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Difference between revisions of "2004 Pan African MO Problems/Problem 6"

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Revision as of 00:14, 3 February 2023

Let $ABCD$ be a cyclic quadrilateral such that $AB$ is a diameter of it's circumcircle. Suppose that $AB$ and $CD$ intersect at $I$, $AD$ and $BC$ at $J$, $AC$ and $BD$ at $K$, and let $N$ be a point on $AB$. Show that $IK$ is perpendicular to $JN$ if and only if $N$ is the midpoint of $AB$.

Solution

We use the notation $\omega_{XYZ}$ to represent the circumcircle of $XYZ$ for any three points $X, Y, Z$, and use $\Omega$ to represent the circumcircle of $ABCD$. Without loss of generality, assume that $AB$ and $CD$ are not parallel and that $K$ is closer to $B$ than to $A$. In this solution, we use $N'$ to represent the midpoint of $AB$ and prove that $IK\perp JN'$.

Note that $\angle JDK = 180^\circ - \angle ADK = 180^\circ - \angle ADB = 90^\circ$. Similarly, note that $\angle JCK = 180^\circ - \angle BCK = 180^\circ - \angle BCA = 90^\circ$. Therefore, $JK$ is a diameter of $\omega_{DKC}$.

Define points $E, E'$ such that $\omega_{AKB}\cap\omega_{DKC} = \{E, K\}, \omega_{ADN'}\cap\omega_{BCN'} = \{E', N'\}$. We claim that $E' = E$, and we prove this by showing that $E'$ lies on $\omega_{DKC}$ and that it lies on $\omega_{AKB}$.

To prove that $E'\in\omega_{DKC}$, we angle chase: $\angle DE'N' = 180^\circ - \angle DAN' = 180^\circ - \angle DAB$, $\angle CE'N' = 180^\circ - \angle CBN' = 180^\circ - \angle CBA$, so that $\angle DE'C = 360^\circ - \angle DE'N' - \angle CE'N' = 360^\circ - (180^\circ - \angle DAB) - (180^\circ - \angle CBA) = \angle DAB + \angle CBA$; similarly, $\angle DEC = \angle DKC = 180^\circ - \angle BKC = 180^\circ - (\angle KAB + \angle KBA) = (90^\circ - \angle KAB) + (90^\circ - \angle KBA) = \angle CBA + \angle DAB$. Therefore, $E'\in\omega_{DKC}$.

To prove that $E'\in\omega_{AKB}$, we make a similar angle chase: $\angle AE'N' = \angle ADN' = \angle DAN' = \angle DAB$, $\angle BE'N' = \angle BCN' = \angle CBN' = \angle CBA$, so that $\angle AE'B = \angle AE'N' + \angle BE'N' = \angle DAB + \angle CBA$; similarly, we have $\angle AKB = \angle DKC = \angle CBA + \angle DAB$. Therefore, $E'\in\omega_{AKB}$.

Now, the radical center of $\omega_{ADN'}, \omega_{BCN'}, \Omega$ must be $J$, so that the radical axis of $\omega_{ADN'}, \omega_{BCN'}$, which is $E'N'$, must also pass through $J$. But $E' = E$, so $JN'$ passes through $E$.

Therefore, the angle between $JN'$ and $IK$ is also the angle between $JE$ and $KE$, which is trivially $90^\circ$ because $E$ is part of the circle with diameter $JK$, and we are done.

See also

2004 Pan African MO (Problems)
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Problem
All Pan African MO Problems and Solutions
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