Can't use Thales' theorem!

Geometry Level 4

Let a circle with radius 1 have centre $O$ , and let $AB$ be its diameter . $C$ , $D$ and $E$ are collinear points in that order such that $C$ is on $AB$ extended closer to $A$ , $D$ and $E$ are on the circle and $BE=ED=DC$ . If the value of $\cos \angle EOB$ can be represented as $\dfrac {a}{b}$ , where $a$ is an integer , $b$ is a positive integer and both are coprime, find the value of $a+b$ .

The answer is 5.

1 solution

Wen Z
Aug 30, 2016

$EO$ is an anble bisector (of $\angle DEB$ ) so $\frac{CE}{EB}=\frac{CO}{OB}$ . Because $AO=OB$ , $CA=AO=OB=OE=r$ where $r$ is the radius of the circle. Let $BE=ED=DC=x$ Now by power of a point , we have

$CD\cdot CE=CA\cdot CB$

which is the same as

$2x^2=3r^2$

Now by the cosine rule on $\bigtriangleup EOC$ we have

$EB^2=OE^2+OB^2-2OE\cdot OB \cos \angle EOB$

i.e

$x^2=2r^2=2r^2 \cos \angle EOB$

Multiplying by two yields

$2x^2=4r^2(1-\cos \angle EOB)$

Lastly equating this with the power of a point equation we have

$3r^2=4r^2(1-\cos \angle EOB)$

which yields

$1-\cos \angle EOB=\frac{3}{4}$ ,

which yields

$\cos \angle EOB=\frac{1}{4}$

Therefore $a+b=\fbox{5}$

Just a remark, since you're given that the circle is a unit circle you could say $r=1$ . But that doesn't really affect the proof

Wen Z - 4 years, 9 months ago

Yep. You nailed the proof. Instead of using angle bisector theorem, you could instead show $\angle CEB = \angle COE$ , so $\Delta COE \sim \Delta CEB$ , and determine the value of $CA$ as a result to use in the power of a point.

Sharky Kesa - 4 years, 9 months ago

I had that in my diagram but for the proof I realised that I could cut out a lot of redundant stuffs.

Wen Z - 4 years, 9 months ago

Can't use Thales' theorem!

The answer is 5.

1 solution

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