distance wanted, eye, tangents to circle (2000 AMQ Concours College p1)

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High School Math Grades 9-12, Ages 13-18

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distance wanted, eye, tangents to circle (2000 AMQ Concours College p1)

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#1 h Mar 12, 2021, 5:19 AM

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An "eye" has the following form:

https://cdn.artofproblemsolving.com/attachments/f/2/470da6fa2c12b4f2a6ac6a9720015a163a0fcc.png

where $AB, AC, A 'B', A 'C'$ are tangent to the circle of quadrilateral $B B' C'C$ with center $O$ . Points $A, O$ and $A '$ are collinear. Knowing that $AB = AC = A 'B' = A 'C' = 1$ cm and that the area of the circle is $\pi / 4$ cm $^2$ , determine the distance between points $A$ and $A'$ .

This post has been edited 1 time. Last edited by parmenides51, Mar 12, 2021, 5:19 AM

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#2 h Mar 12, 2021, 5:19 AM

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#3 h Mar 12, 2021, 5:56 AM

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The area of the circle is $\frac{\pi}{4}$ , so the radius is $\frac{1}{2}$ . Using Pythagorean theorem on $ABO$ , we get $AO^2 = 1 + \frac{1}{4} = \frac{5}{4} \implies AO = \frac{\sqrt 5}{2}$ . Since $AO = A'O,$ we have $AA' = 2\cdot \frac{\sqrt{5}}{2} = \sqrt {5}$ $\blacksquare$

This post has been edited 2 times. Last edited by JustinLee2017, Mar 12, 2021, 6:01 AM
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