Polish MO Finals 2015, Problem 5

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Polish MO Finals 2015, Problem 5

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Source: Polish MO Finals 2015

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#1 h Jul 27, 2016, 10:42 PM • 2 Y

Y by Adventure10, Mango247

Prove that diagonals of a convex quadrilateral are perpendicular if and only if inside of the quadrilateral there is a point, whose orthogonal projections on sides of the quadrilateral are vertices of a rectangle.

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#2 h Jul 28, 2016, 4:34 PM • 2 Y

Y by Adventure10, Mango247

We can bash this problem with applications of Pythagorean theorem and Law of Cosines, but here is a nicer way and more instructive

Let $ABCD$ be the quadrilateral. Let $AB\cap CD=Y, AD \cap BC = Z, AC \cap BD = X$ . First suppose that the diagonals are perpendicular. Let the perpendicular from $X$ to $AB$ meet $AB, CD$ at $M, M'$ . Let the perpendicular from $X$ to $BC$ meet $BC, DA$ at $N, N'$ . Let the perpendicular from $X$ to $CD$ meet $CD, AB$ at $P,P'$ . Let the perpendicular from $X$ to $DA$ meet $DA,BC$ at $Q,Q'$ .

Lemma: $MNPQM'N'P'Q'$ is cyclic.
Proof: Angle chase. First, note $\angle XQP = \angle XDP = \angle PXC=\angle PNC$ , thus $QPNQ'$ is cyclic. Similarly we know $QPMQ'$ is cyclic, thus $Q' \in (MNPQ)$ . Similarly we know $M',N',P' \in (MNPQ)$ as desired.

It's clear that $M'N'P'Q'$ is a rectangle because $M'P',N'Q'$ are both diameters of $(MNPQ)$ .

Now, let $X'$ be a point with $X'P' \perp AB, X'M' \perp CD$ . From the properties of the pedal circles of isogonal conjugates, it's easy to deduce that $X,X'$ are isogonal conjugates w.r.t. $\triangle YAD$ , thus $X'N' \perp AD$ . Similarly by looking at $\triangle YBC$ we know that $X'Q' \perp BC$ , thus $X'$ is a point so that its projections onto the sides of $ABCD$ form a rectangle, namely $M'N'P'Q'$ .

Next, we suppose we are given the quadrilateral $ABCD$ , a point $X'$ with the property that its projections on the sides form a rectangle, and let the projections be $M',N',P',Q'$ similar to earlier (so $P'\in AB, Q'\in BC, M'\in CD, N' \in DA$ ). Define $Y,Z$ the same as before. Let $(M'N'P'Q')$ cut $AB,BC,CD,DA$ again at $M,N,P,Q$ . Similarly to before, let $X$ be a point so $XM\perp AB, XP \perp CD$ . Then we know $X,X'$ are isogonal conjugates w.r.t. $\triangle YAD$ , thus $XQ \perp AD$ , and similarly by looking at $\triangle YBC$ we know $XN \perp BC$ .

Now we angle chase: $\angle DXC = \angle DXP + \angle CXP = \angle DQP + \angle CNP$ . This in turn equals $\angle DM'N' + \angle CM'Q' = 180 - \angle N'M'Q' = 90$ . Therefore $X \in (CD)$ . Similarly, $X \in (AD)$ so $\angle AXD + \angle CXD = 180 \implies X\in AC$ and similarly we know $X\in BD$ . Then it's easy to see $AC,BD$ are perpendicular at $X$ , so we are done.

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#3 h Jul 31, 2016, 2:34 PM • 2 Y

Y by guptaamitu1, Adventure10

See ISL 2008 G6-it's not too hard to see these are equivalent.

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#4 h Mar 10, 2017, 4:00 AM • 3 Y

Y by guptaamitu1, Adventure10, Mango247

Using a well fact can prove the problem fast:
A point $P$ has a isogonal conjugation wrt $ABCD$ point if and only if $\measuredangle APD=\measuredangle BPC$

This post has been edited 3 times. Last edited by ltf0501, Mar 10, 2017, 12:58 PM

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