Thread: mad hatter practice problems

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...?

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



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:



Now calculate the ration.

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:



The perimeter of the triangle is:



The area of this right triangle can be calculated by:

because it is a half square:

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:



Solve for n. I've got n = 15

Hello, jmmsan!

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

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

Its base is 12.

And we can find . . . right?