Problem on Area of Triangle | SMO, 2010 | Problem 32

Try this beautiful problem from Singapore Mathematics Olympiad based on area of triangle.

Given that ABCD is a square . Points E and F lie on the side BC and CD respectively, such that BE = \(\frac {1}{3} AB\) = CF. G is the intersection of BF and DE . If

\(\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}\) is in its lowest term find the value of m+n.

Here is the 1st hint for them who got stuck in this problem:

At 1st we will join the points BD and CG .

Then to proceed with this sum we will assume the length of AB to be 1. The area of \(\triangle {BGE} \) and \(\triangle {FGC}\) are a and b respectively.

Try to find the area of \(\triangle {EGC} \) and \(\triangle {DGF} \)..........................

Now for the 2nd hint let's start from the previous hint:

So the area of \(\triangle {EGC} \) and \(\triangle {DGF} \) are 2a and 2b .From the given value in the question we can say the area of \(\triangle {BFC} = \frac {1}{3}\)(as BE = CF = 1/3 AB\).

We can again write 3a + b = \(\frac {1}{6}\)

Similarly 3b + 2a = area of the \(\triangle DEC = \frac {1}{3}\).

Now solve this two equation and find a and b..............

In the last hint :

2a + 3b = 1/3........................(1)

3a + b = 1/6...............(2)

so in \( (1) \times 3\) and in \((2) \times 2\)

6a + 9b = 1

6a + 2b = 1/3

so , 7 b = 2/3 (subtracting above 2 equation)

b = \(\frac {2}{21}\) and a = \(\frac {1}{42}\)

So,\(\frac {Area of ABGD}{Area of ABCD} = 1-3(a+b) = 1- \frac {15}{42} = \frac {9}{14}\)

Comparing the given values from the question , \(\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}\)

m = 9 and n = 14

and m+n = 23(answer)

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