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6 geometry problems out of 8 at our TST so far..

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Source: Moldova 2007 IMO-BMO TST II problem 3

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481 posts

iura

#1 h Mar 23, 2007, 2:55 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $M, N$ be points inside the angle $\angle BAC$ usch that $\angle MAB\equiv \angle NAC$ . If $M_{1}, M_{2}$ and $N_{1}, N_{2}$ are the projections of $M$ and $N$ on $AB, AC$ respectively then prove that $M, N$ and $P$ the intersection of $M_{1}N_{2}$ with $N_{1}M_{2}$ are collinear.

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1198 posts

Megus

#2 h Mar 24, 2007, 10:06 AM • 2 Y

Y by Adventure10, Mango247

It's just an outline of a proof: Let $E$ be intersection of $M_{1}M$ and $N_{2}N$ and $F$ be intersection of $MM_{2}$ and $NN_{1}$ . Angle chase to get that $M_{1}M_{2}N_{1}N_{2}$ is cyclic then prove that $E$ and $F$ are isogonally conjugated. To prove that $M,N,P$ are collinear use Menelaos for triangle $EM_{1}N_{2}$ . To count wanted ratios use Menelaos for triangle $M_{1}N_{2}A$ and line $N_{1}M_{2}P$ , power of point ( $AN_{2}\cdot AM_{2}=AN_{1}\cdot AM_{1}$ ) and sine theorem to get $\frac{EM}{MM_{1}}=\frac{AE sin \angle{EAM}}{AM_{1}sin \angle{MAM_{1}}}$ and $\frac{EN}{NN_{2}}=\frac{AE sin \angle{EAN}}{AN_{2}sin \angle{NAN_{2}}}$ .

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312 posts

#3 h Apr 2, 2007, 8:16 AM • 2 Y

Y by Adventure10, Mango247

First of all let us prove that
$X_{1}X_{2}Y_{1}Y_{2}$ is concyclic. Observe that $AX_{1}UX_{2}$ and $AY_{1}VY_{2}$
are concyclic, it follows that

$\angle X_{2}X_{1}A = \angle X_{2}UA = 90^{\circ}-\angle BAU ; \angle AY_{2}Y_{1}=\angle AVY_{1}=90^{\circ}-\angle CAV.$

Therefore $\angle X_{2}X_{1}Y_{1}+\angle X_{2}Y_{2}Y_{1}= 180^{\circ}, \text{ }\Rightarrow$ $X_{1}X_{2}Y_{1}Y_{2}$ is
concyclic.

Second thing that we want to prove is that $O$ , the
circumcenter of $X_{1}X_{2}Y_{1}Y_{2}$ lies on $UV$ . Indeed, circumcenter
lies on the intersections of perpendiculars to the angle sides in
the midpoints of $X_{1}Y_{1}$ and $X_{2}Y_{2}$ . As $UX_{1}Y_{1}V$ and $UX_{2}Y_{2}V$
are trapezoids, we get these perpendiculars intersect in the
midpoint of $UV$ .

Let $M$ be point of intersection $X_{1}X_{2}$ and $Y_{1}Y_{2}$ .
Observe that $M$ is the radical center of $AX_{1}UX_{2}$ , $AY_{1}VY_{2}$ ,
and $X_{1}X_{2}Y_{1}Y_{2}$ . It follows that $AM$ is the radical axe for
$AX_{1}UX_{2}$ , $AY_{1}VY_{2}$ and is perpendicular to the center line. As
$AU, AV$ are diameters, we get that $AM$ is perpendicular to the
midline of $AUV$ . Hence $AM \perp UV$ .

Using the properties of projective geometry we have

(a) $AW$ is a polar of $M$ with respect to the circle
$X_{1}X_{2}Y_{1}Y_{2}$ .

(b) $MW$ s a polar of $A$ with respect to the circle
$X_{1}X_{2}Y_{1}Y_{2}$ . \newline Thus $AM$ is polar of $W$ with respect to
the circle $X_{1}X_{2}Y_{1}Y_{2}, \text{ }\Rightarrow AM \text{ }\perp WO$ .

Finally, adding the above facts we get that $W \in UV$ .

Attachments:

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246 posts

radio

#4 h Apr 2, 2007, 4:03 PM • 2 Y

Y by Adventure10, Mango247

Megus wrote:

then prove that $E$ and $F$ are isogonally conjugated.

How to prove it?

Also, in the end I'm left to prove $\frac{M_{2}N_{2}}{M_{1}N_{1}}=\frac{\sin FAN}{\sin EAN}$ . But I don't know how it is true.

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552 posts

TomciO

#5 h Apr 3, 2007, 11:38 PM • 2 Y

Y by Adventure10, Mango247

Hey, I think that after proving the concyclity you can end the problem easily with using Pascal two times directly (proving that $U, W, O$ and $W, O, V$ are collinear). You've proved Pascal Theorem for this special case probably

.

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246 posts

radio

#6 h Apr 4, 2007, 1:45 PM • 1 Y

Y by Adventure10

TomciO wrote:

Hey, I think that after proving the concyclity you can end the problem easily with using Pascal two times directly (proving that $U, W, O$ and $W, O, V$ are collinear). You've proved Pascal Theorem for this special case probably

.

Pascal how? I see only four points on the circle, with no tangents.

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552 posts

TomciO

#7 h Apr 4, 2007, 3:09 PM • 2 Y

Y by Adventure10, Mango247

Ok, it's almost direct application of Pascal. Just extend the lines $UX_{1}$ and $UX_{2}$ to meet the circle again at $X_{1}'$ and $X_{2}'$ respectively. Then notice that $O$ is a intersection of $X_{1}'Y_{1}$ and $X_{2}'Y_{2}$ . Is it clear now

?

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7054 posts

#8 h Apr 4, 2007, 5:34 PM • 2 Y

Y by Adventure10, Mango247

An interesting generalization.

In $\triangle ABC$ let $\{M,N\}\subset (BC)$ so that $M\in (BN)$ , $\widehat{BAM}\equiv\widehat{CAN}$ and

$\{\begin{array}{ccc}\{M_{1},N_{1}\}\subset (AB) & , & \frac{BM_{1}}{BM}=\frac{M_{1}N_{1}}{MN}=m\\\\ \{M_{2},N_{2}\}\subset (AC) & , & \frac{CN_{2}}{CN}=\frac{N_{2}M_{2}}{NM}=n\end{array}$ .

For $P\in M_{1}N_{2}\cap M_{2}N_{1}$ prove that $\boxed{P\in BC\Longleftrightarrow nb-mc=\frac{b^{2}-c^{2}}{a}}$ .

Particularly, for $\{\begin{array}{c}m=\cos B\\\ n=\cos C\end{array}$ , i.e. $\{\begin{array}{ccc}MM_{1}\perp AB & , & MM_{2}\perp AC\\\ NN_{1}\perp AB & , & NN_{2}\perp AC\end{array}$ obtain

the proposed problem : $P\in BC$ $\Longleftrightarrow$ $b^{2}-c^{2}=a(b\cos C-c\cos B)$ (true !).

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1327 posts

#9 h Apr 4, 2007, 8:43 PM • 2 Y

Y by Adventure10, Mango247

Because of the quadrilateral $X_{1}X_{2}Y_{1}X_{2}$ is cyclic $($ ''prowler’s'' wording $),$ we can also complete the solution, based on the well known theorem which it has already been proved in the last replies at

http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

Kostas Vittas.

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312 posts

#10 h Apr 5, 2007, 10:30 PM • 2 Y

Y by Adventure10, Mango247

Your ideas are all nice.
Sincerely I proposed this problem for Moldavian TST.
I stole it from the fact that Pascal's line for projections of two
isogonal points on triangle's sides passes through these two points.

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6555 posts

#11 h Apr 6, 2007, 10:14 AM • 2 Y

Y by Adventure10, Mango247

Actually, we had something related to this on ML three years ago:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=13816#13816]http://www.mathlinks.ro/Forum/viewtopic.php?t=4322 post #5[/url] wrote:

Let the lines YZ' and ZY' meet at R. [...] so that the point R lies on PQ.

Here, P and Q are two isogonally conjugate points with respect to a triangle ABC, but the idea of the proof works for any angle BAC and two points M and N satisfying < MAB = < CAN (with directed angles).

darij

This post has been edited 1 time. Last edited by darij grinberg, Jul 3, 2021, 6:08 PM
Reason: Hereby -> Here

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7054 posts

#12 h Apr 7, 2007, 9:35 PM • 2 Y

Y by Adventure10, Mango247

An interesting generalization.

In $\triangle ABC$ let $\{M,N\}\subset (BC)$ so that $M\in (BN)$ , $\widehat{BAM}\equiv\widehat{CAN}$ , $E\in (AB)$ , $F\in (AC)$ and

$\{\begin{array}{ccc}\{M_{1},N_{1}\}\subset (AB) & , & MM_{1}\parallel NN_{1}\parallel CE\\\\ \{M_{2},N_{2}\}\subset (AC) & , & MM_{2}\parallel NN_{2}\parallel BF\end{array}$ .

For $P\in M_{1}N_{2}\cap M_{2}N_{1}$ prove that $\boxed{P\in BC\Longleftrightarrow AE\cdot AB=AF\cdot AC}$ .

Particular cases. Denote the intersection $X\in BF\cap CE$ .

$1.\blacktriangleright$ For $X: =H$ - the orthocenter of $\triangle ABC$ obtain the proposed problem : $P\in BC$ $\Longleftrightarrow$ $c\cdot b\cos A=b\cdot c\cos A$ (true !).

$2.\blacktriangleright$ For $X: =A$ obtain a new problem : $\{\begin{array}{c}MM_{1}\parallel NN_{1}\parallel AC\\\\ NN_{2}\parallel MM_{2}\parallel AB\\\\ P\in M_{1}N_{2}\cap M_{2}N_{1}\end{array}\|$ $\Longrightarrow$ $P\in BC$ .

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