Question

The diagram below shows twelve 30-60-90 triangles placed in a circle so that the hypotenuse of each triangle coincides with the longer leg of the next triangle. The fourth and last triangle in this diagram are shaded. The ratio of the perimeters of these two triangles can be written as m/n where m and n are relatively prime positive integers. Find m + n

Answer 1

Answer:

m+n = 337

Step-by-step explanation:

Lets start from the first triangle, It is given to be as 30-60-90.

The hypotenous of first rectangle be 2x, then other sides are by default x and

$\sqrt{3}(x)$ , using laws of trigonometry.

(sin(30) = 0.5  and cos(30) = $\frac{\sqrt{3}}{2}$)

The perimeter of first triangle is ,

= $2x + x + \sqrt{3}(x) = x(3+\sqrt{3}) = \sqrt{3}x(1+\sqrt{3})$

Now, for second triangle, the longer leg is 2x, and similarly again,

other 2 sides are $\frac{2x}{\sqrt{3}} and \frac{4x}{\sqrt{3}}$.

Again the perimeter of triangle comes out as,

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

Thus, the repeating pattern is identified. The consecutive perimeters differ by, multiplying by factor $\frac{2}{\sqrt{3}}$

Thus, we can say that perimeter of 4th triangle is,

= $(\frac{2}{\sqrt{3}})^{3}(X)$, where X is the repeating constant.

And of 12th triangle is,

= $(\frac{2}{\sqrt{3}})^{11}(X)$,

Evaluating the above ratio, we get,

= $\frac{81}{256}$

Thus, m =256 and n=81.

Thus, m+n = 256+81 = 337.