### Bang's Theorem

If all the faces of a tetrahedron have the same perimeter then show that they are all congruent.

### 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?

### Medallions

I keep three circular medallions in a rectangular box in which they just fit with each one touching the other two. The smallest one has radius 4 cm and touches one side of the box, the middle sized one has radius 9 cm and touches two sides of the box and the largest one touches three sides of the box. What is the radius of the largest one?

##### Stage: 4 Challenge Level:

Thank you for your excellent solutions Tan Chor Kiang, Raffles Institution, Singapore, Robert Goudie, Madras College, Fife, Scotland and James. Chor Kiang's and James' methods were very similar, and involved solving quadratic equations, but James used logarithms.

You have to find all real solutions of the equations $$(x^2-7x+11)^{(x^2-11x+30)} = 1.$$ This is Chor Kiang's solution: The left hand side can only be 1 if the base is 1, or -1 or the index is 0.

Hence, you have 3 cases.

Case 1 $x^2-11x+30 = 0$.

Conditions: None

\begin{eqnarray}x^2-11x+30 &=& 0 \\ (x-6)(x-5) &=& 0 \\ x - 6 &=& 0\; \mathrm{or} \; x-5&=&0\\ x &=& 6\; \mathrm{or} \; x &=& 5\end{eqnarray}

Case 2 $x^2-7x+11 = 1$

Conditions: None

\begin{eqnarray} x^2-7x+11 &=& 1 \\ x^2-7x+10 &=& 0 \\ (x-2)(x-5) &=& 0 \\ x - 2 &=& 0 \; \mathrm{or} \; x - 5 &=& 0 \\ x &=& 2\; \mathrm{or} \; x&=&5 \end{eqnarray}.

Case 3 $x^2-7x+11 = -1$

Conditions: Index must be an even number. Proof: This is already true, since the index is $(x-6)(x-5)$. Hence, one of the numbers must be an even number. When and even number multiplies an odd number, an even number results

\begin{eqnarray} x^2-7x+11 &=& -1 \\ x^2-7x+12 &=& 0 \\ (x-3)(x-4) &=& 0 \\ x - 3 &=& 0\; \mathrm{or} \; x - 4 &=& 0 \\ x &=& 3\; \mathrm{or} \; x &=& 4 \end{eqnarray}
. Therefore the possible solutions are $x = 2, \; 3, \; 4, \; 5$ or $6$.