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Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

Triangle Midpoints

You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle?

Fermat's Poser

Find the point whose sum of distances from the vertices (corners) of a given triangle is a minimum.

3-4-5 Circle

Age 14 to 16 Short Challenge Level:
Using area
Imagine drawing lines from each vertex to the centre of the circle, as shown below.
 
This gives us 3 triangles all of height $r$ and of bases $3, 4$ and $5$ respectively (the angle between the radius and the tangent is $90^\text{o}$, so these can be considered the bases and heights). Thus, the areas of the triangles are $\frac{1}{2}\times 3r$, $\frac{1}{2}\times 4r$ and $\frac{1}{2}\times 5r$ respectively.
The whole triangle has area $\frac{1}{2} \times 3 \times 4 = 6$.
We know that the areas of the smaller triangles must sum to give the area of the whole triangle so we have that:
$\frac{1}{2}\times 3r + \frac{1}{2}\times 4r + \frac{1}{2}\times 5r = 6$
$\Rightarrow \frac{1}{2}\times 12r = 6$
$\Rightarrow 6r=6$
$\Rightarrow r=1$.

Using perimeter
The angle between a radius and a tangent is  $90^\circ$, which is shown on the diagram below.
 
Three of the angles in the quadrilateral at the bottom left are right angles, so the fourth angle must be a right angle too. This means that it must be a square, since two adjacent sides both have length $r$.

This is shown in the diagram below, and the other parts of the sides of length 3 and 4 are also shown in terms of $r$. 

Using symmetry, or the circle theorem that the lengths of two tangents drawn from a point to the circumference are equal, or by drawing a line from the vertex to the centre and observing similar triangles, these lengths can also be labelled on the side of length 5, as shown below.
 
This means that  $4-r$ and $3-r$ must add up to 5, so
$4-r+3-r=5$, so $7-2r=5$, so $2r=2$, so $r=1$. 
You can find more short problems, arranged by curriculum topic, in our short problems collection.