Difference between revisions of "2001 AMC 12 Problems/Problem 24"
(New page: == Problem == In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB...) |
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− | We start with the observation that <math>\angle | + | We start with the observation that <math>\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ</math>, and <math>\angle ADC = 15^\circ + 45^\circ = 60^\circ</math>. |
We can draw the height <math>CE</math> from <math>C</math> onto <math>AD</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</math>. | We can draw the height <math>CE</math> from <math>C</math> onto <math>AD</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</math>. |
Revision as of 13:57, 23 March 2009
Problem
In ,
. Point
is on
so that
and
. Find
.
Solution
We start with the observation that , and
.
We can draw the height from
onto
. In the triangle
, we have
. Hence
.
By the definition of , we also have
, therefore
. This means that the triangle
is isosceles, and as
, we must have
.
Then we compute , thus
and the triangle
is isosceles as well. Hence
.
Now we can note that , hence also the triangle
is isosceles and we have
.
Combining the previous two observations we get that , and as
, this means that
.
Finally, we get .
See Also
2001 AMC 12 (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 12 Problems and Solutions |