Brazilian Mathematical Olympiad 2007

April 9, 2009

Problem:

Problem 5 on the Brazilian Mathematical Olympiad 2007. Let $ABCD$ be a convex quadrangle, $P$ the intersection of lines $AB$ and $CD$, $Q$ the intersection of lines $AD$ and $BC$ and $O$ the intersection of diagonals $AC$ and $BD$. Show that if $\angle POQ=9{0}^{\circ}$, then $PO$ is the bisector of $\angle AOD$ and $QO$ is the bisector of $\angle AOB$.

Solution:

Embed the problem in a coordinate system, with $O$ at the origin, $P$ on the $x$-axis at $\left(p,0\right)$ and $Q$ on the $y$ axis at $\left(0,q\right)$. Let the equation of the line through $A$ and $B$ passing through $P$ be $y={m}_{AB}\left(x-p\right)$. Likewise, let the line passing through $C$ and$D$ be $y={m}_{CD}\left(x-p\right)$. The line through $A$ and $D$ passing through $Q$ is $y={m}_{AD}+q$ and the line through $B$ and $C$ passing through $Q$ is $y={m}_{BC}+q$

Taking these equations in pairs and solving for the intersection point we obtain the coordinates for $A$, $B$, $C$ and $D$ as:

- $A:\left(\frac{q+{m}_{AB}p}{-{m}_{AD}+{m}_{AB}},\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{-{m}_{AD}+{m}_{AB}}\right)$
- $B:\left(\frac{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}{-{m}_{BC}+{m}_{AB}},\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{BC}\phantom{\rule{0em}{0ex}}p+q\right)}{-{m}_{BC}+{m}_{AB}}\right)$
- $C:\left(-\frac{q+{m}_{CD}\phantom{\rule{0em}{0ex}}p}{{m}_{BC}-{m}_{CD}},-\frac{{m}_{CD}\phantom{\rule{0em}{0ex}}\left({m}_{BC}\phantom{\rule{0em}{0ex}}p+q\right)}{{m}_{BC}-{m}_{CD}}\right)$
- $D:\left(-\frac{q+{m}_{CD}\phantom{\rule{0em}{0ex}}p}{{m}_{AD}-{m}_{CD}},-\frac{{m}_{CD}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{{m}_{AD}-{m}_{CD}}\right)$

Then the slope of the line through the origin $O$ and $A$ is: $\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}$ and the slope of the line through $O$ and $C$ is $\frac{{m}_{CD}\phantom{\rule{0em}{0ex}}\left({m}_{BC}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{CD}\phantom{\rule{0em}{0ex}}p}$. These two slopes must be equal, hence

$$\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}=\frac{{m}_{CD}\phantom{\rule{0em}{0ex}}\left({m}_{BC}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{CD}\phantom{\rule{0em}{0ex}}p}.$$

Likewise, the slope of the line through the origin $O$ and $B$ is $\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{BC}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}$ and the slope of the line through the origin $O$ and $D$ is $\frac{{m}_{CD}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{CD}\phantom{\rule{0em}{0ex}}p}$ and

$$\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{BC}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}=\frac{{m}_{CD}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{CD}\phantom{\rule{0em}{0ex}}p}.$$

Rearranging each of the slope equations and then equating them yields

$$\frac{{m}_{AB}}{{m}_{CD}}\frac{{m}_{AD}p+q}{{m}_{BC}p+q}\frac{{m}_{CD}p+q}{{m}_{AB}p+q}=1=\frac{{m}_{AB}}{{m}_{CD}}\frac{{m}_{BC}p+q}{{m}_{AD}p+q}\frac{{m}_{AB}p+q}{{m}_{CD}p+q}.$$

Hence

$$\frac{{\left({m}_{AD}p+q\right)}^{2}}{{\left({m}_{BC}p+q\right)}^{2}}=1$$

or

$$\left({m}_{AD}p+q\right)=\pm \left({m}_{BC}p+q\right)$$

In order for the quadrilateral to be a non-degenerate figure the slopes of the lines through $O$ and $A$ and through $O$ and $B$ must be equal but of opposite sign

Then the slope of the line through the origin $O$ and $A$ is: $\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}$ and the slope of the line through the origin $O$ and $B$ is $\frac{{m}_{AB}\phantom{\rule{0em}{0ex}}-\left({m}_{AD}\phantom{\rule{0em}{0ex}}p+q\right)}{q+{m}_{AB}\phantom{\rule{0em}{0ex}}p}.$ Recall that the slope of the line is tangent of the angle with the $x$-axis. The equality of the magnitudes of the slopes (even though they are opposite in sign) means the angles that $\overline{OA}$ and $\overline{OB}$ make with the $x$-axis are equal. The $y$-axis containing the line $OQ$ is then the angle bisector for $\angle AOB$. Returning to the original notation, $QO$ is the bisector of $\angle AOB$. The proof that $PO$ is the bisector of $\angle AOD$ is similar.

Alternate Solution from Carlos Yuzo Shine

Extend $QO$ as a line and let $R$ be the intersection of lines $QO$ and $CD$. The lines $AC$, $DB$, and $PR$ are concurrent at $O$. This is one of the hypotheses of the problem. Applying Ceva’s Theorem on $\u25b3QDC$ obtain $DR\cdot CB\cdot QA=RC\cdot BQ\cdot AD$. Considering overlapping $\u25b3QDC$ and $\u25b3PAD$, and Applying Menelaus’ Theorem $PD\cdot CB\cdot QA=PC\cdot BQ\cdot AD$. Then dividing the second equality by the first,

$$\frac{PD}{DR}=\frac{PC}{RC}.$$

We say the points $P$, $D$, $R$, $C$ are such that $D$ and $C$ are harmonic conjugate points to $P$ and $R$. In Geometry Revisited, page 108, Coxeter and Greitzer say that $R$ and $C$ divide the interval $DC$ internally and externally. Then the circle with diameter $\overline{RD}$ is the locus of points whose distance from $D$ and $C$ are in the ratio above. In Geometry Revisited, page 108, Coxeter and Greitzer call this the circle of Apollonius. Recall from the hypotheses of the problem that $\angle POR$ is a right angle, Then $\angle POR$ is also a right angle which subtends a diameter of the circle. Therefore $O$ is on this circle of Apollonius as well. This uses the other hypothesis of the problem. This means that

$$\frac{OD}{DR}=\frac{OC}{RC}$$

or

$$\frac{OD}{OC}=\frac{DR}{RC}.$$

Then by the Angle Bisector Theorem, $\overline{QR}$ is the bisector of $\angle COD$. Equivalently, the line $QO$ is the bisector of $\angle AOB$.

Commentary:

My solution given first is an analytic geometry solution in contrast to the alternate synthetic geometry solution given by Carlos Yuzo Shine. Usually, coordinate geometry or analytic geometry proofs are not considered as elegant as coordinate free or synthetic geometry proofs. See for example, the comment at the beginning of Section 2.2, page 31, in Geometry Revisited by H.S. Coxeter and S. L. Greitzer, repeated from E. T. Bell’s Men of Mathematics. However, here analytic geometry was the only reasonable way I could find to solve the problem. Furthermore, the fact that two lines intersect at a right angle suggests that using these as coordinate axes would be a good strategy to express the relations in the problem.

Carlos Yuzo Shine, the Deputy Leader of the 2008 Brazilian IMO team is the creator of this problem. In email correspondence of March 28, 2009, he told me that this problem was inspired by a 2005 or 2006 Bulgarian problem solved with several application of the Law of Sines. Exploring this idea, Carlos came up with this problem by combining the Law of Sines and Menelaus’ Theorem. A friend of Carlos found the synthetic geometry solution, and after the Brazilian IMO, another friend found a coordinate geometry solution which is essentially the same solution as presented here.

For this particular problem, GeoGebra neither suggested nor aided my solution. Drawing the diagram with GeoGebra allows one to see that the problem is true but the dynamic geometry diagram did not give me any insight into a solution.