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A nice problem. Olympic VietNam 2007-2008 (New).

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Source: Vietnam MO 2008, Problem 7

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#1 h Jan 29, 2008, 6:26 AM • 3 Y

Y by braveheartlong, Adventure10, Mango247

Let $AD$ is centroid of $ABC$ triangle. Let $(d)$ is the perpendicular line with $AD$ . Let $M$ is a point on $(d)$ . Let $E, F$ are midpoints of $MB, MC$ respectively. The line through point $E$ and perpendicular with $(d)$ meet $AB$ at $P$ . The line through point $F$ and perpendicular with $(d)$ meet $AC$ at $Q$ . Let $(d')$ is a line through point $M$ and perpendicular with $PQ$ . Prove $(d')$ always pass a fixed point.

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#2 h Jan 29, 2008, 11:16 AM • 2 Y

Y by Adventure10, Mango247

To: Thanhnam2902

Herebelow an easy approach to this nice problem:
Take B' and C' on (d) as the orthogonal projections of B and C respectively onto (d).
When M comes to B', P goes to B, and Q to A, while when it comes to C', P goes to A and Q to C, therefore the fixed point must be the common point of the perpendiculars from B' onto AB and from C' onto AC respectively, let it be F, and it's fixed.
We have now to show that, for any position of M on (d), the perpendicular from M to PQ, got as required, passes through F.
But there is a well-known theorem stating that: The perpendiculars from the vertices of a triangle onto the sides of another triangle concur iff the perpendiculars from the vertices of the latter onto the sides of the first one concur. (Carnot??).
For our case we take the

Quote:

triangle

B'MC' and the triangle APQ. Since the perpendiculars from A, P and Q onto (d) are parallel (we assume they concur somewhere), we get that the perpendiculars from B' to AP. C' to AC and M to PQ will be concurrent as well and we are done.

We can see from here that when (d) is moving, F describes a straight line, parallel to AD.

Best regards,
sunken rock

A blind man sees better the details

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#3 h Jan 29, 2008, 1:48 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

This is full picture of the problem:

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#4 h Feb 10, 2008, 11:17 AM • 2 Y

Y by Adventure10, Mango247

sunken rock wrote:

The perpendiculars from the vertices of a triangle onto the sides of another triangle concur iff the perpendiculars from the vertices of the latter onto the sides of the first one concur. (Carnot??).

Two triangles satisfying this property are called orthologic.
For more information, see here
http://www.mathlinks.ro/Forum/viewtopic.php?t=4337

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#5 h Mar 15, 2008, 8:59 AM • 5 Y

Y by Hantaehee, lmht, Trafalgar0246, Adventure10, Mango247

My solution:
Image not found
We will prove that $d';d_1;d_2$ are concurrent (*)
We have $B'A=C'A;B'P=MP;C'Q=MQ$ so
$B'A^2-B'P^2+MP^2-MQ^2+C'Q^2-C'A^2=0$
By Carnot's theorem with triangle $APQ$ we have $d';d_1;d_2$ are concurrent
Done!

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#6 h Aug 27, 2008, 9:28 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Problem stated by mr.danh:

Let ABC be a triangle. The line $d\parallel BC$ . A point M moves on $d$ . The perpendicular line from the midpoints of MB,MC to BC meets AB,AC at D,E respectively. Prove that the perpendicular line from M to DE passes through a fixed point when M moves on $d$ .

Approach by mr.danh:

My opinion:
D,E lie on a perp bisector of M'B,M'C where M' is the orthogonal projection of M on BC.
N is the point symmetric to M' w.r.t DE. Easy to show that E lie on the circumcircle (ABC), which easily gives us M'N passes through the fixed point F', which is the intersection of altitude AH and the circumcircle (ABC). Hence, if F is the point satisfying $\vec{F'F} = \vec{M'M}$ , F is the fixed point that the perp line from M to DE passes through.
The problem was proven.

Approach by Nemion:

What popped first to my mind was an analytic solution:
$A = (0,a)$ , $B = (b,0)$ $C = (c,0)$ $M = (t,m)$ with $m$ fixed. midpoints $P$ , $Q$ of $BM$ , $CM$ are: $P = (\frac {t + b}{2},\frac {m}{2})$ and $Q = (\frac {t + c}{2},\frac {m}{2})$ equation of $AB$ : $y = \frac {a}{ - b}x + a$ eq $AC$ : $y = \frac {a}{ - c} + a$ then $E = (\frac {t + b}{2},(\frac {a}{ - b})(\frac {t + b}{2}) + a))$ and $F = (\frac {t + c}{2},(\frac {a}{ - c})(\frac {t + c}{2}) + a))$ slope of $EF$ (Reducing terms and making the computations) is: $\frac {at}{bc}$ then the slope of the perpendicular to $EF$ trough $M$ is: $- \frac {bc}{at}$ then the equation of the perpendicular to $EF$ trough $M$ is: $y = - \frac {bc}{at}(x - t) + m$ the intersection of this line with the $y$ axis is: $y = m + \frac {bc}{a}$ (point: $(0,\frac {bc}{a} + m)$ ) but this last point doesn't depend on $t$ (but on $m$ ) therefore it is fixed for $m$ fixed.

Approach by yetti:

As long as $d \parallel BC,$ the line $DE$ does not depend on $d.$ Assume the proposition is true for the line $BC$ itself. Let $M' \in BC$ be the foot of normal from $M \in d,$ let $p \parallel p'$ be the perpendiculars from $M, M'$ to $DE,$ let $F'$ be the fixed point of $p'$ and let a parallel to $MM'$ through $F'$ cut $p$ at $F.$ $MM'F'F$ is a parallelogram, $FF' = MM'$ and $(FF' \parallel MM') \perp BC.$ Since $F' \in p'$ is fixed, $F \in p$ is also fixed. Therefore, it is sufficient to prove the proposition for $BC$ itself, hence assume $M \in BC.$

Let $MD, ME$ cut perpendiculars to $BC$ at $B, C$ at points $U, V.$ $D, E$ are midpoints of $MU, MV,$ $DE \parallel UV$ is a midline of the $\triangle MUV.$ Since $D, E$ are on perpendicular bisectors of $MB, MC,$ $\angle UMB = \angle B,$ $\angle VMC = \angle C.$ Parallels to $UM, VM$ through $C, B,$ respectively, intersect at the reflection $A'$ of $A$ in the perpendicular bisector of $BC.$ Define 3 points on the line at infinity: $X^* \equiv MU \cap CA',$ $Y^* \equiv MV \cap BA',$ $Z^* \equiv BU \cap CV.$ Consider the line $MBC$ and the line at infinity $X^*Y^*Z^*.$ By Pappus theorem, the intersections $U \equiv MX^* \cap BZ^*,$ $V \equiv MY^* \cap CZ^*,$ $A' \equiv CX^* \cap BY^*$ are collinear.

Let $W$ be the foot of a normal from $M$ to $UV,$ i.e., $MW \perp DE$ is the line supposedly passing through a fixed point. $MUWB$ is cyclic because of the right angles $\angle MBU, \angle MWU,$ hence $\angle A'WB = \angle UMB = \angle B.$ Since $A', B$ are on the circumcircle of the $\triangle ABC$ and $\angle A'CB = \angle ABC = \angle B,$ it follows that $W \in (O).$ Let $MW$ cut $(O)$ again at $F.$ Since the angle $\angle A'WF$ is right, $A'F$ is a diameter of $(O)$ from a fixed point $A',$ so that $F$ is also fixed. Obviously, the angle $\angle A'AF$ is also right and $F$ is the other intersection of the A-altitude of the $\triangle ABC$ with it circumcircle $(O).$

Approach by vittasko:

The orthopole $K$ of the triangle $\bigtriangleup ADE,$ with respect to the sideline $BC,$ lies on the midperpendicular of the segment $HA',$ where $H$ is the orthocenter of $\bigtriangleup ABC$ and $A'$ is the orthogonal projection of $A$ on $BC$ $($ $AA'$ = $A$ -altitude of $\bigtriangleup$ $ABC$ $).$

The line segment $HM',$ where $M'$ is the orthogonal projection of $M$ on $BC,$ always passes through the point $K.$

So, the line through the point $M$ and perpendicular to $DE$ as the problem states, as parallel to $DK,$ intersects the line segment $AA'$ at one point so be it $T,$ such that $A'T = HA' - MM'$ $($ in my drawing, the point $M$ is inwardly to $\bigtriangleup ABC$ and $MM' < HA'$ $)$ and so, it is a fixed point.

I will check my notes and if I am not mistaken, I will post here next time the details about the proof of this result.

Kostas Vittas.

Approach by vittasko:

Let $P,\ Q$ be, the midpoints of the segments $MB,\ MC$ respectively and we denote as $D'\ M'\ E',$ the orthogonal projections of $D,\ M,\ E$ respectively, on $BC.$

It is well known (and easy to prove by Carnot theorem ), that the lines through the points $D',\ A',\ E',$ where $A'$ is the orthogonal projection of $A$ on $BC$ $($ $AA',$ is the $A$ -altitude of $\bigtriangleup ABC$ $),$ are concurrent at one point so be it $K,$ as the orthopole of the triangle $\bigtriangleup ADE,$ with respect to the sideline $BC$ of $\bigtriangleup ABC.$

Because of now, $M'D' = D'B$ and $M'E' = E'C$ and $D'K\parallel BH$ and $E'K\parallel CH,$ we conclude that the points $M',\ K,\ H,$ are collinear and $M'K = KH$ $,(1)$

It is easy to show that the line through the point $M$ and perpendicular to $DE,$ as parallel to $A'K,$ intersects the line segment $AA',$ at one point so be it $T,$ such that $A'T = MZ,$ where $Z\equiv A'K\cap M'M,$ as the fourth vertex of the rectangle $HA'M'Z.$

It is clear that $T$ is a fixed point on the line segment $AA',$ because of $A'T = MZ = M'Z - M'M = HA' - M'M$ $,(2)$

Hence, the line through the point $M$ and perpendicular to $DE$ as the problem states, passes through the fixed point $T$ and the proof is completed.

Kostas Vittas.

PS. It is easy to show that the line through the point $M'$ and perpendicular to $DE,$ passes through the point $H',$ as the intersection point of the circumcircle of $\bigtriangleup ABC,$ from its altitude $AA'.$

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#7 h Aug 28, 2008, 11:09 AM • 2 Y

Y by Adventure10, Mango247

An extension of this problem

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#8 h Oct 8, 2019, 2:51 AM • 3 Y

Y by Tafi_ak, Adventure10, Mango247

Move $M$ linearly on $d$ , in terms of the variable $t$ denoting time. Observe that $P, Q$ also move linearly on $AB, AC$ respectively. As a result of the above, for a particular point $S$ in the plane, the function $f(t) = MP^2 - MQ^2 + SQ^2 - SP^2$ is a quadratic in $t.$ Hence, if it is zero for $3$ points, it is identically $0$ and we'd be done.

We therefore just need to show that the lines $d'$ pass three a fixed point for some three choices of $M$ . We will take the feet of the altitude of $B, D, C$ respectively to $d$ as our three choices. It's easy to check that these three choices of $M$ indeed lead to an intersection point for $S$ .

$\square$

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