Those are fine circles!

Geometry Level 4

A circumcentre $O_{1}$ of the first circle of radius $r$ , is located on a circumference of the second circle of radius $2r$ ( $O_{2}$ is its circumcentre). Points $A$ and $B$ are common points of circumferences. Segments $AB$ and $O_{2}O_{1}$ intercept in point $X$ . An appropriate image of all this is provided above.

If $\frac{O_{2}X}{O_{2}O_{1}} = \frac{a}{b}$ , where $a$ and $b$ are coprime positive integers, find $a + b$ .

The answer is 15.

2 solutions

Drex Beckman
Jan 17, 2016

I didn't use Heron's formula. I'm not clever enough. I knew the perpendicular bisector of course divides two equal radii in half, so it is reasonable to assume that a radius of half would be divided into $\frac{1}{2^{2}}=0.25$ . So we know from the information given that $O_{2} X=1.75$ , and the ratio is $\frac{1.75}{2}$ . If you multiply 1.75 by 4, You get 7 and 4 multiplied by 2 is 8. You get the two desired numbers.

Interesting logic. I didn't think that way at all.

Milan Milanic - 5 years, 4 months ago

Milan Milanic
Jan 17, 2016

Solution:

I will be using labels from the image.

Area of $\triangle O_{1}AO_{2}$ is $\frac{r^{2}}{4}\sqrt{15}$ (using Heron's formula ).

Length of $AX = \frac{r}{4}\sqrt{15}$ . Using Pythagorean theorem we get that $O_{2}X = \frac{7}{4}r$ . Therefore, $a = 7$ and $b = 8$ .

Solution is $\boxed{15}$ .

Hey, nice problem. Interesting property of the perpendicular bisector. Took me a while to figure it out, though. Keep it up! :)

Drex Beckman - 5 years, 4 months ago

Thanks! ;)

Milan Milanic - 5 years, 4 months ago

Those are fine circles!

The answer is 15.

2 solutions

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