# Chaitanya's Random Pages

## July 25, 2013

### Solving a tricky geometry problem using cross ratios

Filed under: mathematics — ckrao @ 12:17 pm

I recently had fun playing around with the following problem, based on the last question of the 2003 AIME I.

In triangle $ABC$ let be the midpoint of $CA$, and let $D$ be the point on $CA$ such that $BD$ bisects angle $ABC$. Let $F$ be the point on $BC$ such that $DF$ and $BD$ are perpendicular. If $DF$ meets $BM$ at $E$, find the ratio $DE/EF$ in terms of the side lengths of $ABC$.

The interested reader might like to have an attempt at this problem before reading further. 🙂

Initially I solved this using vector geometry applying facts such as $\vec{BM} = (\vec{BA} + \vec{BC})/2$, $\vec{BD} = (|BC|\vec{BA} + |AB|\vec{BC})/(|BC| + |AB|)$ and $\vec{DB}. \vec{DF} = 0$ (see below). After seeing it led to a relatively simple answer I started looking for a more elegant approach and after seeing this page that shows methods using mass point geometry and applications of Menelaus’ theorem, came up with the following nice solution.

Extend $FD$ to meet $BA$ in $G$ as shown below. As $BD$ is an angle bisector perpendicular to $GF$, $F$ reflects in line $BD$ to $G$ so $D$ is the midpoint of $GF$.

Also the lines $ADMC$ and $GDEF$ are both perspective from the point $B$, so the cross ratios $(AM/MC)/(AD/DC)$ and $(GE/EF)/(GD/DF)$ are equal (indeed, by using the triangle area formula $\frac{1}{2}ab\sin C$ these ratios can be shown to be equal to $(\sin \angle ABM/ \sin \angle MBC)/(\sin \angle ABD / \sin \angle DBC)$). Since $M$ and $D$ are midpoints of $AC$ and $GF$ respectively, $AM/MC = GD/DF = 1$, so we are left with

$\displaystyle \frac{GE}{EF} = \frac{DC}{AD}.$

By the angle bisector theorem, this ratio is equal to $a/c$, where $a = BC, c = AB$. We finally have

$\displaystyle \frac{DE}{EF} = \frac{GE-EF}{2EF} = \frac{GE/EF - 1}{2} = \frac{a-c}{2c},$

as required. It’s interesting that the answer has no dependence on the length of side $AC$.

Here is the original way I had solved the problem. Let $B$ be the origin and define vectors $\boldsymbol{a} = \vec{BA}$ and $\boldsymbol{c} = \vec{BA}$ of length $c$ and $a$ respectively. Then as $BD$ is the angle bisector,  we can write $\vec{BD} = \lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}$, where $\lambda_1 = a/(a+c)$ and $\lambda_2 = c/(a+c)$.

Also let $\vec{BF} = k \boldsymbol{c}$ where $k = BF/BC$. We find $k$ by using the fact that $DF \perp DB$:

$\displaystyle \left(kc - (\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})\right).(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}) = 0.$

From this,

$\displaystyle k = \frac{(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}).(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}{\boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})} = \frac{BD^2}{\boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}.$

Since $E$ is on $DF$, for some $\mu$ between 0 and 1,

$\displaystyle \vec{BM} = \mu \vec{BD} + (1-\mu) \vec{BF} = \mu \lambda_1 \boldsymbol{a} + (\mu \lambda_2 + (1-\mu) k )\boldsymbol{c}.$

Since $E$ is also on the median $BM$, it and have equal components of $a$ and $c$ (being a median).

Hence, $\mu \lambda_1 = \mu \lambda_2 + (1-\mu) k$ from which

$\displaystyle \frac{\lambda_1 - \lambda_2}{k} = \frac{1-\mu}{\mu} = \frac{(\lambda_1 - \lambda_2) \boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}{BD^2}. \quad \quad (1)$

We wish to find the ratio $DE/EF = (1-\mu)/\mu$, which is as above.

Here is one easy way to find $k = BF/BC$ that I found only afterwards. If $P$ is the midpoint of $BF$ then since $BDF$ is a right-angled triangle, $BP = PD = PF$ and so $\angle PDB = \angle DPE = \angle DBA$, from which $AB \parallel DP$.

Then $\triangle BCA$ and $\triangle PCD$ are similar, from which

$\displaystyle \frac{BP}{BC} = \frac{AD}{AC} = \frac{c}{a+c}.$

Hence $k = BF/BC = \frac{2c}{a+c}$ and so

$\displaystyle \frac{DE}{EF} = \frac{\lambda_1 - \lambda_2}{k} = \frac{(a-c)}{(a+c)} \frac{(a+c)}{2c} = \frac{a-c}{2c}.$

The original way I had found $k$ was without the benefit of this construction and by pure computation instead. Referring back to (1), we have $\boldsymbol{c}.\boldsymbol{c} = a^2$ and by the cosine rule, $\boldsymbol{c}.\boldsymbol{a} = (a^2 + c^2 - b^2)/2$. Recalling that $\lambda_1 = a/(a+c)$ and $\lambda_2 = c/(a+c)$,

\begin{aligned} \boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}) &= \lambda_1 \boldsymbol{c}.\boldsymbol{a} + \lambda_2 \boldsymbol{c}.\boldsymbol{c}\\ &= \frac{a(a^2 + c^2 - b^2) + 2ca^2}{2(a+c)} \\ &= \frac{a(a+b+c)(a-b+c)}{2(a+c)}. \quad \quad (2) \end{aligned}

Secondly the denominator of (1) is

\begin{aligned} BD^2 &= (\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}).(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})\\ &= \lambda_1^2 c^2 + \lambda_2^2 a^2 + 2\lambda_1 \lambda_2 \boldsymbol{a}.\boldsymbol{c} \\ &= \frac{1}{(a+c)^2} \left[a^2 c^2 + c^2 a^2 + ac(a^2 + c^2 - b^2) \right]\\ &= \frac{ac}{(a+c)^2} \left[2ac + a^2 + c^2 - b^2 \right] \\ &= \frac{ac(a+b+c)(a-b+c)}{(a+c)^2}. \quad \quad (3)\end{aligned}

Combining (2) and (3) in (1) leads to the following feast of cancellation:

\begin{aligned} \frac{DE}{EF} &= \frac{(\lambda_1 - \lambda_2) \boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}{BD^2} \\&= \frac{(a-c)}{(a+c)} \frac{a(a+b+c)(a-b+c)}{2(a+c)} \frac{(a+c)^2}{ac(a+b+c)(a-b+c)} \\ &= \frac{a-c}{2c},\end{aligned}

as was found before. Note that expressions (2) and (3) are respectively the square of the length of the angle bisector $BD^2$ and the dot product $\vec{BC}.\vec{BD}$ in terms of the side lengths of any triangle, and we have found them to be similar in form.