Difference between revisions of "1996 AHSME Problems/Problem 30"
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Furthermore, <math>\angle BAD</math> and <math>\angle BED</math> subtend the same arc, as do <math>\angle ABE</math> and <math>\angle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</cmath> It follows that <cmath>\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad | Furthermore, <math>\angle BAD</math> and <math>\angle BED</math> subtend the same arc, as do <math>\angle ABE</math> and <math>\angle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</cmath> It follows that <cmath>\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad | ||
− | \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.</cmath> Solving the two equations simultaneously yields <math>AD=360/49,</math> so <math>m+n=\boxed{409} | + | \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.</cmath> Solving the two equations simultaneously yields <math>AD=360/49,</math> so <math>m+n=\boxed{409}. \blacksquare</math> |
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==Solution 2== | ==Solution 2== |
Revision as of 20:07, 27 September 2016
Contents
[hide]Problem
A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to , where and are relatively prime positive integers. Find .
Solution 1
In hexagon , let and let . Since arc is one third of the circumference of the circle, it follows that . Similarly, . Let be the intersection of and , that of and , and that of and . Triangles and are equilateral, and by symmetry, triangle is isosceles and thus also equilateral.
Furthermore, and subtend the same arc, as do and . Hence triangles and are similar. Therefore, It follows that Solving the two equations simultaneously yields so
Solution 2
Let and . Since (add up the angles), and thus . Therefore, . as well.
Now I focus on triangle . By the Law of Cosines, , so . Seeing and , we can now use the Law of Sines to get:
Now I focus on triangle . and , and we are given that , so We know , but we need to find . Using various identities, we see Returning to finding , we remember Plugging in and solving, we see . Thus, the answer is , which is answer choice .
Solution 2
Let be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides the circumradius satisfies where is the semiperimeter. Applying this to the trapezoid with sides , we see that many terms cancel and we are left with Similar canceling occurs for the trapezoid with sides , and since the two quadrilaterals share the same circumradius, we can equate: Solving for gives , so the answer is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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All AHSME Problems and Solutions |
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