Difference between revisions of "1997 AHSME Problems/Problem 19"
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Therefore, <math>2BF = 2 + \sqrt{3} - 1</math>, and we get <math>BF = \frac{1}{2} + \frac{\sqrt{3}}{2}</math> | Therefore, <math>2BF = 2 + \sqrt{3} - 1</math>, and we get <math>BF = \frac{1}{2} + \frac{\sqrt{3}}{2}</math> | ||
− | The radius of the circle is <math>AD</math>, which is <math>BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2 | + | The radius of the circle is <math>AD</math>, which is <math>BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}</math> |
Using decimal approximations, <math>r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37</math>, and the answer is <math>\boxed{D}</math>. | Using decimal approximations, <math>r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37</math>, and the answer is <math>\boxed{D}</math>. |
Revision as of 18:00, 9 August 2015
Contents
Problem
A circle with center is tangent to the coordinate axes and to the hypotenuse of the -- triangle as shown, where . To the nearest hundredth, what is the radius of the circle?
Solution
Draw radii and to the axes, and label the point of tangency to triangle point . Let the radius of the circle be . Square has side length .
Because and are tangents from a common point , .
Similarly, , and we can write:
Equating the radii lengths, we have
This means
by the 30-60-90 triangle.
Therefore, , and we get
The radius of the circle is , which is
Using decimal approximations, , and the answer is .
Solution 2
From the diagram above, it is more direct to note that BC = CF + BF = r - + r - 1 = 2
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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