Day 18: An Incredible Result

It is possible to use trigonometry or calculus but here is an algebraic proof:

As we enlarge the diagram the ratio remains constant so we can assume that $AB=2$ without loss of generality. Now name some points as follows:

$ABC$ is similar to $EFC$ and $EF=1$ . Now let $FC=a$ . Then $BC=2a$ and $EC =2a-1$ . Now by Pythagoras' Theorem on $EFC$ , $(2a-1)^2=a^2 +1^2$ Solving gives $a=\frac{4}{3}$ and therefore $\frac{AB}{BC} = \frac{3}{4}$ which gives the answer $3 + 4 = \boxed{7}$ as required.

If you can't see the image here, go to https://i.imgur.com/yyJ0BeC

Sorry if the solution is not legible, but this seems easier than LaTeX :P

Let the center of the semicircle be $O$ , where it touches $AC$ at $P$ and its radius be $OP = 1$ . Also let $DC = x$

We note that $\triangle ABC$ and $\triangle OPC$ are similar. Therefore,

$\dfrac {AC}{AB} = \dfrac {OP}{OC}\quad \Rightarrow \dfrac {\sqrt{AB^2+ BC^2}}{AB} = \dfrac {1+x}{1}$

$\Rightarrow \dfrac {\sqrt{2^2+ (2+x)^2}}{2} = 1+x\quad \Rightarrow \sqrt{4+ 4+4x+x^2} = 2(1+x)$

$\Rightarrow x^2 + 4x + 8 = 2^2(1+x)^2 \quad \Rightarrow x^2 + 4x + 8 = 4x^2 + 8x + 4$

$\Rightarrow 3x^2 + 4x - 4 = 0\quad \Rightarrow (3x-2)(x+2)=0\quad \Rightarrow x = \frac {2}{3}$ as $x > 0$ .

Now, $\dfrac {AB}{BC} = \dfrac {2}{2+x} = \dfrac {2}{2+\frac {2}{3}} = \dfrac{3}{4}$

$\Rightarrow a = 3$ , $b = 4$ and $a+b = 3+4 = \boxed {7}$ .

To make things easier to understand, I will refer to the diagram in Michael Ng's answer. If we reflect the semicircle across $BD$ , segment $AB$ would be tangent to the resulting circle. This means that $AB=AF$ since $AF$ is also tangent to the resulting circle. Now $AB=AF=BD=a$ , $ED=EF=\frac{a}{2}$ , $DC=b$ , and $FC=c$ .

Pythagorean Theorem over triangle ABC gives us: $a^2+(a+b)^2=(a+c)^2$ which simplifies to: $a^2+2ab+b^2=2ac+c^2$

Pythagorean Theorem over triangle EFC gives us: $\left(\frac{a}{2}\right)^2+c^2=\left(\frac{a}{2}+b\right)^2$ which simplifies to: $c^2=ab+b^2$

Plug in for c in the first equation and cancel to get: $a^2+ab=2a \sqrt{ab+b^2}$

This cancel a's and square both sides to get: $(a+b)^2=4(a+b)b$

Cancel (a+b) and solve for be to get $b=\frac{a}{3}$

This means that $\frac{AB}{BC}=\frac{3}{4}$ so the answer is 7.