### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

### Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?

### Our Ages

I am exactly n times my daughter's age. In m years I shall be ... How old am I?

# Are You Kidding

##### Stage: 4 Challenge Level:

An excellent solution from Sheila Norrie of Madras College, who not only managed to find that the area was 48 square units, but was also able to say that there are an infinite number of isosceles triangles with a whole number perimeter and a whole number area. Andrei Lazanu of School 205, Bucharest was also able to establish that the area was 48 square units using a computer program, given below. Both solutions involved exhaustive methods! Well done.

Here is part of Sheila's solution:

 We know from Pythagoras theorem that $h^2 + x^2 = c^2$ From this we can say that $c =\sqrt{( h^2 + x^2 )}$ So the perimeter must equal $\sqrt{( h^2 + x^2 )} + \sqrt{( h^2 + x^2 )} + 2x$ We know the perimeter is 32 so: $$\sqrt{( h^2 + x^2 )} + \sqrt{( h^2 + x^2 )} + 2x = 32$$ and we know $h = 8$ $$\sqrt{( 8^2 + x^2 )} + \sqrt{( 8^2 + x^2 )} + 2x = 32$$ $$\sqrt{( 8^2 + x^2 )} + x = 16$$

To find a solution Sheila then used trial and improvement - testing for values of x from 1 to 6.

If x = 6: $\sqrt{( 8^2 + 6^2 )} + 6 = 16$

x = 6 is the only solution because nothing works for under 6. If x was greater, the perimeter would have to increase which it can't.

We know the area of the triangle = 1/2 bh

For this triangle

• h = 8
• b = 12 (2 x 6)

So area = 48

Part 2

For whole number areas and perimeters, the lengths h, x and c have to be whole numbers. In addition h 2 + x 2 must equal c 2

So we are looking for Pythagorean triples. Since there are an infinite number of Pythagorean triples there must be an infinite number of triangles.

Andrei's program in Matlab that he used to calculate the Pythagorean Triples up to a perimeter of 100 units.

Here is the program:

 Pythagorean Triples i=1; for a=1:100; a1=a^2; for b=1:100; b1=b^2; for c=1:100; c1=c^2; if c1-b1-a1=0 s=a*b; p=2*(c+a); if p< =100 v(i,1:5)=[a b c p s]; i=i+1; end end end end end