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Challenge: Yin and Ying addendum

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Challenge: Yin and Ying addendum [#permalink] New post 18 Oct 2003, 00:46
Hi:

Someone solved Stolyar's original YIN-YAN problem too quickly so I decided to add a little twist to make it more "interesting."

In the attached diagram, the assumptions are the same as in Stolyar's problem (A, B, C are on diameter, center of circle is B, and all arcs are semicircles). Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded area region above the red line to the area of the shaded region below the red line? (Diagram is not drawn to scale and angles drawn are not accurate).

A) 3/4
B) 5/6
C) 1
D) 7/5
E) 9/7
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ying.jpg
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ying.jpg
ying.jpg [ 11.69 KiB | Viewed 673 times ]


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Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: Challenge: Yin and Ying addendum [#permalink] New post 18 Oct 2003, 05:40
AkamaiBrah wrote:
Hi:

Someone solved Stolyar's original YIN-YAN problem too quickly so I decided to add a little twist to make it more "interesting."

In the attached diagram, the assumptions are the same as in Stolyar's problem (A, B, C are on diameter, center of circle is B, and all arcs are semicircles). Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded area region above the red line to the area of the shaded region below the red line? (Diagram is not drawn to scale and angles drawn are not accurate).

A) 3/4
B) 5/6
C) 1
D) 7/5
E) 9/7



stolyar, what if the info YXA = 105 is not given.

can we still solve the problem.

I tried using the property that an angle inscribed in a semicircle is 90.

So i get AYC = 90 and AXC = 90 . Also ABC,BYC and XYB are

isosceles triangles.

I couldnt really go further , any ideas?

thanks
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 [#permalink] New post 18 Oct 2003, 07:29
I do not think so.
If an angle is not given, then point Y can slide anywhere around the circle, making parts of the shaded region change. We need to fix point Y somehow.
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 [#permalink] New post 18 Oct 2003, 11:30
stolyar wrote:
I do not think so.
If an angle is not given, then point Y can slide anywhere around the circle, making parts of the shaded region change. We need to fix point Y somehow.


Stolyar was able to solve this very nicely, but I deleted his post so that everyone else would have a chance to solve it.

P.S.: Stolyar -- everyone knows you are a whiz. Since you are a moderator, please give everyone a chance to solve before solving! (you can always message your solution to me privately and I will let you know how you did and give you credit if you get it right).
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 18 Oct 2003, 23:15
The answer is D. 7/5

If angle axy=105 then sector acy=210. obviously sector ac=180 right. Thus, sector yc=30 (210-180). Sector yc is enclosed by the central angle ybc, making angle ycb=30 degrees. 30/360 = 1/12. so the area of that portion is pi/12.

the area of the semi circle is .5*(.5)^2*pi or pi/8

so the area below the red line is pi/12 + pi/8 = 5pi/24

the area of the shaded region is pi/2 (see original ying yang problem).

thus the area above the red line is pi/2 -pi/12 - pi/8 = 7pi/24

the ratio requested in the problem is (7pi/24)/(5pi/24) = 7/5

Great problems akamai. If the teachers at that math work shop are half as good as you then i'm sure i would have gotten my money's worth, too bad they're filled up.
Keep em coming
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 [#permalink] New post 24 Oct 2003, 05:37
how is sector acy = 210?
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 [#permalink] New post 24 Oct 2003, 20:18
Soumala wrote:
how is sector acy = 210?


the problem states that axy = 105. For any non-central angle in a circle the sector enclosed by that angle is always twice that angle. For a central angle the sector is the same as the angle. For example if xby = 80 then sector xy would also equal 80.
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 [#permalink] New post 25 Oct 2003, 09:11
Thanks for the explanation Sandoval.Looks like I need to brush up my fundamentals.
  [#permalink] 25 Oct 2003, 09:11
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Challenge: Yin and Ying addendum

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