My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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
Attached Images
File Type: jpg Find the radius of circle.jpg (94.2 KB, 9 views)
Shen is offline  
 
April 13th, 2014, 09:25 PM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,470
Thanks: 2499

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.
v8archie is offline  
April 13th, 2014, 10:34 PM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,470
Thanks: 2499

Math Focus: Mainly analysis and algebra
I haven't written anything down, but $\frac{\pi}{25}$ is my current guess.
v8archie is offline  
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
Shen is offline  
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
Attached Images
File Type: jpg Find the radius of circle.jpg (95.4 KB, 8 views)
Shen is offline  
April 14th, 2014, 07:41 AM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,470
Thanks: 2499

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.
v8archie is offline  
April 14th, 2014, 09:13 AM   #7
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,879
Thanks: 1087

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 $\displaystyle \frac{\pi}{25}$.
greg1313 is offline  
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
Shen is offline  
April 14th, 2014, 11:42 AM   #9
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,470
Thanks: 2499

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.
v8archie is offline  
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
Shen is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
arc, chord, circle, congruent, equal, find, length, radius



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
find radius if a circle is inscribed in quadrilateral amitdixit Algebra 1 August 25th, 2012 08:49 PM
Equal areas defined by parallel chords in a circle ballensr Algebra 0 February 9th, 2012 09:51 AM
how do I calculate radius of a circle, from chord length? ,, boki-san Algebra 1 February 13th, 2011 10:34 AM
Find points on a line of equal distance from certain point horees Algebra 3 December 18th, 2009 01:22 PM
Equal transformation on non-equal sets honzik Applied Math 0 June 9th, 2009 10:38 AM





Copyright © 2018 My Math Forum. All rights reserved.