Shape and Territory

If for any triangle ABC tan(A - B) + tan(B - C) + tan(C - A) = 0 what can you say about the triangle?

Napoleon's Hat

Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?

The Root Cause

Prove that if a is a natural number and the square root of a is rational, then it is a square number (an integer n^2 for some integer n.)

Mind Your Ps and Qs

Stage: 5 Short Challenge Level:

Leo, Lily, Pippa and Lewis from Fakenham College sent us this entertaining solution

We transferred the equations onto cards so we could move them about easily. Good idea!

These are the ones we got to match, then we coloured them in and added glitter.Glittery maths? Excellent idea .

$x< 0 \Leftrightarrow x\int^x_0 ydy < 0$

$x=0 \Rightarrow \int^x_0 \cos y dy =0$

$0< x< 1\Rightarrow \cos(x/2)> \sin(x/2)$

$x> 2\Rightarrow x^3> 1$

$x> 4\Rightarrow 2\int^{x^2}_0ydy> x^2$

$x=-2\Rightarrow |x|> 1$

$x=1\Rightarrow x^2+x-2=0$

We didn't manage to get four of < => , but were pleased we got this :)

You should be pleased! This was good logical thinking. These seven were all correct, but there was a mistake in an eighth one you included: $x^2+4x+4 =0 \Rightarrow x> 1$. Can you see why?

The full solution that we obtained was

$x^3> 1 \Leftrightarrow x> 1$
$2\int^{x^2}_0ydy> x^2 \Leftrightarrow |x|> 1$
$x=-2 \Leftrightarrow x^2+4x+4 =0$
$x=1\Rightarrow x^2+x-2=0$
$0< x< 1 \Rightarrow \cos(x/2)> \sin(x/2)$
$x> 4\Rightarrow x> 2$
$x< 0 \Leftrightarrow x\int^x_0 ydy < 0$
$x=0 \Rightarrow \int^x_0 \cos y dy =0$