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Bang's Theorem

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

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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?

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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?

Converse

Stage: 4 Challenge Level: Challenge Level:1

Clearly if $a$, $b$ and $c$ are the lengths of the sides of a triangle and the triangle is equilateral then \[a^2 + b^2 + c^2 = ab + bc + ca.\]

Is the converse true, and if so can you prove it? That is if $a^2 + b^2 + c^2 = ab + bc + ca$ is the triangle with side lengths $a$, $b$ and $c$ necessarily equilateral?

Again you don't require much mathematical knowledge to do this, just the ability to use elementary algebra. Here is a very neat solution from Koopa Koo, Boston College, USA.

\[a^2 + b^2 + c^2 -ab - bc - ca = 0\]

implies that

\[(1/2)[(a - b)^2 + (b - c)^2 + (c - a)^2] = 0\]

which implies $a = b = c$. So, the converse is also true.

Another, rather clumsier, method is to consider the expression

\[a^2 + b^2 + c^2 -ab - bc - ca = 0\]

as a quadratic equation for $a$ in terms of $b$ and $c$, namely:

\[a^2 - a(b + c) + (b^2 + c^2 - bc) = 0,\]

and then the condition for this quadratic equation to have real roots requires that $a = b = c.$