Ratio is rational

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Source: Indian IMOTC 2004 Day 5 Problem 1

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1592 posts

Rushil

#1 h Sep 23, 2005, 12:48 PM • 1 Y

Y by Adventure10

Let $ABC$ be an acute-angled triangle and $\Gamma$ be a circle with $AB$ as diameter intersecting $BC$ and $CA$ at $F ( \not= B)$ and $E (\not= A)$ respectively. Tangents are drawn at $E$ and $F$ to $\Gamma$ intersect at $P$ . Show that the ratio of the circumcentre of triangle $ABC$ to that if $EFP$ is a rational number.

This post has been edited 1 time. Last edited by Rushil, Oct 16, 2005, 5:00 AM

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6555 posts

#2 h Sep 26, 2005, 12:47 AM • 2 Y

Y by Adventure10, Mango247

I have tried to search for this problem, since I am sure it already appeared on this forum, but it turned out easier to solve it again.

The core of the problem is the following "theorem":

Theorem 1. Let ABC be a triangle and X, Y, Z the feet of its altitudes from the vertices A, B, C, respectively. Also, let H be the point of intersection of these altitudes, i. e. the orthocenter of triangle ABC. Since < AXB = 90° and < AYB = 90°, the points X and Y lie on the circle with diameter AB. Then, the tangents to this circle at the points X and Y both pass through the midpoint R of the segment CH.

Proof of Theorem 1. The center of the circle with diameter AB is the midpoint M of the segment AB.

It is well-known that the feet X, Y, Z of the altitudes of triangle ABC, the midpoint M of its side AB and the midpoint R of the segment CH all lie on the nine-point circle of triangle ABC. Since < RZM = 90°, the segment RM is a diameter of this nine-point circle. Hence, we also have < RYM = 90°, so that $RY\perp MY$ . Now, since M is the center of the circle with diameter AB, and Y is a point on this circle, the relation $RY\perp MY$ shows that the line RY is the tangent to the circle with diameter AB at the point Y (since the tangent to a circle at a point is perpendicular to the radius through this point). Hence, the point R lies on the tangent to the circle with diameter AB at the point Y. Similarly, the point R lies on the tangent to the circle with diameter AB at the point X. Hence, Theorem 1 is proven.

Now, your points E and F are the points Y and X from Theorem 1, and your point P is the point of intersection of the tangents to the circle with diameter AB at the points X and Y, i. e. the midpoint R of the segment CH (after Theorem 1). Hence, the problem requires to prove that the ratio between the circumradii of triangles ABC and YXR is a rational number. But this is trivial: Since the points Y, X and R lie on the nine-point circle of triangle ABC, the circumcircle of triangle YXR is the nine-point circle of triangle ABC, and thus its radius is $\dfrac12$ of the circumradius of triangle ABC (this is a trivial property of the nine-point circle); in other words, the ratio between the circumradii of triangles ABC and YXR is 2, and since 2 is a rational number, the problem is solved.

darij

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xirti

#3 h Apr 8, 2006, 10:01 AM • 2 Y

Y by Adventure10, Mango247

Since $\angle OFP=\angle OEP=90^o$ the quadrilateral $EOFP$ is cyclic. And it's circumcenter have the diameter $PO$ . It's easy to see by angle chasing, and using the cyclic quadrilaterals $OEFP,BCEF$ , (Where $O$ is the midpoint of $BC$ ) that $\angle OPE=\angle A$ . Hence, $2R\sin A= BC=2 OE=\frac 1 2 OP \sin A=r\sin A$ ,where $R$ is the circumradius of $ABC$ , and $r$ the one of the $EFP$ . Hence the ratio of the two radius is $\frac 1 2$ .

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803 posts

#4 h Oct 14, 2013, 5:10 AM • 2 Y

Y by Adventure10, Mango247

$R(EFP)=\frac{EF}{2sin2C}=\frac{OE}{2sinC}=c/4sinC=R(ABC)/2$

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339 posts

#5 h Jun 6, 2014, 6:02 AM • 2 Y

Y by Adventure10, Mango247

Lol, this is probably the trolliest problem ever posed in IMOTC (yeah, I know, I just made that word up).

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jayme

#6 h Jun 7, 2014, 3:49 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
with harmonic division, we have the nature of the point of intersection of the tangent and the C-altitude... the C-Euler's point...
Very nice problem which permit us to consider the Euler's circle of ABC..
Sincerely
Jean-Louis

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