contest-mathgeometrypuzzle
Given: in the figure below, ABCD is a square, and DH=CG=FB=AE. Areas of quadrilaterals green, red and blue are given in the figure.
Find: area of the orange quadrilateral.
I'm struggling with this problem. I can see that we can find congruent triangles HDG, GCF, FBE and AEH. And most likely this is useful in the solution, but I don't see how.
Hints and solutions welcomed. Sorry if this is a dup.
The following solution is not very elegant, but it gets the job done. Then again, the desired answer is not itself very pretty, so this will do.
Let the vertices of the triangle adjacent to the green, yellow, and blue areas be $A$, $B$, and $C$ respectively. Suppose the barycentric coordinates of the point $P$ are $\alpha$, $\beta$, and $\gamma$, so that $$\begin{align} \alpha A + \beta B + \gamma C &= P, \\ \alpha + \beta + \gamma &= 1. \end{align}$$ Let the area of the whole triangle be $\triangle$. Then the area of the red region is $\alpha\triangle$, because the height of $P$ from $BC$ is $\alpha$ times that of $A$. The yellow and red together form a triangle whose third vertex divides $BC$ in the ratio $\alpha:\beta$, so its area is $\triangle\alpha/(\alpha+\beta)$. Similarly, the area of the blue and red together is $\triangle\alpha/(\alpha+\gamma)$. (All this should become clear from looking at the diagrams in the MathWorld article on barycentric coordinates.) Solving, you get $$\begin{align} \beta &= \frac B{B+R}, \\ \gamma &= \frac Y{Y+R}, \\ \alpha &= 1 - \beta - \gamma = \frac{R^2-BY}{(B+R)(Y+R)}, \\ \triangle &= \frac R\alpha = \frac{R(B+R)(Y+R)}{R^2-BY}. \end{align}$$ The remaining green area is of course $\triangle - R - B - Y$, which simplifies to the desired result.
Quadrilaterals are not rigid: you can fix a side and move one of the other vertices freely while respecting all side lengths. The area changes when you move the vertex.
(image from http://www.mathsisfun.com/definitions/rigid.html)
Best Answer
Answer: $93+69-60=102$.
Solution: