## A Geometry Problem from the 2007 Bay Area Mathematical Olympiad

January 27, 2009

Problem: This is problem 3 on the 2007 Bay Area Mathematical Olympiad, located at  http://www.bamo.org/attachments/bamo2007examsol.pdf  In $△ABC$, $D$ and $E$ are two points in the interior of side $\overline{BC}$ such that $BD=CE$ and $\angle BAD=\angle CAE$. Prove that $△ABC$ is isosceles.

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Solution: Without loss of generality, assume that the points $B$, $D$ $E$ and $C$ are in that order on $\overline{BC}$. (Otherwise exchange the labels on points $D$ and $E$ so the former $E$ becomes ${D}^{\prime }$ and the former $D$ becomes ${E}^{\prime }$. Then recognize that the common length $ED$ can be subtracted from both lengths i.e. $BD-DE=BE=B{D}^{\prime }$ and $CE-ED=CD=C{E}^{\prime }$. Likewise the common angle $\angle DAE$ can be subtracted from both angles to preserve the hypotheses.)

From vertex $A$ to point $F$ on the opposite side $\overline{BC}$, let segment $\overline{AF}$ be the angle bisector of $\angle DAE$. Let $\alpha =\angle BAD=\angle CAE$. Let $\beta =\angle DAF=\angle EAF$. I will use $\gamma =\angle AFC$ as a parameter and express several quantities in terms of this parameter and the fixed length $AF$.

Note that $0<\alpha <\pi ∕2$, $0<\beta <\pi ∕2$ and $\alpha +\beta <\pi ∕2$. Note that if $\gamma =\alpha +\beta$ the line $BC$ is parallel to the line $AB$, an impossibility. Furthermore if $\gamma <\alpha +\beta$, the intersection of line $BC$ and line $AB$ has “flipped” to the other side of $A$ and now $E$ and $D$ are exterior to $\overline{BC}$. Thus $\gamma >\alpha +\beta$. Likewise $\pi -\gamma >\alpha +\beta$ or a similar argument applies to lines $BC$ and $AC$. Summarizing: $\alpha +\beta <\gamma <\pi -\left(\alpha +\beta \right)$.

Apply the Law of Sines to $△FAC$ to obtain

$\frac{FC}{sin\left(FAC\right)}=\frac{AF}{sin\left(ACF\right)}.$

Therefore

$FC=AF\frac{sin\left(FAC\right)}{sin\left(ACF\right)}$

and likewise

$\begin{array}{llll}\hfill FB& =AF\frac{sin\left(FAB\right)}{sin\left(ABF\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill FD& =AF\frac{sin\left(FAD\right)}{sin\left(ADF\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill FE& =AF\frac{sin\left(FAE\right)}{sin\left(AEF\right)}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then

$\begin{array}{llll}\hfill CE& =AF\left[\frac{sin\left(FAC\right)}{sin\left(ACF\right)}-\frac{sin\left(FAE\right)}{sin\left(AEF\right)}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill BD& =AF\left[\frac{sin\left(FAB\right)}{sin\left(ABF\right)}-\frac{sin\left(FAD\right)}{sin\left(ADF\right)}\right].\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Now using that $\alpha =\angle BAD=\angle CAE$ and since $\overline{AF}$ bisects $\angle DAE$, so $\angle FAD=\angle FAE=\beta$ and $\angle FAB=\angle FAC=\alpha +\beta$. Let $\angle AFE=\angle AFC=\gamma$ and $\angle AFB=\angle AFD=\pi -\gamma$. Then

$\begin{array}{llll}\hfill CE& =AF\left[\frac{sin\left(\alpha +\beta \right)}{sin\left(\pi -\left(\alpha +\beta +\gamma \right)\right)}-\frac{sin\left(\beta \right)}{sin\left(\pi -\left(\beta +\gamma \right)\right)}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill BD& =AF\left[\frac{sin\left(\alpha +\beta \right)}{sin\left(\pi -\left(\pi -\gamma \right)-\left(\alpha +\beta \right)\right)}-\frac{sin\left(\beta \right)}{sin\left(\pi -\left(\pi -\gamma \right)-\beta \right)}\right].\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Simplifying

$\begin{array}{llll}\hfill CE& =AF\left[\frac{sin\left(\alpha +\beta \right)}{sin\left(\alpha +\beta +\gamma \right)}-\frac{sin\left(\beta \right)}{sin\left(\beta +\gamma \right)}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill BD& =AF\left[\frac{sin\left(\alpha +\beta \right)}{sin\left(-\alpha -\beta +\gamma \right)}-\frac{sin\left(\beta \right)}{sin\left(-\beta +\gamma \right)}\right].\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Now equate these expressions since $BD=CE$

$\frac{sin\left(\alpha +\beta \right)}{sin\left(\alpha +\beta +\gamma \right)}-\frac{sin\left(\beta \right)}{sin\left(\beta +\gamma \right)}=\frac{sin\left(\alpha +\beta \right)}{sin\left(\gamma -\alpha -\beta \right)}-\frac{sin\left(\beta \right)}{sin\left(\gamma -\beta \right)}$

or

$\frac{sin\left(\alpha +\beta \right)}{sin\left(\alpha +\beta +\gamma \right)}-\frac{sin\left(\alpha +\beta \right)}{sin\left(\gamma -\alpha -\beta \right)}=\frac{sin\left(\beta \right)}{sin\left(\beta +\gamma \right)}-\frac{sin\left(\beta \right)}{sin\left(\gamma -\beta \right)}.$

Adding over a common denominator, and reducing trigonometric products to sums, obtain

$\frac{4cos\left(\gamma \right){sin}^{2}\left(\alpha +\beta \right)}{cos\left(2\gamma \right)-cos\left(2\left(\alpha +\beta \right)\right)}=\frac{4cos\left(\gamma \right){sin}^{2}\left(\beta \right)}{cos\left(2\gamma \right)-cos\left(2\beta \right)}$

Recalling that $\alpha +\beta <\gamma <\pi -\left(\alpha +\beta \right)$, neither of the denominator expressions is $0$.

The only way these two expressions can be equal is for $cos\left(\gamma \right)=0$ or equivalently $\gamma =\pi ∕2$, hence $\angle FCA=\angle FBA$ and $△ABC$ is isosceles.

Commentary: Constructing this problem with GeoGebra showed me that the varying the base $\overline{BC}$ was the key to understanding the problem. That suggested rotating the base $\overline{BC}$ about the point $F$ using the angle $\gamma$ as a parameter to measure the rotation was a good way to organize the proof. The rest of the proof then organizes itself around clearly expressing the lengths $BD$ and $CE$ in terms of the parameter $\gamma$. Expressing the lengths yielded expressions such as $sin\left(\gamma -\alpha -\beta \right)$ in denominators. Again rotating segment $\overline{BC}$ about $F$ shows that if $\gamma =\alpha +\beta$ the line $BC$ is parallel to the line $AB$, so this parameter condition must be ruled out.

The official source at  http://www.bamo.org/attachments/bamo2007examsol.pdf  gives $4$ distinct solutions, each shorter than the proof above. The solution above is a combination of their Solution 2 and Solution 3. However, each of their solutions gives a proof by contradiction, while the proof above is essentially direct. The proof above is dynamic in nature, based on the dynamic geometry view of the problem from GeoGebra by rotating the base. This indicates how the technology used to view the problem can influence the thinking and the derived proof.