Another of Alison’s Ace Puzzles

A nice puzzle this week, via NRICH's magnificent @ajk44: a semicircle is inscribed in a 3-4-5 triangle as shown. Find $X$.

Insemicircle puzzle

I think it's a nice puzzle because Alison's way of doing it was entirely different to mine, but thankfully got the same answer. You might like to try it yourself.


My way

My way was to consider similar triangles. Draw a perpendicular at the point of tangency, T; that crosses the '4' side at the centre, C. Now, the triangle ATC is similar to the original triangle (it also has a right angle and shares the angle at A, so all three angles are the same.)

Supporting triangle

In particular, that means $\frac{r}{r+X} = \frac{3}{5}$. We also know that $2r+X = 4$, because of how the base is split up.

Writing the first fraction as $\frac{2r}{2r+2X}$ to make the substitution easier, we get $\frac{4-X}{4+X} = \frac{3}{5}$, and it's clear that $X=1$.

Alison's way

Alison's way was neater, I think. She used the Equal Tangents theorem (the one I mentioned last week) to spot that $T$ splits the hypotenuse in a ratio of 3:2.

Using the same triangle as me, she got $\frac{r}{2} = \frac{3}{4}$, so $r = \frac{3}{2}$ and $X = 1$.

A third way

If you're a coordinate geometry sort of person, you might wonder about the equation of the hypotenuse and of the semi-circle. The hypotenuse is $4y+3x=12$, assuming the right angle is at the origin, and the semicircle is $(x-r)^2 + y^2 = r^2$, for some unknown value of $r$.

Those two can be combined into a quadratic – and for the right value of $r$, it will have only one solution.

Multiplying out the semicircle gives $x^2 – 2rx + y^2 = 0$. I'd multiply that through by 16 to give $16x^2 – 32rx + (4y)^2 = 0$, reasoning that I can then easily substitute $4y = 12 – 3x$ into it.

Expanding, $16x^2 – 32rx + (12-3x)^2 = 16x^2 – 32rx + 144 – 72x + 9x^2$.

Simpifying: $25x^2 – (32r + 72)x + 144 = 0$.

This needs to have a repeated root, so its discriminant must be 0: here, $b^2 – 4ac = (32r + 72)^2 – 14,400$

So, $(32r + 72)^2 = 14,400$, which means $32r + 72 = 120$, or $32r = 48$. Aha! $r = \frac{3}{2}$ as before.


I'm sure there are other, equally nice, approaches. What other ways might you attack it?

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

2 comments on “Another of Alison’s Ace Puzzles

  • Stephen Cavadino

    I started down the co-ordinate geometry route then aborted when I realised i could draw similar triangles and ended up with the same solution as you.

  • Stephen Cavadino

    Oddly, I have just seen the other post on this puzzle, forgot that id solved it before and did it again using an entirely different method to the two I tried here.

Leave a Reply

Your email address will not be published. Required fields are marked *

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter