Difference between revisions of "1967 AHSME Problems/Problem 32"
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Since <math>AO \cdot OC = BO \cdot OD</math>, <math>ABCD</math> is cyclic through power of a point. From the given information, we see that <math>\triangle{AOB}\sim \triangle{DOC}</math> and <math>\triangle{BOC} \sim \triangle{AOD}</math>. Hence, we can find <math>CD=\frac{9}{2}</math> and <math>AD=2 \cdot BC</math>. Letting <math>BC</math> be <math>x</math>, we can use Ptolemy's to get | Since <math>AO \cdot OC = BO \cdot OD</math>, <math>ABCD</math> is cyclic through power of a point. From the given information, we see that <math>\triangle{AOB}\sim \triangle{DOC}</math> and <math>\triangle{BOC} \sim \triangle{AOD}</math>. Hence, we can find <math>CD=\frac{9}{2}</math> and <math>AD=2 \cdot BC</math>. Letting <math>BC</math> be <math>x</math>, we can use Ptolemy's to get | ||
<cmath>6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}</cmath> | <cmath>6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}</cmath> |
Revision as of 21:12, 29 May 2021
Contents
[hide]Problem
In quadrilateral with diagonals and , intersecting at , , , , , and . The length of is:
Solution 1
After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of , but we want to find the value of AD. We can apply stewart's theorem now, letting , and we have , and we see that ,
Solution 2
(Diagram not to scale)
Since , is cyclic through power of a point. From the given information, we see that and . Hence, we can find and . Letting be , we can use Ptolemy's to get Since we are solving for
- PhunsukhWangdu
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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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