Geometry is so cool

Geometry Level 4

Taken on the sides $AB$ and $BC$ of $\triangle ABC$ are points $K$ and $P$ such that $AK:BK=1:2$ and $CP:BP=2:1$ . The striaght lines $AP$ and $CK$ intersect at point $E$ . Find the area of $\triangle ABC$ if it is known that the area of $\triangle BEC$ is equal to $4\text{ cm}^2$ .

Give your answer in $\text{cm}^{2}$ .

The answer is 7.

4 solutions

Niranjan Khanderia
Oct 15, 2015

$Given ~~\Delta BEC=4,~~\implies~~\Delta BEP=\dfrac 4 {1+2}=\dfrac 4 3,~~\Delta CEP=\dfrac 83.\\ 2*\Delta ACK=\Delta BCK,~~\implies~~2*(X+Z)=2X+\dfrac 4 3 + \dfrac 8 3 .\\ \therefore~Z=2.\\ 2*\Delta ABP=\Delta ACP,~~\implies~~2*(X+2X+\dfrac 4 3 )=\dfrac 8 3 + Z.\\ \therefore~6X=Z=2.\\ \Delta ABC=3X + 4 + Z=1 + 4 + 2 = \Large ~~~~~\color{#D61F06}{7}$

Also, area of BKC=2 (area of AKC) So we can say 2 ( X+Z) =2X + 4 Z=2 X=1/3 Area of ABC= 3X +Z+4 =1+2+4 =7

Saloni Baweja - 5 years, 8 months ago

Is there a typo? Some thing is not correct. Please correct it.
2(X+Z)=2X+4,......Z=2,........X=1/3*ABE then ????

Niranjan Khanderia - 5 years, 8 months ago

Alan Yan
Oct 16, 2015

We will use mass points. Assign masses 4, 2,1 to A, B, C respectively. This implies that P has a mass of 3 which implies that $\frac{PE}{AP} = \frac{4}{7} = \frac{4}{[ABC]} \implies [ABC] = \boxed{7}$

which method is this?

Dev Sharma - 5 years, 7 months ago

Mass Points. It basically utilizes that every system has one center of mass. This originates from Archimedes's idea of a lever. If you have a segment $AB$ with $C$ on the segment between them such that AC = 4 and BC = 3, you can make $C$ the center of mass by assigning a mass of 3 to A and a mass of 4 to B. In general, if you have a lever $AB$ with a fulcrum $C$ such that $AC = m$ and $BC = n$ , in order to balance this you need to put a mass of $x , y$ on $A,B$ respectively such that $mx = ny$ .

Additionally, the center of mass is the sum of the two masses. Once you know this concept, ratio questions become extremely easy since it becomes easy arithmetic. For more information, you can read about it here:

Mass points .

Alan Yan - 5 years, 7 months ago

Pranav Jangir
Nov 9, 2015

This was my way of the solution.Although it's out of category of classical geometry I guess,but it works and this method came to my mind first.I thought maybe a diversity in solution can't harm anyone.And sorry for the bad handwriting and even more worse lighting,I don't know how to use latex or any other software.Also,I didn't get those other methods like what Niranjan Khanderia did and also the concept of mass points.Alan Yan I would be really glad if you people can explain me those methods.If you have any problem in understanding the solution,I apologize once more,but please do ask me...

Rakhi Bhattacharyya
Jan 1, 2016

Draw a cevian BM. By Ceva's theorem find CM/AM. Then find the areas of AEC and AEB in terms of BEC through the ratio obtained in the last step. Add to get the result.

Geometry is so cool

The answer is 7.

4 solutions

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