Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they subtend the same arc, <math>\overarc{AC}</math>. Furthermore, <math>\angle BAE \cong \angle EAC</math> because <math>\overline{AE}</math> is an angle bisector, so <math>\triangle ABE \sim \triangle ADC</math> by <math>\text{AA}</math> similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By the [[Angle Bisector Theorem]], <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math>, so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math>. This in turn gives <math>BE = \frac{21}{5}</math>. Plugging this into the similarity proportion gives: <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they subtend the same arc, <math>\overarc{AC}</math>. Furthermore, <math>\angle BAE \cong \angle EAC</math> because <math>\overline{AE}</math> is an angle bisector, so <math>\triangle ABE \sim \triangle ADC</math> by <math>\text{AA}</math> similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By the [[Angle Bisector Theorem]], <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math>, so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math>. This in turn gives <math>BE = \frac{21}{5}</math>. Plugging this into the similarity proportion gives: <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | ||
− | ==Solution 3== | + | ==Solution 3 == |
− | By the Angle Bisector Theorem, we have <math>DC = \frac{24}{5}.</math> | + | By the Angle Bisector Theorem, we have <math>DC = \frac{24}{5}</math> and <math>BD = \frac{21}{5}.</math> |
+ | Let <math>AD =x.</math> By Stewart's Theorem, we have <math>7^2 \cdot \frac{24}{5} +8^2 \cdot \frac{21}{5} = 9(x^2 + \frac{24}{5} \cdot \frac{21}{5}),</math> which gives x=\frac{ | ||
== See Also == | == See Also == |
Revision as of 22:55, 1 February 2021
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
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
Let . Observe that because they subtend the same 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
By the Angle Bisector Theorem, we have and Let By Stewart's Theorem, we have which gives x=\frac{
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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