1/2 and 1/3 points

Geometry Level 3

A B C D ABCD is a square, E E and F F are midpoints of sides A D AD and B C BC respectively,

and D I = I J = J C DI= IJ=JC ,

and A B = 1 AB=1 ,

If area of shaded region is a b \dfrac{a}{b} , where a a and b b are positive co-prime Integers,

Then find the value of a + b a+b .


The answer is 17.

Vaibhav Priyadarshi by Vaibhav Priyadarshi , 17, India

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2 solutions

Noel Lo
Jul 12, 2018

Considering that E is the midpoint of AD and F is the midpoint of BC, the length of GH is the average of the length of IJ and AB which is 1 3 + 1 2 = 2 3 \dfrac{\frac{1}{3}+1}{2}=\dfrac{2}{3} . Note that the perpendicular height of the trapezium is 1 2 \dfrac{1}{2} . Then the area of the trapezium is as follows:

1 2 × 1 2 × ( 2 3 + 1 ) = 1 4 × 5 3 = 5 4 × 3 = 5 12 \dfrac{1}{2} \times \dfrac{1}{2} \times (\dfrac{2}{3}+1)=\dfrac{1}{4} \times \dfrac{5}{3}= \dfrac{5}{4\times 3}=\dfrac{5}{12}

Therefore as required, a + b = 5 + 12 = 17 a+b=5+12=\boxed{17} .

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Since △ I D A ∼ △ G E A \triangle IDA \sim \triangle GEA ,

E G D I = E A A D \dfrac{EG}{DI}=\dfrac{EA}{AD} ⟹ \implies E G 1 3 = 1 2 1 \dfrac{EG}{\dfrac{1}{3}}=\dfrac{\dfrac{1}{2}}{1} ⟹ \implies E G = 1 6 EG=\dfrac{1}{6}

G H = E F − E G − H F = E F − 2 E G = 1 − 2 ( 1 6 ) = 2 3 GH=EF-EG-HF=EF-2EG=1-2\left(\dfrac{1}{6}\right)=\dfrac{2}{3}

The required area is,

A = 1 2 ( G H + A B ) ( E A ) = 1 2 ( 2 3 + 1 ) ( 1 2 ) = 5 12 A=\dfrac{1}{2}(GH+AB)(EA)=\dfrac{1}{2}\left(\dfrac{2}{3}+1\right)\left(\dfrac{1}{2}\right) = \boxed{\dfrac{5}{12}}

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