Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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− | In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>AD | + | ==Problem== |
− | <math>A | + | |
− | <math> | + | In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>\frac{AD}{CD}</math>? |
− | <math> | + | |
− | <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> |
− | <math> | + | |
+ | == 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 [[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== | ||
+ | <asy> | ||
+ | import graph; | ||
+ | import geometry; | ||
+ | import markers; | ||
+ | |||
+ | unitsize(0.5 cm); | ||
+ | |||
+ | pair A, B, C, D, E, I; | ||
+ | |||
+ | A = (11/3,8*sqrt(5)/3); | ||
+ | B = (0,0); | ||
+ | C = (9,0); | ||
+ | I = incenter(A,B,C); | ||
+ | D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); | ||
+ | E = extension(A,D,B,C); | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(circumcircle(A,B,C)); | ||
+ | draw(D--A); | ||
+ | draw(D--B); | ||
+ | draw(D--C); | ||
+ | |||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE); | ||
+ | label("$D$", D, S); | ||
+ | label("$E$", E, NE); | ||
+ | |||
+ | markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true))); | ||
+ | markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true))); | ||
+ | markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true))); | ||
+ | markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true))); | ||
+ | markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); | ||
+ | </asy> | ||
+ | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they both subtend 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 - Angle Bisector Theorem== | ||
+ | We know that <math>\overline{AD}</math> bisects <math>\angle BAC</math>, so <math>\angle BAD = \angle CAD</math>. Additionally, <math>\angle BAD</math> and <math>\angle BCD</math> subtend the same arc, giving <math>\angle BAD = \angle BCD</math>. Similarly, <math>\angle CAD = \angle CBD</math> and <math>\angle ABC = \angle ADC</math>. | ||
+ | |||
+ | These angle relationships tell us that <math>\triangle ABE\sim \triangle ADC</math> by AA Similarity, so <math>AD/CD = AB/BE</math>. By the angle bisector theorem, <math>AB/BE = AC/CE</math>. Hence, | ||
+ | <cmath>\frac{AB}{BE} = \frac{AC}{CE} = \frac{AB + AC}{BE + CE} = \frac{AB + AC}{BC} = \frac{7 + 8}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}.</cmath> | ||
+ | |||
+ | --vaporwave | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2004|ab=B|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:13, 20 August 2024
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.