Another of Alison’s Ace Puzzles, Revisited

This is a guest post from @ImMisterAl, who prefers to remain anonymous in real life. It refers to the problem in this post: a semi-circle is inscribed in a 3-4-5 triangle as shown; find $X$.

Insemicircle puzzle

As with any mathematical problem, my first thought was to sort out exactly what I know or can easily find, and what I want to know.

I know:

  • we have a right angled triangle with all the side lengths given, and I could use trigonometry to find the angles if I wanted to
  • we have a (semi)circle, so I could possibly use circle theorems for something

I want:

  • the length $x$, which is the base of the triangle minus the diameter of the circle

Hmm. So, if I find the radius of the circle I can easily double that to get the diameter, from which I can find $x$. But how to get to that radius?

Since I know that I can find any angle in the triangle, that's what I'll focus on. Let’s call the top vertex of the triangle point $B$ and the vertex at the right angle point $D$. Let the centre of the circle be $C$ and the point of tangency be $T$.

Drawing the line $BC$ forms two triangles, $BCD$ and $BTC$ which look as though they might be congruent. Are they congruent? Yes they are, since both these triangles are right-angled, and the Equal Tangents theorem tells us that $BD = BT$, and also $CD$ must equal $CT$ since they’re both just radii of the circle. That means the angle $CBD$ is half of angle $ABD$. That’s got to be useful, right?

(Confession time: At this point I reached for my calculator – yes, I know, and I’m sorry1 – and I got it to tell me angle $ABD$, which I divided by 2 to give angle $CBD$. I then considered the triangle $BCD$ and used the trigonometry buttons on my calculator again to tell me the radius of the circle. But when I saw how simple the final answer was, I realised there must be a neater way.)

Let’s call the angle $CBD$ $\theta$, which would make angle $ABD$ equal to $2\theta$. Triangle $CBD$ then tells us that $\tan(\theta)=\frac{r}{3}$ and from triangle $ABD$ we get $\tan(2\theta)=\frac{4}{3}$. Ah, but I know an identity that links those two things. Exciting stuff! Let’s piece it all together. Here’s the identity:

$\tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$

Substituting what we have into that gives us the equation:

$\frac{4}{3} = \frac{ 2 \left(\frac{r}{3}\right)}{1-\left(\frac{r}{3}\right)^2}$

This rearranges into the quadratic equation:

$2r^2 + 9r - 18 = 0$,

which has solutions $r=-6$ and $r=\frac{3}{2}$, and since the radius can’t be negative, $\frac{3}{2}$ must be our answer. From here it’s a simple step to show that $x$ equals 1.

Of course, I kicked myself when I later saw that using similar triangles gave a far more elegant solution, but I was pleased to use the double angle identity for $\tan$ in an actual problem. It’s always the identity that tends to get overlooked in my opinion.

Mister Al

  1. I want your maths badge on my desk by 9am tomorrow - Ed. []

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One comment on “Another of Alison’s Ace Puzzles, Revisited

  • Stephen Cavadino

    I drew a line as you did from c to t and considered the RAT cta (as angle cta is 90 as its a radius meeting a tangent).

    BT = BB (Tangents from a point are equal) hence TA =2.

    Pythagoras’s Theorem on cta gives

    (r+x)^2 = r^2 + x^2

    which simplifies to

    x^2 + 2rx – 4 =0 (1)

    from AD we know 2r + x = 4 hence 2r = 4- x (2)

    Subbing 2 into 1 gives 4x – 4 =0 and so x = 1 as required.

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