don't know if it true for InMO.

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#8 h Jan 29, 2010, 12:45 PM • 2 Y

Y by Adventure10, Mango247

My goodness - Dr Rijul has posted his solution already.

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Shravu

#9 h Jan 20, 2012, 12:26 PM • 2 Y

Y by Adventure10, Mango247

we have not used that triangle is acute in rijul's solution

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#10 h Jan 25, 2022, 2:01 PM

Y by

Let E be midpoint of AC and D be midpoint of BH. Note that BH = 2ON so DH = ON and we also have DH || ON because they're both perpendicular to AC so DHEO is parallelogram. Let S be midpoint of OH we know S is midpoint of ED and OH and PQ so DQEP is parallelogram so EQ || PD || BC so Q lies on the line made of midpoints of AC and AB.
we're Done.

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