2 Kites & 1 Triangle Riddle

Draw diagonals $DG$ and $EF$ intersecting at $H$ , diagonals $CF$ and $BG$ intersecting at $I$ , and a circle centered at $G$ with a radius of $17$ passing through $C$ , $E$ , and $F$ .

Since $CE$ is a diameter of the circle, by Thales' Theorem $\angle CFE = 90°$ . Since all side lengths between any points within the kites are whole integers, $EF$ and $CF$ are integers; therefore, the sides of $\triangle CEF$ are a Pythagorean triple with a hypotenuse of $34$ , which could only be $(16, 30, 34)$ . Therefore, $EF = 16$ and $CF = 30$ .

By properties of diagonals of kites, $\angle FIB = 90°$ and $FI = CI$ ; and since $CF = 30$ , $FI = CI = 15$ . Similarly, $\angle DHF = 90°$ and $FH = HE = 8$ .

$\triangle DHF$ and $\triangle FIB$ are similar by AA similarity, so $\frac{DF}{BF} = \frac{DH}{FI}$ . If $x = DF$ , then $BF = 35 - x$ and by Pythagorean's Theorem on $\triangle DHF$ , $DH = \sqrt{x^2 - 8^2}$ . Now the similarity ratio is $\frac{x}{35 - x} = \frac{\sqrt{x^2 - 8^2}}{15}$ , which has one integer solution of $x = 10$ .

Using Pythagorean's Theorem and the similarity ratio, $DE = 10$ , $DH = 6$ , $HG = 15$ , $GI = 8$ , $BI = 20$ , and $BF = BC = 25$ ; and combining the area of each right triangle gives trapezoid $BCED$ an area of $A_{BCED} = 588$ .

$\triangle ADE$ and $\triangle ABC$ are similar by AA similarity, and has an area ratio of $\frac{DE^2}{BC^2} = \frac{10^2}{25^2} = \frac{4}{25}$ . If $A_{\triangle ADE}$ is the area of $\triangle ADE$ , then $\frac{4}{25} = \frac{A_{\triangle ADE}}{A_{\triangle ADE} + 588}$ , which solves to $A_{\triangle ADE} = \boxed{112}$ .

Let us add the diagonals to the kites stated in the question and designate the intersection points as $H$ within the green kite and $I$ within the red one, as shown above.

Since $DE || BC$ , the triangles $ADE$ and $ABC$ are similar, and so we will attempt to find the ration between $BC:DE$ . Suppose $\dfrac{BC}{DE} = k$ for some constant $k$ . Then the area of $ABC$ is $k^2$ times that of $ADE$ .

In order to find such ratio, we will determine the desired lengths by investigating the diagonals in both kites. First, the diagonals $BG$ and \GD) both bisect the red and the green kites respectively. Thus, $\angle CGB = \angle BGF$ and $\angle FGD = \angle DGE$ , and since these four angles add up to a straight angle, $180 = 2\angle BGF + 2\angle FGD$ . Hence, we can prove that $90 = \angle BGF + \angle FGD = \angle BGD$ , the right angle.

Then due to the kite properties, the diagonals within will be perpendicular to each other, making both $\angle FIG$ and $\angle FHG$ right angles as well. As a result, $FIGH$ is a rectangle with the diagonal $FG = 17$ , as in the question.

Then let $FH = x$ and $FI = y$ and $FD = DE = z$ . According to Pythagorean theorem, $x^2 + y^2 = 17^2$ . With $FIGH$ as the rectangle, $BG || FE$ and $FC || DG$ , and so the triangles $BIF$ and $FHD$ are similar. Also, because, $BC = BF$ and $DE = DF$ , the ratio of $BF:FD$ is also $k$ as well. Here fore, $k = \dfrac{35-z}{z}$ . Then considering the triangle $BIF$ , $(kx)^2 + y^2 = (kz)^2$ , and for $FHD$ , $x^2 + (\dfrac{y}{k})^2 = z^2$ .

Now we will arrange the system of equations for these variables:

$x^2 + y^2 = 17^2$ $(\dfrac{x(35-z)}{z})^2 + y^2 = (35-z)^2$ $x^2 + (\dfrac{y(z)}{35-z})^2 = z^2$

Solving the equations, we will obtain $x = 8$ ; $y=15$ ; and $z = 10$ . Then $k = \dfrac{35-10}{10} = \dfrac{5}{2}$ , and so $BC = BF = 25$ . Consequently, $BI = \dfrac{8\times 5}{2} = 20$ , and $DH = \dfrac{15\times 2}{5} =6$ .

Furthermore, the area of the red kite $BCGF$ = $\dfrac{1}{2}\times 2 \times 15 \times (20+8) = 420$ .

The area of the green kite $DEGF$ = $\dfrac{1}{2}\times 2 \times 8 \times (15+6) = 168$ .

Then let $S$ be the solution area of the triangle $ADE$ . We can write $k^2 S = S + 168 + 420 = S + 588$ .

Thus, $(k^2 -1)S = 588 = (\dfrac{25-4}{4}) S = (\dfrac{21}{4}) S$ .

Finally, $S = \dfrac{4\times 588}{21} = \boxed{112}$ .