Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I found this question a while ago on a SAT practice exam or something, can't quite remember. So given an acute triangle $ABC$ with $P$ a point inside it and $AP$, $BP$, and $CP$ meeting the opposite sides at $D$, $E$, and $F$ respectively:

alt text

How can you find the area of triangle $ABC$ given the areas of triangles $x$, $y$, and $z$?

share|improve this question
add comment

2 Answers 2

up vote 4 down vote accepted

This was not likely to have been an SAT practice problem, though it is a typical contest problem.

AE:EC = x:y (since those two triangles have the same altitude to AC, the ratios of their areas is the ratios of their bases with respect to that altitude) and AP:PD = (x+y):z (same idea as AE:EC). Knowing these two ratios, apply the technique of mass points, putting masses zy at A, zx at C (gives the ratio x:y for AE:EC), and y(x+y) at D (gives (x+y):z for AP:PD). This results in a mass of y(x+y)-zx at B, so the ratio BD:DC = zx:(y(x+y)-zx). This must also be the ratio of the areas of △ABD to △ADC (common altitude again), so (area of △ABD):(x+y+z) = zx:(y(x+y)-zx). Solving from there is a matter of bashing out the algebra.

share|improve this answer
    
Bashing out the algebra: (Total area)/(x+y+z) = 1 + (△ABD)/(x+y+z) = (y(x+y)-zx+zx)/(y(x+y)-zx), so the area is $(x+y+z)y(x+y)/(y(x+y)-zx))$. This is the same as the other answer, which is great. –  ShreevatsaR Jul 31 '10 at 6:31
1  
Wow, mass point geometry is cool –  Casebash Jul 31 '10 at 6:56
add comment

Method 1: We can perform an affine transformation to make ADC any triangle we wish, while preserving colinearality and the ratios of areas. In particular, this allows us to assign exact coordinates to A, D and C. We can then use the ratio of x and y to find E and the ratio of z and x+y+z to find P. We can then use these coordinates to find B. Once we know B, we know the ratio of BC to DC and hence the ratio of the area of the whole triangle to the known areas.

Method 2: Let Q be the area of the whole triangle

AE:EC=x:y=|AEB|:|BEC|
|APB|=xQ/(x+y)-x
|BPD|=yQ/(x+y)-y-z
Now AP:PD=x+y:z
z(xQ/(x+y)-x)=(x+y)(yQ/(x+y)-y-z)
Q(zx/(x+y)-y)=xz-xy-y^2-xz-yz=-xy-y^2-xz
Q(zx-xy-y^2)/(x+y)=(-xy-y^2-xz)
Q=(-xy-y^2-xz)*(x+y)/(zx-xy-y^2)
share|improve this answer
    
So what's the area? :-) –  ShreevatsaR Jul 31 '10 at 6:18
    
@ShreevataR: Tried to calculate the area. Hopefully its correct, although I'd expect a problem like this to have a neater solution –  Casebash Jul 31 '10 at 6:22
    
In method 2, for clarity, you probably should use something other than P for the area (I'd use k), since P is also a point. –  Isaac Jul 31 '10 at 6:30
    
@Isaac: Thanks, that always gets me –  Casebash Jul 31 '10 at 6:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.