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07 Oct 2020, 02:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (02:41) correct
40%
(02:55)
wrong
based on 35
sessions
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A semicircle with radius a is inscribed in a rectangle with its diameter along one side of the rectangle. A circle with radius b is then inscribed so it is tangent to the rectangle and the semicircle, as shown above. What is the value of a/b?
A. \(\frac{\sqrt{2}}{\sqrt{2} - 1}\)
B. \(\frac{\sqrt{2} + 1}{\sqrt{2}}\)
C. 2:1
D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)
E. 6:1
Attachment:
Untitled.png [ 22.09 KiB | Viewed 2953 times ]
Let the rectangle be ABCD (A and D being the corners of the semi circle, B being the vertex enclosing the smaller circle and C being the other one). O be the centre of the semi circle, M be the centre of smaller circle, E be the point of intersection of Semi circle and the rectangle and F be the intersection of smaller circle and BC
We know that the length of the rectangle (AB) will be 2a (diameter of the semi circle)
OA = OD = a
In the smaller semi circle
We need to find the distance of centre to the vertex enclosing the small circle i.e vertex B
Since the point of intersection of the small circle with the vertex is tangent we can safely deduce that the angle will be 90
In the triangle formed with the vertex MFB
\(MB ^2 = b^2+b^2\)
Diagonal of smaller circle = MB = \(\sqrt{2}b\)
In the triangle OEB, right angled at E
\(OB^2 = OE^2 + BE^2\)
We know that OE=BE=a
subsituting we get
OB = \(\sqrt{2}a\)
OB = radius of the semicircle + radius of the smaller circle + MB
\(\sqrt{2}a = a + b + \sqrt{2}b\)
rearranging we get
\(a(\sqrt{2} - 1) = b(\sqrt{2} + 1)\)
Simplifying we get
\(\frac{a}{b} = \frac{(\sqrt{2} + 1) }{ (\sqrt{2} - 1)}\)
Hence IMO D
We know that the length of the rectangle (AB) will be 2a (diameter of the semi circle)
OA = OD = a
In the smaller semi circle
We need to find the distance of centre to the vertex enclosing the small circle i.e vertex B
Since the point of intersection of the small circle with the vertex is tangent we can safely deduce that the angle will be 90
In the triangle formed with the vertex MFB
\(MB ^2 = b^2+b^2\)
Diagonal of smaller circle = MB = \(\sqrt{2}b\)
In the triangle OEB, right angled at E
\(OB^2 = OE^2 + BE^2\)
We know that OE=BE=a
subsituting we get
OB = \(\sqrt{2}a\)
OB = radius of the semicircle + radius of the smaller circle + MB
\(\sqrt{2}a = a + b + \sqrt{2}b\)
rearranging we get
\(a(\sqrt{2} - 1) = b(\sqrt{2} + 1)\)
Simplifying we get
\(\frac{a}{b} = \frac{(\sqrt{2} + 1) }{ (\sqrt{2} - 1)}\)
Hence IMO D
Attachments
9140c2f1-04f3-43d4-9d1d-7aaaf442ea9e.JPG [ 40.84 KiB | Viewed 2582 times ]
07 Oct 2020, 06:57
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Hi
Could u pls upload a diagram - it’d be v helpful to understand
Posted from my mobile device
