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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
3 years, 10 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 \(\lambda :1\)

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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 - 3 years, 10 months ago

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Excellent solution! 5 points to you!

Cody Johnson - 3 years, 10 months ago

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Thanks :)

Jatin Yadav - 3 years, 10 months ago

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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 - 3 years, 10 months ago

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Whoops, fixed.

Cody Johnson - 3 years, 10 months ago

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