AIME Problem 4

Geometry Level 4

In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$ .

This problem is part of this set .

The answer is 539.

3 solutions

Chew-Seong Cheong
Mar 21, 2015

We note that $\triangle CDE \equiv \triangle JFC \equiv \triangle HBJ \equiv \triangle NKH \equiv \triangle MAN$ are similar triangles with side lengths in the ratio of $1:2:\sqrt{5}$ .

Now, let the side lengths of square $ABCD$ , $KLMN$ and $FGHJ$ be $a$ , $b=\sqrt{99} = 3\sqrt{11}$ and $x$ respectively. Then, we have:

$a = \overline{AB} = \overline{AN} + \overline{NH} + \overline{HB} = \frac {1}{\sqrt{5}}b + \frac {\sqrt{5}}{2}b + \frac {2}{\sqrt{5}}x$

$a = \overline{BC} = \overline{BJ} + \overline{JC} = \frac {1}{\sqrt{5}}x + \frac {\sqrt{5}}{2}x$

Equating the two equations, we have:

$\begin{aligned} \frac {1}{\sqrt{5}}b + \frac {\sqrt{5}}{2}b + \frac {2}{\sqrt{5}}x & = \frac {1}{\sqrt{5}}x + \frac {\sqrt{5}}{2}x \\ \frac {1}{\sqrt{5}}x + \frac {\sqrt{5}}{2}x - \frac {2}{\sqrt{5}}x & = \frac {1}{\sqrt{5}}b + \frac {\sqrt{5}}{2}b \\ \frac {\sqrt{5}}{2}x - \frac {1}{\sqrt{5}}x & = \frac {1}{\sqrt{5}}b + \frac {\sqrt{5}}{2}b \\ \left(\frac {5-2}{2\sqrt{5}}\right) x & = \left(\frac {2+5}{2\sqrt{5}}\right) b \\ 3x & = 7 b \\ \Rightarrow x & = 7\sqrt{11} \\ \end{aligned}$

The area of square $FGHJ = x^2 = 49 \times 11 = \boxed{539}$

$All ~~\triangle s ~~have~~ sides~~in ~~ratio~~of~~1 : 2: \sqrt5.\\ Let~AB=BC=p,~~BJ=n,~~AM=m.\\\therefore HJ=\sqrt5n.....(1)~~~JG=JC* \dfrac{2}{\sqrt5}=( p-n )*\dfrac{2}{\sqrt5} ...(2)\\(1)~ =~ (2),~~{\sqrt5} n=( p-n )*\dfrac{2}{\sqrt5} ,~~\therefore~n=\dfrac{2p}{7}...(3) \\\therefore~HB=2n=\dfrac{4p}{7}....(4)\\Also~~NM=\sqrt5 m.........(5)\\Using ~~(4), ~~NK=NH*\dfrac{2}{\sqrt5}=(AB-AN-HB)*\dfrac{2}{\sqrt5}\\=(p-m-\dfrac{4p}{7})*\dfrac{2}{\sqrt5} =(\dfrac{3p}{7} -m)*\dfrac{2}{\sqrt5}....(6) \\(5)~ =~ (6),~~{\sqrt5} m=((\dfrac{3p}{7} -m)*\dfrac{2}{\sqrt5} ,~~\therefore~m=\dfrac{6p}{7^2}...(7) \\From ~~(3)~and~(7)~~Area~~FGHJ= \dfrac{5*\dfrac{(2p)^2}{7^2}}{5*\dfrac{(6p)^2}{49^2} }*99=\boxed {\color{#D61F06}{539} }\\\large \text{You method is better. I have given above just little }\\\text{ different method. We can use analytical geometry also.}$

Niranjan Khanderia - 6 years, 2 months ago

Nice and simple

Ivan Martinez - 6 years, 2 months ago

Ahmad Saad
Dec 4, 2016

Arnab Banerjee
Mar 30, 2015

An excellent one!!!!!

AIME Problem 4

The answer is 539.

3 solutions

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