Difference between revisions of "2001 AMC 12 Problems/Problem 24"
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Using Law of Sines on triangle <math>ADB</math>, we find that <math>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}</math>. Since we know that <math>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}</math>, <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, and <math>\sin 120^\circ = \frac{\sqrt{3}}{2}</math>, we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>. | Using Law of Sines on triangle <math>ADB</math>, we find that <math>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}</math>. Since we know that <math>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}</math>, <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, and <math>\sin 120^\circ = \frac{\sqrt{3}}{2}</math>, we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>. | ||
− | Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <math>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)</math>. Simplifying the | + | Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <math>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)</math>. Simplifying the right side, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>. |
Now, we apply Law of Sines to triangle <math>ABC</math> to see that <math>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}</math>. After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <math>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}</math>. | Now, we apply Law of Sines to triangle <math>ABC</math> to see that <math>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}</math>. After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <math>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}</math>. |
Revision as of 21:53, 25 May 2019
Problem
In ,
. Point
is on
so that
and
. Find
.
Solution 1
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 .
Solution 2
Draw a good diagram! Now, let's call , so
. Given the rather nice angles of
and
as you can see, let's do trig. Drop an altitude from
to
; call this point
. We realize that there is no specific factor of
we can call this just yet, so let
. Notice that in
we get
. Using the 60-degree angle in
, we obtain
. The comparable ratio is that
. If we involve our
, we get:
. Eliminating
and removing radicals from the denominator, we get
. From there, one can easily obtain
. Now we finally have a desired ratio. Since
upon calculation, we know that
can be simplified. Indeed, if you know that
or even take a minute or two to work out the sine and cosine using
, and perhaps the half- or double-angle formulas, you get
.
Solution 3
Without loss of generality, we can assume that and
. As above, we are able to find that
and
.
Using Law of Sines on triangle , we find that
. Since we know that
,
, and
, we can compute
to equal
and
to be
.
Next, we apply Law of Cosines to triangle to see that
. Simplifying the right side, we get
, so
.
Now, we apply Law of Sines to triangle to see that
. After rearranging and noting that
, we get
.
Dividing the RHS through by , we see that
, so
is either
or
. Since
is not a choice, we know
.
Note that we can also confirm that by computing
with Law of Sines.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.