When perpendicular becomes concurrent

The Asian Pacific Mathematical Olympiad (APMO) is a long-running regional olympiad that challenges young mathletes across the region with a handful of hard, original problems each year.¹ Problem 4 from this year’s edition is a nice showcase of a common trick: restating an unfriendly definition into a more symmetric one turns the perpendicularity we want into a friendlier concurrence, and the solution follows naturally.

Problem 4. Let \(ABCD\) be a quadrilateral with an incircle \(\omega\) of center \(I\). The diagonals \(AC\) and \(BD\) intersect at \(E\). Let \(J\) be the incenter of triangle \(ABD\). The extension of the ray \(EJ\) intersects \(\omega\) at \(P\). Prove that \(PI \perp BD\).

Solution. First, let’s modify the statement. It seems more intuitive to let \(P\) be the endpoint of the diameter of \(\omega\) perpendicular to \(BD\) that lies in the interior of \(\bigtriangleup ABD\). Thus, the problem reduces to showing that the points \(P,\ J\), and \(E\) are collinear.

Another key point. The triangle \(\bigtriangleup ABD\) appears to be an arbitrary choice. In other words, the statement should remain valid for \(\bigtriangleup BCD\) too. In this spirit, define \(F\) and \(P'\) to be the incenter of \(\bigtriangleup BCD\) and the intersection of \(PI\) with \(\omega\) that lies within \(\bigtriangleup BCD\) (i.e., the point diametrically opposite to \(P\) in \(\omega\)), respectively. Then, it should also be true that \(P', F\), and \(E\) are collinear. With these preparations, we’re ready to tackle the problem at hand. Consider the following

Claim 1. The line \(JF\) is perpendicular to \(BD\).

Proof. Let \(R\) and \(R'\) be the points where the incircles of \(\bigtriangleup ABD\) and \(\bigtriangleup BCD\) touch \(BD\). It’s known that

\[DR = s_{ABD}-AB = \frac{AB+DA-BD}{2},\quad DR' = s_{BCD} = \frac{BC+CD-BD}{2}\]

where \(s_{XYZ}\) stands for the semiperimeter of \(\bigtriangleup XYZ\). Pitot’s theorem implies straightforwardly that \(\frac{AB+DA-BD}{2}=\frac{BC+CD-BD}{2}\). Hence, \(DR=DR'\), so \(R\equiv R'\) and thus \(J,\ R\), and \(F\) are collinear with \(JF\perp BD\). \(\square\)

With this result, we have already used the fact that \(ABCD\) has an incircle and the definition of $J$. We need to exploit the redefinition of \(P\). Let’s now argue that

Claim 2. The lines \(AP'\), \(CP''\) and \(BD\) concur.

Proof. Let \(R'\) be the point diametrically opposite to \(R\) in the incircle of \(\bigtriangleup ABD\). As \(A\) is the exsimilicenter of this circle and \(\omega\), it turns out that \(R'\) lies on \(AP\). More importantly, it’s known that the point \(T\equiv \overline{AR'}\cap \overline{BD}\) is the contact point of the \(A\)-excircle of \(\bigtriangleup ABD\) with \(BD\) and that \(BT=DR\). Thus, this point \(T\) lies on \(AP\) and satisfies that \(BT=DR\). Analogously, we can prove that \(CP'\) meets \(BD\) at a point \(T'\) where \(BT'=DR\), which forces \(T\equiv T'\), implying the desired concurrence. \(\square\)

We’re ready to conclude. By Claim 1, observe that \(PP'\parallel JF\). Together with Claim 2, we find

\[\frac{PS}{SP'} = \frac{R'R}{RR''} = \frac{JR}{RF}\]

with \(S\equiv \overline{PP'}\cap\overline{BD}\). This result and Thales’ theorem jointly imply that \(PJ\), \(P'F\), and \(BD\) concur at \(E\). The required collinearity follows. \(\square\)

1. For more information and the rest of this year’s problems, go to apmo-official.org. ↩︎