A geometry problem by A Former Brilliant Member

Geometry Level 3

Triangle $ABC$ is an isosceles right angled triangle with $\angle C = 90^\circ$ and $|AC| = |BC| = 12$ . Point $M$ is the midpoint of side $AB.$ Point $D$ on side $AC$ is such that $|CD|=3$ . Point $E$ is the intersection point of line segments $CM$ and $BD$ .

If the area of quadrilateral $AMED$ is $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a +b$ .

The answer is 167.

1 solution

Chew-Seong Cheong
Dec 8, 2018

Drop an altitude from $E$ to $AC$ at $F$ . Let $|EF| = |CF| =x$ . We note that $\triangle EFD$ is similar to $\triangle BCD$ . Then $\dfrac {|FD|}{|EF|} = \dfrac {|CD|}{|BC|} = \dfrac 3{12} = \dfrac 14$ $\implies \dfrac {|FD|}x = \dfrac 14$ $\implies |FD| = \dfrac 14x$ . Since $|CF|+|FD| = |CD|$ or $x+\dfrac 14x = 3$ $\implies x = \dfrac {12}5$ .

Then the area of quadrilateral $AMED$ is given by:

$\begin{aligned} [AMED] & = [AMC] - [CED] \\ & = \frac 12 [ABC] - \frac 12 x \times 3 \\ & = \frac 12 \times \frac {12\times 12}2 - \frac 12 \times \frac {12}5 \times 3 \\ & = 36 - \frac {18}5 = \frac {162}5 \end{aligned}$

Therefore, $a+b = 162+5 = \boxed{167}$ .

I think it is $|EF| = |CF|=x$ not $AF$ .

A Former Brilliant Member - 2 years, 6 months ago

Thanks. It is "It is" or "it's" and not "its". Just use "it is". It is a cat and the cat has its tail.

Chew-Seong Cheong - 2 years, 6 months ago

Oh sure I will:)

A Former Brilliant Member - 2 years, 6 months ago

Or, Sir, we could set up the co-ordinate axes about BC and CA............and then simply bash it out using co-ordinate geometry...........!!!

Aaghaz Mahajan - 2 years, 6 months ago

Yes, I did that first. But I think the solution I gave here is more trigonometrical.

Chew-Seong Cheong - 2 years, 6 months ago

A geometry problem by A Former Brilliant Member

The answer is 167.

1 solution

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