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April 13th, 2014, 11:25 AM  #1 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1  Find the radius of the congruent circle from equal arc and equal chord length
Dear All I have been trying this state math contest for several times. Please see the attachment. In the problem, they gave equal arc length and equidistant chord from center. As far i did, this inner triangle will be equilateral. But how can i know the chord length? Please give me a solution of this problem. Thanks in advance 
April 13th, 2014, 09:25 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra 
I haven't solved it yet myself, but that equilateral triangle seems to be key to me. A radius of the centre circle is perpendicular to one side, forming a small triangle with a radius of the large circle that passes through the centre of one of the circles near the edge. Of this triangle, we know two angles and a side, so we can compute the distance from O to the intersection of two chords. This is the same as the distance from the intersection to the centre of the circle near the edge. We can now compute the radius of the large circle in terms of the radius of the small ones. But we are given that the radius of the large circle is 1, so we can find the radius of the small circles and hence their area. 
April 13th, 2014, 10:34 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra 
I haven't written anything down, but $\frac{\pi}{25}$ is my current guess.

April 14th, 2014, 06:41 AM  #4 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Thanks.... But can you please draw a figure of the small triangle that is formed. I can not draw it. Also mark the two side and angle which are equal please. Thanks in advance and Regards 
April 14th, 2014, 06:55 AM  #5 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Hi V8archie Does the small triangle look like the attachment? So in between triangle OQP and QRS we know that angle OQP = angle RQS, angle OPQ = angle RSQ = 90 degree and OP = RS equidistant chord So OQ= RQ. But from here how can we be sure that 5*small radius= big radius= 1? Thanks in advance and Regards 
April 14th, 2014, 07:41 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra 
That's it. You can easily extend OR to to become a radius of the large circle, by adding a radius of the small circle. OR is twice the hypotenuse of the small triangle (the triangles are congruent). You know the angle POQ.

April 14th, 2014, 09:13 AM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond 
Let $x$ be the measure of the side of the equilateral triangle containing the circle centered at the origin with radius $r$. Then $\displaystyle r=\frac{\sqrt3 x}{6}$. The height of our equilateral triangle is $\displaystyle \frac{\sqrt3x}{2}$ so the distance from the origin to an apex of our triangle is $\displaystyle \frac{\sqrt3x}{3}$. $\displaystyle \frac{\sqrt3x}{3}+\frac{\sqrt3x}{3}+\frac{\sqrt3 x}{6}=1$ implies $\displaystyle x=\frac{6}{5\sqrt3}$ thus $\displaystyle r=\frac15$ so the area of a smaller circle is $\displaystyle \frac{\pi}{25}$. 
April 14th, 2014, 09:19 AM  #8 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Yeah Thanks Now i got it. Thanks again 
April 14th, 2014, 11:42 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra 
I just noted that the angle POQ is $\frac{\pi}{6}$ (half of the angle of the equilateral triangle), so $ \vec{OQ} = \vec{QR} = \frac{r}{sin{\frac{\pi}{6}}} = 2r$ (OQ is the hypotenuse, an the opposite of length $r$ is a radius of the small circle). Then the radius of the large circle is $2(2r) + r = 5r$. It's essentially the same calculation as that by greg1313, but considering only the radius of the large circle, not the sides of the triangles. 
April 14th, 2014, 01:21 PM  #10 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Hi Greg1313 I think the height of the equilateral triangle is x/2√3. But we don't need this for solving. Only the hypotenuse is sufficient. This method is also good and short. Thanks again 

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