Difference between revisions of "2004 AMC 10B Problems/Problem 24"
(→Solution 3) |
(→Solution 3) |
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Line 57: | Line 57: | ||
P.S | P.S | ||
− | We get AB by the numbers given in the problem. | + | We get <math>AB</math> by the numbers given in the problem. |
− | We get | + | We get <math>BE</math> by setting up a systems of equations. |
Using the Angle Bisector Theorem: | Using the Angle Bisector Theorem: | ||
− | 7 | + | <math>\frac{7}{8}=\frac{EB}{EC} |
− | We also know that EB and EC add up to 9 (using the numbers given in the problem) | + | We also know that </math>EB<math> and </math>EC<math> add up to 9 (using the numbers given in the problem) |
− | EB+EC=9 | + | </math>EB+EC=9$ |
We then solve. | We then solve. |
Revision as of 18:50, 6 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
Set 's length as
.
's length must also be
since
and
intercept arcs of equal length (because
). Using Ballemy's Theorem,
. The ratio is
Solution 2
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
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
P.S
We get by the numbers given in the problem.
We get
by setting up a systems of equations.
Using the Angle Bisector Theorem:
$\frac{7}{8}=\frac{EB}{EC}
We also know that$ (Error compiling LaTeX. Unknown error_msg)EBEC
EB+EC=9$
We then solve.
--rosebuddy_vxd
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.