quadrilateral plane geometry

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Source: Romanian Nationals RMO 2005 - grade 7, problem 3

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Valentin Vornicu

7301 posts

Valentin Vornicu

#1 h Mar 31, 2005, 10:13 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a quadrilateral with $AB\parallel CD$ and $AC \perp BD$ . Let $O$ be the intersection of $AC$ and $BD$ . On the rays $(OA$ and $(OB$ we consider the points $M$ and $N$ respectively such that $\angle ANC = \angle BMD = 90^\circ$ . We denote with $E$ the midpoint of the segment $MN$ . Prove that

a) $\triangle OMN \sim \triangle OBA$ ;
b) $OE \perp AB$ .

Claudiu-Stefan Popa

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549 posts

mna851

#2 h Apr 1, 2005, 12:52 AM • 2 Y

Y by Adventure10, Mango247

This is unrelated, but does anyone else find it odd that advanced American math students struggle with problems 7th grade Romanian students can solve.

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probability1.01

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probability1.01

#3 h Apr 1, 2005, 1:11 AM • 2 Y

Y by Adventure10, Mango247

Awww... it's not that bad... but way harder than our MathCounts, and that includes 8th graders.

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#4 h Apr 1, 2005, 1:15 AM • 2 Y

Y by Adventure10, Mango247

mna851 wrote:

This is unrelated, but does anyone else find it odd that advanced American math students struggle with problems 7th grade Romanian students can solve.

Unrelated too, I also noticed that, then, I am trying to discuss about that in this thread, if you want to give your opinion

. By the way, Valentin (and also any Romanian who is either a 12th grader or already in college), I would like to know your opinion about the thread topic I am mentioning!

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549 posts

mna851

#5 h Apr 2, 2005, 4:16 AM • 2 Y

Y by Adventure10, Mango247

Probability1.01 I know the problem isn't that bad I can solve it and I'm a freshmen, but there is no chance I could have done it when I was in 7th grade.

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Valentin Vornicu

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Valentin Vornicu

#6 h Apr 2, 2005, 11:08 AM • 2 Y

Y by Adventure10, Mango247

As far as the stats go, this problem was solved by 20 contestants out of a pool of 97 students. Anyway let's stop the off-topic now and concentrate on solving this very nice problem

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#7 h Apr 2, 2005, 11:36 PM • 2 Y

Y by Adventure10, Mango247

a) $\triangle OMN \sim \triangle OBA$ ;

Set OA=a, OB=b, and by similar triangles AOB and COD, set OC=ka, OD=kb. Then triangles AON and NOC are similar, so a/ON=ON/ka, i.e. $ON=a\sqrt k$ , and similarly, $OM=b\sqrt k$ . Then OM/ON=b/a=OB/OA, so the triangles are similar.

b) $OE \perp AB$ .

Let OE and AB intersect at X. Then

<XOA = <EOM = <EMO (since MEO is isosceles)
=<XBO (by similar triangles in part a)
=90-<XAO (since AOB is right),

so triangle AXO is a right triangle.

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nr1337

#8 h Apr 3, 2005, 12:05 AM • 2 Y

Y by Adventure10, Mango247

mna851 wrote:

This is unrelated, but does anyone else find it odd that advanced American math students struggle with problems 7th grade Romanian students can solve.

These problems look intimidating at first, but just stare hard at them for a couple minutes and solutions quickly emerge....

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#9 h Apr 3, 2005, 6:16 AM • 2 Y

Y by Adventure10, Mango247

I'm just curious if there's some way to do this only with angles i.e. angle chasing.

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Valentin Vornicu

7301 posts

Valentin Vornicu

#10 h Apr 3, 2005, 8:33 AM • 2 Y

Y by Adventure10, Mango247

polymorphic wrote:

I'm just curious if there's some way to do this only with angles i.e. angle chasing.

None that I know of.

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2643 posts

yetti

#11 h Apr 3, 2005, 9:00 AM • 2 Y

Y by Adventure10, Mango247

The triangles $\triangle OMN \sim \triangle OBA$ are similar regardless of whether the trapezoid diagonals $AC, BD$ are perpendicular or not. Moreover, the angles $\measuredangle ANC = \measuredangle BMD$ have to be only equal, not necessarily right. From similarity of the triangles $\triangle OBA \sim \triangle ODC$ , it follows that $\frac{OA}{OC} = \frac{OB}{OD}$ . Rotate the the triangle $\triangle ACN$ by the angle $\measuredangle AOB$ around the point $O$ counter-clockwise and then reflect the rotated triangle in the line $BD$ into a congruent triangle $\triangle A'C'N'$ . The corresponding sides of the triangles $\triangle A'C'N', \triangle BDM$ become parallel and the triangles themselves centrally similar with homothety center $O$ and homothety coefficient $h = \frac{OA}{OC} = \frac{OB}{OD}$ . Consequently, the angles $\measuredangle MAN \equiv \measuredangle CAN = \measuredangle DBM \equiv \measuredangle NBM$ are equal. This means that the quadrilateral $ABNM$ is cyclic and the sums of its opposite angles are equal to $180^o$ . Thus

$\measuredangle OMN = 180^o - \measuredangle AMN = \measuredangle ABN \equiv \measuredangle ABO$

$\measuredangle ONM = 180^o - \measuredangle BNM = \measuredangle BAM \equiv \measuredangle BAO$

As a result, the triangles $\triangle OMN \sim \triangle OBA$ are similar. This implies that the triangles $\triangle OMN \sim \triangle ODC$ are also similar and consequently, the quadrilateral $MCDN$ is cyclic. Until now, no right angles have been used. If the diagonals $AC \perp BD$ are perpendicular, the diagonals $MC \perp ND$ of the cyclic quadrilateral $MCDN$ are also perpendicular. By one of Brahmagupta's theorems, the diagonal intersection of a cyclic quadrilateral with perpendicular diagonals is identical with its anticenter - the intersection of its maltitudes, i.e., normals from midpoints of its sides to the opposite sides. Since $E$ is the midpoint of the side $MN$ , the line $EO$ is one of the maltitudes and therefore, $OE \perp CD$ . Since the trapezoid bases $AB \parallel CD$ are parallel, $OE \perp AB$ as well.

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