CMC - Problem 8

Problem 8. (5 points) Let $\triangle ABC$ be a triangle with area $5040$. Let $B_1,B_2,C_1,$ and $C_2$ be the trisectors of $\overline{AC}$ and $\overline{AB}$, respectively, such that $AB_1<AB_2$ and $AC_1<AC_2$. Now let $P=BB_1\cap CC_1$, $Q=BB_2\cap CC_1$, $R=BB_1\cap CC_2$, and $S=BB_2\cap CC_2$. Find the area $[PQRS]$.

#Geometry #CMC #MathProblem #Math

Easy Math Editor

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Comments

I solved using vectors:

Any point $W$ will be shown as $\vec{w}$

Say there is a point $M$ on $AC$ dividing it in the ratio $\lambda :1$ Alternate text

Then using section formula,

$\vec{m} = \frac{ \vec{a} + \lambda \vec{c}}{\lambda +1}$

$\Rightarrow (\lambda + 1)\vec{m} = \vec{a} + \lambda \vec{c}$ .... $(i)$

Similarly, if a point $N$ divides $AB$ in the ratio $\mu :1$ ,

$(\mu + 1) \vec{n} = \vec{a} + \mu \vec{b}$ ... $(ii)$

From $(i)$ and $(ii)$ , we get :

$\frac{(\lambda + 1)\vec{m} + \mu \vec{b}}{1 + \lambda + \mu} = \frac{(\mu + 1) \vec{n} + \lambda \vec{c}}{1 + \lambda + \mu} = \frac{\vec{a} + \lambda \vec{c} + \mu \vec{b}}{1 + \lambda + \mu} = \vec{l}$ (say)

Hence, by section formula, $L$ is a common point on $BM$ and $CN$ ,i.e. $BM \cap CN$

Now, let's come to our case, we know all the ratios, hence , we can easily find $\vec{p}, \vec{q}, \vec{r}, \vec{s}$ as:

$\vec{p} = \frac{2 \vec{a} + \vec{b} + \vec{c}}{4}$ , ( $\lambda = \mu = \frac{1}{2})$

$\vec{q} = \frac{ 2\vec{a} + 4\vec{b} + \vec{c}}{7} , (\lambda = \frac{1}{2} , \mu = 2)$

$\vec{r} = \frac{2\vec{a} + \vec{b} + 4\vec{c}}{7}, (\lambda = 2, \mu =\frac{1}{2})$

$\vec{s} = \frac{\vec{a} + 2\vec{b} + 2\vec{c}}{5}$ , $(\lambda = \mu = 2)$

Now, we can assume $\vec{a} = 0$ to simplify further part, note that it won't affect the answer

Also, in the question $P,S$ are opposite , and $Q,R$ are opposite,

Area of $PQRS$ = $\frac{1}{2}|\vec{d_{1}} \times \vec{d_{2}}|$ , $\vec{d_{1}}, \vec{d_{2}}$ are diagonal vectors.

Hence area = $\frac{1}{2}|(\vec{p} - \vec{s}) \times (\vec{q} - \vec{r})|$

= $\frac{1}{2}|\frac{3}{20}(\vec{b} + \vec{c}) \times (\frac{3}{7}(\vec{b} - \vec{c})) |$

= $\frac{9}{70} (\frac{1}{2}|\vec{b} \times \vec{c}|)$

= $(\frac{9}{70})$ Ar( $ABC$ ) = $\fbox{648}$

jatin yadav - 7 years, 6 months ago

Excellent solution! 5 points to you!

Cody Johnson - 7 years, 6 months ago

Thanks :)

jatin yadav - 7 years, 6 months ago

First of all P,Q,R,S are not adjacent vertices and secondly, my answer comes $\frac{9 \triangle}{70}$ , which is out of $0-999$ , you sure its 13370

jatin yadav - 7 years, 6 months ago

Whoops, fixed.

Cody Johnson - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$