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 May 12th, 2013, 11:32 PM #1 Newbie   Joined: May 2013 Posts: 26 Thanks: 0 interesting!! hello, here's the question If A,D are ends of diameter of circle of radius r and B and C are on semicircle such that AB=BC=r/2 then find the ratio of CD to radius
May 13th, 2013, 12:03 AM   #2
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Re: interesting!!

Quote:
 Originally Posted by benten hello, here's the question If A,D are ends of diameter of circle of radius r and B and C are on semicircle such that AB=BC=r/2 then find the ratio :BD to radius
I'm getting

$\frac{\sqrt{15}}{2}$

but i didn't use BC at all so i'm wondering if you have something else in mind.

 May 13th, 2013, 12:11 AM #3 Newbie   Joined: May 2013 Posts: 26 Thanks: 0 Re: interesting!! I am really sorry; it's CD to radius ratio.
 May 13th, 2013, 04:44 AM #4 Newbie   Joined: May 2013 Posts: 26 Thanks: 0 Re: interesting!! I got the answer to be 4*(2)^1/2, but I doubt 'cuz I don't know the answer.
 May 13th, 2013, 05:22 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: interesting!! Let the radius of the circle be 1. Let the center of the circle be O. Then angle COD, in degrees, is $180\,-\,4\sin^{\small{-1}}$$\frac14$$$ Using the Law of Cosines we calculate CD: \begin{*align}\bar{CD}\,&=\,\sqrt{2\,-\,2\cos$$180\,-\,4\sin^{\small{-1}}\(\frac14$$\)} \\ &=\,\sqrt{2\,+\,2\cos$$4\sin^{\small{-1}}\(\frac14$$\)} \\ &=\,\sqrt{2\,+\,2$$1\,-\,2\sin^2\(2\sin^{\small{-1}}\(\frac14$$\)\)} \\ &=\,\sqrt{2\,+\,2$$1\,-\,2\,\cdot\,4\,\cdot\frac{1}{16}\,\cdot\,\frac{15} {16}$$} \\ &=\,sqrt{2\,+\,2$$1\,-\,\frac{15}{32}$$} \\ &=\,\sqrt{\frac{98}{32}} \\ &\,=\,\frac{7}{4}\end{*align} Hence the ratio of CD to the radius of the circle is 7:4.
 May 13th, 2013, 05:28 AM #6 Newbie   Joined: May 2013 Posts: 26 Thanks: 0 Re: interesting!! Thank you and how did you get the angle 180-4*sin^-1(1/4)?
 May 13th, 2013, 05:29 AM #7 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: interesting!! [color=#000000]I think that the result should be $\frac{7}{4}$, I will post later.[/color]
 May 13th, 2013, 05:32 AM #8 Newbie   Joined: May 2013 Posts: 26 Thanks: 0 Re: interesting!! Okay, I got it!!
 May 13th, 2013, 05:38 AM #9 Newbie   Joined: May 2013 Posts: 26 Thanks: 0 Re: interesting!! but is my approach wrong....my approach is if both chords belong to a regular polygon of n sides then n*(r/2) is greatest value less than 2*pi*r then n must be 12 and internal angle is 150 then answer is different
May 14th, 2013, 02:04 AM   #10
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Re: interesting!!

[color=#000000]I will do the same as greg1313, here is a brief solution.

[attachment=0:33lnkz3n]greg.png[/attachment:33lnkz3n]

Using the law of cosines in the triangle $\triangle AOB$

$\overline{AB}^2=\overline{OA}^2+\overline{OB}^2-2\cdot \overline{OA}\cdot \overline{OB}\cdot \cos\left(A\widehat{O}B\right)\Rightarrow\cos\left (A\widehat{O}B\right)=\frac{7}{8}$

in the right triangle $\triangle OEA$, $\overline{OE}=\cos\left(A\widehat{O}B\right)\cdot r=\frac{7}{8}r$. We can easily prove that the colored angles are equal, which proves that $\overline{OB}\parallel \overline{CD}$. So $\frac{\overline{OE}}{\overline{CD}}=\frac{\overlin e{OA}}{\overline{AD}}=\frac{1}{2}$ and finally $\overline{OE}=\frac{7}{8}r\Rightarrow\frac{\overli ne{CD}}{2}=\frac{7}{8}r$$\Rightarrow\overline{CD}=\frac{7}{4}r$.[/color]
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