# 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 $$[PQRS]$$.

Note by Cody Johnson
4 years, 6 months ago

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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($$ABC$$) = $$\fbox{648}$$

- 4 years, 6 months ago

Excellent solution! 5 points to you!

- 4 years, 6 months ago

Thanks :)

- 4 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

- 4 years, 6 months ago

Whoops, fixed.

- 4 years, 6 months ago