Geometry problem from IMO Past Papers.

Question

Each side of square ABCD is 2 unit in length. P is a point on BD such that
$\angle BCP = 30^{\circ}$. Area of triangle DCP can be expressed as $a-\sqrt{a}$ . Then, a= ?

Answer 1

Drop the height $PQ$ from $P$ onto $CD$. Let $x=DQ=QP$, then $QC=\frac{x}{\sqrt 3}$ and so we have (from $CD=DQ+QC$: $2=x(1+\frac{1}{\sqrt 3})$, i.e.

$$x=\frac{2}{1+\frac{1}{\sqrt 3}}=3-\sqrt 3$$

Because the area of $\triangle DCP$ is $\frac{1}{2}DC\cdot PQ=x$, we have the area equal to $3-\sqrt 3$. This would mean $a=3$.

It is possibly needed to check if there is another solution for $a$: if $a-\sqrt a=3-\sqrt 3$, then $(\sqrt a-\sqrt 3)(\sqrt a + \sqrt 3) = a-3=\sqrt a - \sqrt 3$, i.e. $(\sqrt a-\sqrt 3)(\sqrt a+\sqrt 3-1)=0$. As the second factor is $\gt 0$, the first factor must be $0$, i.e. $\sqrt a=\sqrt 3$, implying $a=3$.

Answer 2

Without trig...

CPQ is an equilateral triangle.

M is the midpoint of PQ

$b = \frac 12 c\\ a^2 + b^2 = c^2\\ a = b\sqrt 3\\ a+b = 2$

the area of $DCP = a$

$a (1 + \frac {\sqrt 3}{3}) = 2\\ a = \frac {6}{3 + \sqrt 3}\\ a = DCP = 3-\sqrt 3$

I am using $a$ differently than you used $a$ in the Original Post. I hope that isn't too confusing.