A Very Simple Plane Geometry Problem From The Spanish Math Olympiad.

This problem from the Spanish Mathematical Olympiad, requires no special knowledge for solving it. Just some basic concepts with a little bit of smart thinking can do the trick.

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Let us extend the line $PQ$ to meet $AB$ at $R$ as shown in the figure below:

Let each side of the square be of length $x$ units.

The area of the trapezium $AQPD$ is

$\dfrac{1}{2} \times PD \times (AD + QP) = \dfrac{1}{2} \times PD \times (x + QP)$

Similarly, the area of the trapezium $BQPC$ is:

$\dfrac{1}{2} \times PC \times (BC + QP) = \dfrac{1}{2} \times PC \times (x + QP)$

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Therefore,

$\dfrac{area[AQPD]}{area[BQPC]} = \dfrac{\cancel{\dfrac{1}{2}} \times PD \times \cancel{(x + QP)}}{\cancel{\dfrac{1}{2}} \times PC \times \cancel{(x + QP)}} = \dfrac{9}{4}$

$\Rightarrow PD:PC = 9:4$

Let $PD = 9k$ and $PC = 4k$

$\therefore AR = 9k$ and $RB = 4k$

Using right angled $\triangle AQB$ and altitude $QR$, we know:

$\triangle ARQ \sim \triangle QRB$

Therefore,

$\dfrac{AR}{QR} = \dfrac{QR}{RB}$

Therefore,

$QR^2 = AR \times RB$

$QR^2 = 9k \times 4k$

$\Rightarrow QR = 6k$

Considering the area of $AQPD$, we have

$area[AQPD] = \dfrac{1}{2} \times PD \times (AD + QP) = 9$

$\Rightarrow area[AQPD] = \dfrac{1}{2} \times 9k \times (13k + 7k)$

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$\therefore 9k \times (13k + 7k) = 18$

$\Rightarrow k \times 20k = 2$

$\Rightarrow k^2 = \dfrac{1}{10}$

$\Rightarrow k = \dfrac{1}{\sqrt{10}}$

The area of $\triangle AQB = \dfrac{1}{2} AB \times QR$

$\Rightarrow area[\triangle AQB] = \dfrac{1}{2} \times 13k \times 6k$

$\Rightarrow area[\triangle AQB] = 39k^2$

$\Rightarrow area[\triangle AQB] = 39 \times \dfrac{1}{10}$

$\Rightarrow ar[\triangle AQB] = \dfrac{39}{10}$

$\therefore a = 39$ and $b = 10$ and $a + b = 49$