Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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<math>\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}</math> | <math>\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}</math> | ||
− | == Solution 1 | + | == Solution 1 (Ptolemy's Theorem) == |
− | Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>\overline{CD}</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length (because <math>\angle BAD=\angle DAC</math>). Using [[ | + | Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>\overline{CD}</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length (because <math>\angle BAD=\angle DAC</math>). Using [[Ptolemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\frac{5}{3}\implies\boxed{\text{(B)}}</math> |
==Solution 2 - Similarity Proportion== | ==Solution 2 - Similarity Proportion== |
Revision as of 12:13, 20 August 2024
Contents
[hide]Problem
In triangle we have , , . Point is on the circumscribed circle of the triangle so that bisects angle . What is the value of ?
Solution 1 (Ptolemy's Theorem)
Set 's length as . 's length must also be since and intercept arcs of equal length (because ). Using Ptolemy's Theorem, . The ratio is
Solution 2 - Similarity Proportion
Let . Observe that because they both subtend arc
Furthermore, because is an angle bisector, so by similarity. Then . By the Angle Bisector Theorem, , so . This in turn gives . Plugging this into the similarity proportion gives: .
Solution 3 - Angle Bisector Theorem
We know that bisects , so . Additionally, and subtend the same arc, giving . Similarly, and .
These angle relationships tell us that by AA Similarity, so . By the angle bisector theorem, . Hence,
--vaporwave
See Also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
All AMC 10 Problems and Solutions |
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