• Feb 29th 2008, 05:07 PM
jmmsan
i'm going over same previous problems from the mad hatter math contest, and i need help with a few.
28. Two congruent 30-60-90 triangles are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is...?
--question: i don't understand how the triangles have coinciding hypotenuses but still overlap slightly---

36. A circle is inscribed in an equilateral triangle, a square is inscribed in the circle. The ratio of the areas of the triangle to the square is...?

38. If the sum of all the angles except one of a convex polygon is 2190 degrees, then the number of sides of the polygon must be...?

44. The perimeter of an isosceles right triangle is 2p, its area is...?
• Feb 29th 2008, 08:28 PM
earboth
Quote:

Originally Posted by jmmsan
...
36. A circle is inscribed in an equilateral triangle, a square is inscribed in the circle. The ratio of the areas of the triangle to the square is...?

...

Let s denote the side of the equilateral triangle. Then it's area is

$a_{\Delta} = \frac12 \cdot s \cdot \frac12 \cdot \sqrt{3} \cdot s = \frac{\sqrt{3}}{4} s^2$

The radius of the circle must be one third of the median = height of the triangle. Therefore the diagonal of the incribed square has the length of 2r. The area of a square can be calculated by:

$a_{square} = \frac12 \cdot d^2~\implies~ a_{square} = \frac12 \cdot \left(2 \cdot \frac13 \cdot \frac12 \cdot \sqrt{3} \cdot s \right)^2 = \frac16 \cdot s^2$

Now calculate the ration.
• Feb 29th 2008, 08:37 PM
earboth
Quote:

Originally Posted by jmmsan
...
44. The perimeter of an isosceles right triangle is 2p, its area is...?

Let s denote the length of one leg of the isosceles right triangle. The the length of the hypotenuse is:

$|h| = s \cdot \sqrt{2}$

The perimeter of the triangle is:

$s+s+s \cdot \sqrt{2} = 2p~\implies~s=\frac{2p}{2+\sqrt{2}}$

The area of this right triangle can be calculated by:

$a = \frac12 \cdot s^2$ because it is a half square:

$a = \frac12 \cdot \left(\frac{2p}{2+\sqrt{2}} \right)^2 = \frac{p^2}{3+2 \sqrt{2}}$
• Feb 29th 2008, 08:56 PM
earboth
Quote:

Originally Posted by jmmsan
...

38. If the sum of all the angles except one of a convex polygon is 2190 degrees, then the number of sides of the polygon must be...?

...

Could it be that there is a typo in your question and that you mean 2160° instead of 2190° ?

Let n denote the number of sides (= number of vertices) of the polygon. Determine a center inside the polygon. Then you can draw a triangle with each side of the polygon and the center. The sum of angles in each triangle is 180°. If you subtract the sum of angles at the center of the polygon (= 360°) then you get the sum of all interior angles of the polygon. But: This sum must be a multiple of 180 and 2190 is not a multiple of 180..

If (and only if) my assumption is correct you can set up the equation:

$n\left(180^\circ - \frac{360^\circ}{n}\right)=2160^\circ + 180^\circ$

Solve for n. I've got n = 15
• Mar 1st 2008, 06:27 AM
Soroban
Hello, jmmsan!

It took a while to figure out the overlapping right triangles . . .

Quote:

28. Two congruent 30-60-90 triangles are placed so that they overlap partly
and their hypotenuses coincide.
If the hypotenuse of each triangle is 12, the area common to both triangles is ... ?

Code:

            *                *           *    *    C    *    *           *        *.*        *         *        *  :  *        *         *    *      :h    *    *       *  * 30°      :          *  *       *  *  *  *  *  *  *  *  *  *  *       A      6              6      B

We want the area of $\Delta ABC$

Its base is 12.

And we can find $h$ . . . right?