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Consider sector of circle $MAB$. $∠AMB = 120◦$.
A circle $S$ touches side $AM$, side $MB$ and arc $AB$ as shown in the figure.
Area of circle $S$ is $75π/(7 + 4√3)$ . Find $4√3$ times the area of $△AMB$.

figure

Here , I know the area of circle , so radius can be calculated. For triangle $AMB$ , it's area is equal to $1/2*AM^2*\sin 120$(degrees). So I just require the length of $AM$, that is the radius of sector.

$AM$ and $MB$ are tangents to the circle so that may be of some help? But I'm stuck here .

Any hints are apreciated . (This is not class-homework , I'm solving sample questions for a competitive exam )

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$$A_S=\frac{75\pi}{7+4\sqrt3}\implies R=5\sqrt{\frac3{7+4\sqrt3}}$$

Let now $\;K\;$ be the intersection point between the circle (with center $\;O\;$ ,say) and the radius $\;AM\;$, and form the $\;30-60-90\;$ straight-angle triangle $\;KOM\;$ ,so that

$$R=KO=\color{red}{\frac{\sqrt3}2}\,MO\implies MO=\color{red}{10}\sqrt{\frac1{7+4\sqrt3}}$$

so finally, the radius $\;r\;$ of the circular sector is

$$r=MO+R=\frac{5(\color{red}2+\sqrt3)}{\sqrt{7+4\sqrt3}}=\color{red}5\;\;\left(\text{because}\;\;(2+\sqrt3)^2=7+4\sqrt3\ldots\right)$$

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  • $\begingroup$ How is the answer that you got is different from the answer that Alijah Ahmed got ( see below) ? $\endgroup$ – A Googler Mar 31 '14 at 7:47
  • $\begingroup$ @AGoogler, I can't tell (I haven't even read the other answer) but mine was the first one...:) And even if it wasn't: I bet you know sometimes several pretty similar answers pop up almost at once. This usually happens with elementary questions, like yours. $\endgroup$ – DonAntonio Mar 31 '14 at 10:40
  • $\begingroup$ No , not that. I mean that you got the radius of the sector as something irrational , while Alijah got it as equal to 5. Who's right? I can't find a flaw in either of you , so I'm confused . Can you take a look at his answer ? $\endgroup$ – A Googler Mar 31 '14 at 10:51
  • $\begingroup$ Oh, I see @AGoogler. Well, let me check for some minutes. $\endgroup$ – DonAntonio Mar 31 '14 at 10:53
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    $\begingroup$ @AGoogler, good you asked me that: it was a stupid mistake of mine. Thanks. I edited my answer and the edited parts are in red now. It all was that in fact $\;R=\frac{\sqrt3}2MO\;$ and not $\;R=\sqrt3 MO\;$ as I wrote. Forgot that $\;1/2\;$ factor. $\endgroup$ – DonAntonio Mar 31 '14 at 11:05
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Denote the centre of the smaller (inscribed) circle as $C$, and let the circle touch tangent $AM$ at point $X$, as illustrated in the figure below.

enter image description here

We are given the area of the inscribed circle, which is $\frac{75\pi}{7+4\sqrt{3}}$. Thus we have $$\frac{75\pi}{7+4\sqrt{3}}=\pi R^2 \Rightarrow R= 5\sqrt{\frac{3}{7+4\sqrt{3}}}$$

Now the line $CM$ can be calculated using the right angled triangle $\triangle MXC$, where $\angle XMC=\frac{\pi}{3}$, $\angle CXM=\frac{\pi}{2}$ ($XM$ is tangent of circle), so that $\angle XCM=\frac{\pi}{6}$, and $CX=R$.

$$CM=\frac{CX}{\sin \frac{\pi}{3}}=\frac{2R}{\sqrt{3}}$$

The radius of the big circle which the sector is part of is $$AM=R+CM=\left(1+\frac{2}{\sqrt{3}}\right)R=\left(\frac{\sqrt{3}+2}{\sqrt{3}}\right)R$$

Substituting in the value of $R$, we have

$$AM=5\sqrt{\frac{3}{7+4\sqrt{3}}}\left(\frac{\sqrt{3}+2}{\sqrt{3}}\right)=5\sqrt{\frac{3(2+\sqrt{3})^2}{(7+4\sqrt{3})(3)}}=5\sqrt{\frac{3(7+4\sqrt{3})}{(7+4\sqrt{3})(3)}}=5$$

Thus the length of $AM$, which is the radius of the sector, is $5$.

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  • $\begingroup$ Wow ! Have you solved all the questions ? $\endgroup$ – A Googler Apr 1 '14 at 14:25
  • $\begingroup$ Maths 2013 - All. Maths 2012 - All. Maths 2011 - Q21 is remained as answer is not given in solutions. Q33 is wrong. Maths 2011 Sample - 45,47,52,53,54 As answers are not given. Maths 2010 - Q55 I didn't got it.Q 58 is wrong. Q67,69,70,73 ans not given. Maths 2009 - Didn't got Q83. Q 87,93,94 no ans is given. Maths 2008 - 95,97,a,b,98,99,100. Maths 2007 - 104,107,108,109,110,112. Just two left from physics I think they are wrong And all from chemistry. $\endgroup$ – Ajay Apr 1 '14 at 14:31
  • $\begingroup$ how much time on an average did it take you to solve one maths question? I'm fearing that I won't be able to complete the question paper in time . $\endgroup$ – A Googler Apr 1 '14 at 15:26
  • $\begingroup$ 3-4 minutes average sometimes it take 6-7 minutes to solve questions like Q49 due to long calculations needed to solve it. What about you? $\endgroup$ – Ajay Apr 2 '14 at 2:26
  • $\begingroup$ Got 55 and 21. I was reading on ray BC and AB instead of in ray BC and AB. Thanks to you I tried to solve them again. $\endgroup$ – Ajay Apr 2 '14 at 3:03

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