Question from IMO

A hoop is resting vertically at stair case as shown in the diagram.AB=12cm and BC=8cm. The radius of the hoop is..

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Comments

If you make a center point, let's call it O, then you know that OA=OC (both radii). If you extend the line of the stair (the one parallel to the ground) out horizontally, it will eventually intersect OA. Label this point of intersection D. Now you have formed a right triangle (triangle ODC). You know that BC=AD=8. Therefore OD=OA-AD=OA-8. You also know that AB=DC=12. Using the Pythagorean theorem, you can determine the length of OC. (OA-8)^{2}+12^{2}=OC^{2}. OA^{2}-18OA+64+144 =OC^{2}. OA^{2]-18OA+208=OC^{2}. Since OC=OA (both radii), we can plug in OA for OC. OA^{2}-18OA+208=OA^{2}. Then it is merely algebra. -18OA+208=0. 18OA=208. OA=208/18. OA=104/9.

Michael Thew - 7 years, 8 months ago

You have everything up to the Pythagorean theorem right. You messed up here: $(OA-8)^{2}+12^{2}=OC^{2}$ $OA^{2}-18OA+64+144 =OC^{2}$

The $-18OA$ should be $-16OA$ . This gives the correct answer $OA=13$ .

Daniel Chiu - 7 years, 8 months ago

Thank you for the correct answer!

neha adepu - 7 years, 8 months ago

Thanks a lot! It was of great help!

neha adepu - 7 years, 8 months ago

O thanks Daniel. I was wondering why it was such a messy fraction.

Michael Thew - 7 years, 8 months ago

Fuck offffff

Vinit Kumar - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$