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.

Inscribed Semicircle

Age 14 to 16 Short Challenge Level:

There are a number of different possible ways of solving this problem, which are presented below.

Solution 1

Since the semicircle touches the line $AB$, it is tangent there, so $\angle DEA = 90^\circ$. But then $\angle EAD = \angle CAB$, as these are the same angle. This means the triangles $ADE$ and $ABC$ are similar.

If the radius of the semicircle is $r$, then $DE = DC = r$ and $AD = 12-r$. Then, similarity says that:
$\frac{AB}{BC} = \frac{AD}{DE}$

Therefore:
$\frac{13}{5}=\frac{12-r}{r}$

Clearing the denominators gives:
$13r = 5(12-r)$

Expanding the brackets and collecting like terms gives:
$18r = 60$

Therefore, $r = 3\frac{1}{3} \text{cm}$.

Solution 2

Again, write $r$ for the radius of the semicircle.

As in solution 1, we can prove that $ADE$ is similar to $ABC$. Since the semicircle touches the line $AB$, it is tangent there, so $\angle DEA = 90^\circ$. But then $\angle EAD = \angle CAB$, as these are the same angle. This means the triangles $ADE$ and $ABC$ are similar.

Then, since $\angle DCB$ is a right angle, also, $CB$ and $EB$ are both tangents of the semicircle. Since they intersect, $EB = CB = 5$, so $AE = 13-5=8$.

Then, similarity says:
$\frac{DE}{AE}=\frac{BC}{AC}$

Therefore:
$\frac{r}{8}=\frac{5}{12}$

Multiplying by $8$ gives:
$r = \frac{40}{12} = 3\frac{1}{3}$

This means the semicircle has radius $3\frac 13 \text{cm}$.

Solution 3

This approach uses Pythagoras' theorem, rather than similarity, to solve the problem. Write $r$ for the radius of the semicircle.

Since $AB$ is tangent to the semicircle at $E$, $\angle AED$ is a right angle.

Also, as $BC$ is tangent to the semicircle at $C$, as $\angle ACD$ is a right angle, the lengths $BC$ and $BE$ are the same length, so both are $5\text{cm}$.

This then means that $AE = 8$, $DE = r$ and $AD = 12-r$.

Now, Pythagoras' theorem says that, since $\angle AED$ is a right angle:
$AD^2 = AE^2 + DE^2$

That is:
$(12-r)^2 = 8^2 + r^2$

Expanding the brackets gives:
$r^2 - 24r+144 = r^2 + 64$

Then, cancelling terms gives:
$24r = 80$

Therefore, $r = \frac{24}{80} = 3\frac 13$, so the radius of the semicircle is $3\frac 13\text{cm}$.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.