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This problem is from IMO $1985$ Problem $1$.

A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.

Let $\angle DAO$ be $\theta$. I want to show that $CG = r(\frac{1-\cos \theta}{\sin \theta})$. It is given that $$\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{1 - \cos \theta}{\sin \theta}$$ for $0 < x < 180^\circ$.
How do I use this fact to prove that $CG = r(\frac{1-\cos \theta}{\sin \theta})$? (without using the fact that $\frac{1-\cos \theta}{\sin \theta} = \tan \frac{\theta}{2}$.)

My observation:

  1. $\angle FOG = \theta$
  2. I also tried reverse engineering: \begin{align} CG = r\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \\ CG^2 = r^2\frac{1 - \cos \theta}{1 + \cos \theta} \\ CG^2 + CG^2\cos \theta = r^2 - r^2\cos \theta \end{align} but I can't find why this is true.
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Solution without trigopnometry.

Let $P\in AB$ such that $AP=AD$.

We can assume also that $P$ is placed between $A$ and $O$ as on your picture.

Thus, $$\measuredangle APD=\frac{1}{2}(180^{\circ}-\measuredangle A)=x,$$ which says that $DPOC$ is cyclic, which gives: $$\measuredangle CPO=\measuredangle CDO=y$$ and since $$\measuredangle B=180^{\circ}-2y,$$ we obtain $$BC=PB$$ and $$AD+BC=AP+PB=AB.$$

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  • $\begingroup$ +1, I did not focus on geometric solution as the question ask was very specific. Please see my edit for another variation :) $\endgroup$ – Math Lover Sep 20 '20 at 17:17
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I am assuming you are trying to prove $AO = AE + CG$ and $OB = DE + BG$.

Coming to your specific question, if you do not want to use $\frac{\theta}{2}$, make a construct with $\theta$.

$OH = r \cos \theta$

$CI = CF \sin \theta = OG - OH$

As $CF = CG$,

$CG \sin \theta = r - r \cos \theta$

$CG = r\frac {1 - \cos \theta} {\sin \theta}$

enter image description here

EDIT: Motivated by Michael Rosenberg's solution without trigonometry but I think somewhat simpler,

Place a point $P$ on line $AD$ (extend if required) such that $AP = AO$.

Then $\, \triangle OPE \cong \triangle OCG$ (Angle-Angle-Angle, one side same)

So, $AO = AP = AE + EP = AE + CG$

Similarly, $OB = BG + DE$

Hence, $AB = BC + AD$

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    $\begingroup$ I want to know how to use the fact that $\sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{1 - \cos x}{\sin x}$, because this is provided as a hint in our exercise. but it is a nice proof why $\frac{1 - \cos x}{\sin x} = \tan \frac{x}{2}$ $\endgroup$ – Learning Mathematics Sep 20 '20 at 6:22
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    $\begingroup$ Yes of course, but I suspect $\sqrt{\frac{1 - \cos x}{1 + \cos x}}$ has some geometric meaning here. $\endgroup$ – Learning Mathematics Sep 20 '20 at 6:34
  • $\begingroup$ ok understood why you want to use this fact. May be that hint is just to take you to $\tan x / 2$ as $cos x = cos^2 {x/2} - sin^2 {x/2}$? $\endgroup$ – Math Lover Sep 20 '20 at 6:37

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