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04 May 2020, 08:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (03:03) correct
69%
(02:27)
wrong
based on 26
sessions
History
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A chord is drawn inside a circle, such that the length of the chord is equal to the radius of the circle. Two circles are drawn, one on each side of the chord, each touching the chord at its midpoint and each tangent to the original circle. What is the ratio of the area of the bigger inscribed circle to the area of the smaller inscribed circle?
A. \((1 + √2) : 1\)
B. \((2 + √3) : 1\)
C. \((7 + 4√3) : 1\)
D. \((9 + 4√3) : 1\)
E. \((97 + 56√3) : 1\)
Are You Up For the Challenge: 700 Level Questions
A. \((1 + √2) : 1\)
B. \((2 + √3) : 1\)
C. \((7 + 4√3) : 1\)
D. \((9 + 4√3) : 1\)
E. \((97 + 56√3) : 1\)
Are You Up For the Challenge: 700 Level Questions
Bunuel
Given: A chord is drawn inside a circle, such that the length of the chord is equal to the radius of the circle. Two circles are drawn, one on each side of the chord, each touching the chord at its midpoint and each tangent to the original circle.
Asked: What is the ratio of the area of the bigger inscribed circle to the area of the smaller inscribed circle?
Attachment:
Screenshot 2020-05-06 at 12.32.20 PM.png [ 35.08 KiB | Viewed 6297 times ]
OE =\( \sqrt{r^2 - (\frac{r}{2})^2} = \frac{r\sqrt{3}}{2}\)
Radius of larger circle = \(\frac{1}{2}(DO + OE) = \frac{1}{2}r + r \sqrt{3}/2 = \frac{r}{2}(1+ \frac{\sqrt{3}}{2})\)
Radius of smaller circle = \(\frac{1}{2}(OC - OE) = \frac{1}{2}r -r \sqrt{3}/2 = \frac{r}{2}(1- \frac{\sqrt{3}}{2})\)
The ratio of the area of the bigger inscribed circle to the area of the smaller inscribed circle = \(\frac{(1+ \frac{\sqrt{3}}{2})^2}{(1- \frac{\sqrt{3}}{2})^2} = \frac{(2+\sqrt{3})^2}{(2-\sqrt{3})^2} = (2+\sqrt{3})^4 = (7+4\sqrt{3})^2 = 97 + 56\sqrt{3}\)
IMO E
04 May 2020, 16:47
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Suppose the radius of circle, OC = R
Given that OB = R; hence, CD = R/2
\(OD = \sqrt{R^2 - (R/2)^2}\) = \(\frac{√3}{2 }R\)
Diameter of smaller inscribed circle, DA = OA-OD = \(R-\frac{√3}{2 }R\) =\(\frac{ 2-√3}{2} R\)
Diameter of smaller inscribed circle, ED = OE+OD = \(R+\frac{√3}{2 }R\) =\(\frac{ 2+√3}{2} R\)
the ratio of the area of the bigger inscribed circle to the area of the smaller inscribed circle = \([\frac{ED}{AD}]^2 = (\frac{ 2+√3}{2-√3})^2\)
[ 2+√3]*[ 2-√3] = 4-3 = 1
Hence, \(\frac{1}{2-√3} = 2+√3\)
\(Ratio = [(2+√3)^2]^2 \)
\(= [4+3+2*2*√3]^2 \)
\(= (7+4√3)^2 \)
\(= 49+48+2*7*4√3\)
\(= 97 +56√3\)
E
Given that OB = R; hence, CD = R/2
\(OD = \sqrt{R^2 - (R/2)^2}\) = \(\frac{√3}{2 }R\)
Diameter of smaller inscribed circle, DA = OA-OD = \(R-\frac{√3}{2 }R\) =\(\frac{ 2-√3}{2} R\)
Diameter of smaller inscribed circle, ED = OE+OD = \(R+\frac{√3}{2 }R\) =\(\frac{ 2+√3}{2} R\)
the ratio of the area of the bigger inscribed circle to the area of the smaller inscribed circle = \([\frac{ED}{AD}]^2 = (\frac{ 2+√3}{2-√3})^2\)
[ 2+√3]*[ 2-√3] = 4-3 = 1
Hence, \(\frac{1}{2-√3} = 2+√3\)
\(Ratio = [(2+√3)^2]^2 \)
\(= [4+3+2*2*√3]^2 \)
\(= (7+4√3)^2 \)
\(= 49+48+2*7*4√3\)
\(= 97 +56√3\)
E
Bunuel
Attachments
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