Difference between revisions of "2004 AMC 10B Problems/Problem 24"
m (→Solution 2) |
Scrabbler94 (talk | contribs) |
||
(20 intermediate revisions by 5 users not shown) | |||
Line 5: | Line 5: | ||
<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 [[Ptolemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\frac{5}{3}\implies\boxed{\text{(B)}}</math> | 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== | + | ==Solution 2 - Similarity Proportion== |
<asy> | <asy> | ||
import graph; | import graph; | ||
Line 44: | Line 44: | ||
markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); | markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); | ||
</asy> | </asy> | ||
− | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they both subtend <math>\overarc{AC} | + | 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> |
− | ==Solution 3== | + | 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>. | 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, | 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{ | + | <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 | --vaporwave |
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.