Tilted Triangles

This problem is similar to this problem .

The three white triangles all have angles that are $60°$ , $\theta$ , and $120° - \theta$ and one side congruent to the side of the blue equilateral triangle, so they are all congruent by the ASA congruency theorem.

Let $A$ and $a$ be the area and side of the green equilateral triangle, $B$ and $b$ be the area and side of the red equilateral triangle, $C$ be the area of the blue equilateral triangle, $D$ be the area of each of the white triangles, and $E$ be the area of the large equilateral triangle. The sides of $E$ are then $a + b$ , and $E = C + 3D$ .

By triangle area equations,

$A = \frac{\sqrt{3}}{4}a^2$

$B = \frac{\sqrt{3}}{4}b^2$

$D = \frac{1}{2}ab \sin 60° = \frac{\sqrt{3}}{4}ab = \sqrt{AB}$

$E = \frac{\sqrt{3}}{4}(a + b)^2 = \frac{\sqrt{3}}{4}a^2 + 2\frac{\sqrt{3}}{4}ab + \frac{\sqrt{3}}{4}b^2 = A + 2D + B = A + B + 2\sqrt{AB}$

$C = E - 3D = (A + B + 2\sqrt{AB}) - 3\sqrt{AB} = A + B - \sqrt{AB}$

In this question, $A = 50$ and $B = 98$ , so $C = 50 + 98 - \sqrt{50 \cdot 98} = \boxed{78}$ .

The ratio of side lengths of the red and green equilateral triangles is $\dfrac {\sqrt{98}}{\sqrt{50}} = \dfrac 75$ . Let the side lengths of the red and green equilateral triangles be $7k$ and $5k$ respectively, and the side length of the blue equilateral triangle be $a$ . Let angle between the right side of the blue triangle and the horizontal line be $\theta$ . Then the angle between the left side of the blue triangle with the horizontal line is $120^\circ - \theta$ . By sine rule , we have: $\dfrac {\sin \theta}{7k} = \dfrac {\sin 60^\circ}a$ and $\dfrac {\sin (120^\circ - \theta)}{5k} = \dfrac {\sin 60^\circ}a$ . Then

$\begin{aligned} \frac {\sin \theta}{7k} & = \frac {\sin (120^\circ - \theta)}{5k} \\ \frac {\sin \theta}7 & = \frac {\frac {\sqrt 3}2\cos \theta + \frac 12 \sin \theta} 5 \\ 5 \sin \theta & = \frac {7\sqrt 3} 2 \cos \theta + \frac 72 \sin \theta \\ 3 \sin \theta & = 7 \sqrt 3 \cos \theta \\ \tan \theta & = \frac 7{\sqrt 3} \\ \implies \sin \theta & = \frac 7{2\sqrt{13}} \end{aligned}$

By sine rule again:

$\begin{aligned} \frac a{\sin 60^\circ} & = \frac {7k}{\sin \theta} \\ \frac {2a}{\sqrt 3} & = \frac {7k \cdot 2\sqrt{13}}7 \\ \implies a & = \sqrt{39}k \end{aligned}$

Since the area of similar triangle is directly proportional to the square of the side length, the area of the blue equilateral triangle is $\dfrac {39k^2}{49k^2} \cdot 98 = \boxed{78}$ .

Red area = 2 x 7² = 98

Green area = 2 x 5² = 50

White + blue area
= 2 x (5 + 7)² = 288

1 White area = 2 x 5 x 7 = 70

Blue area
= {White + blue area} - 3 x {white area}
= 288 - 3 x 70
= 288 - 210
= 78