Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 11"
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From adjacent sides, the following relationships can be derived: | From adjacent sides, the following relationships can be derived: | ||
− | < | + | <cmath>\begin{align*} |
DC &= EC + 1\\ | DC &= EC + 1\\ | ||
AE &= AF + 1\\ | AE &= AF + 1\\ | ||
BD &= BF + 2 | BD &= BF + 2 | ||
− | \end{align*}</ | + | \end{align*}</cmath> |
Since <math>BF = EC</math>, and <math>DC = BF + 1</math>, <math>BD = DC + 1</math>. Thus, <math>BC = BD + DC = BD + (BD - 1)</math>. <math>26 = 2BD</math>. Thus, <math>BD = 13/1</math>. Thus, the answer is <math>\boxed{014}</math>. | Since <math>BF = EC</math>, and <math>DC = BF + 1</math>, <math>BD = DC + 1</math>. Thus, <math>BC = BD + DC = BD + (BD - 1)</math>. <math>26 = 2BD</math>. Thus, <math>BD = 13/1</math>. Thus, the answer is <math>\boxed{014}</math>. |
Latest revision as of 20:02, 29 January 2020
Problem
is inscribed inside such that lie on , respectively. The circumcircles of have centers , respectively. Also, , and . The length of can be written in the form , where and are relatively prime integers. Find .
Solution
From adjacent sides, the following relationships can be derived:
Since , and , . Thus, . . Thus, . Thus, the answer is .
See also
Mock AIME 1 2007-2008 (Problems, Source) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |