Difference between revisions of "2013 AIME I Problems/Problem 12"
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Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | ||
− | ==Solution 4 (Trig)== | + | ==Solution 4 (Special Triangles)== |
+ | [[File:Better_photo_for_2013_aime_i_problem_12.png]] | ||
+ | |||
+ | As we can see, the <math>75^\circ</math> angle of <math>\angle P</math> can be split into a <math>45^\circ</math> angle and a <math>30^\circ</math> angle. This allows us to drop an altitude from point <math>P</math> for <math>\triangle RPQ</math> which intersects <math>\overline{AF}</math> at point <math>a</math> and <math>\overline{RQ}</math> at point <math>b</math>. The main idea of our solution is to obtain enough sides of <math>\triangle RPQ</math> that we are able to directly figure out its area (specifically by figuring out side <math>\overline{RQ}</math> and <math>\overline{Pb}</math>. | ||
+ | |||
+ | We first begin by figuring out the length of <math>\overline{PQ}</math>. This can be easily done, since <math>\overline{AB}</math> is simply <math>1</math> (given in the problem) and <math>\overline{BQ}=1</math> because <math>\triangle BCQ</math> is an equilateral after some simple angle calculations. Now we need to find <math>\overline{PA}</math>. This is when we bring in some simple algebra. | ||
+ | |||
+ | PREPARATION: | ||
+ | <math>\overline{aF}=\overline{Pa}</math> (45-45-90 Right Triangle) | ||
+ | |||
+ | <math>\overline{Pa}=\sqrt{3}\overline{Aa}</math> (30-60-90 Right Triangle) | ||
+ | |||
+ | <math>\overline{PA}=2\overline{Aa}</math> | ||
+ | |||
+ | <math>\overline{Aa}+\overline{aF}=1</math> | ||
+ | |||
+ | SOLVING: | ||
+ | <math>\overline{Aa}+\sqrt{3}\overline{aF}=1</math> | ||
+ | |||
+ | so <math>\overline{Aa}(\sqrt{3}+1)=1</math> | ||
+ | |||
+ | <math>\overline{Aa}=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}</math> | ||
+ | |||
+ | Finally, <math>\overline{PA}=2\cdot\frac{\sqrt{3}-1}{2}=\sqrt{3}-1</math> | ||
+ | |||
+ | |||
+ | Now, we can finally get the length of <math>\overline{PQ}</math> by adding up <math>\overline{PA}+\overline{AB}+\overline{BQ}</math>, which is simply <math>(\sqrt{3}-1)+(1)+(1)=\sqrt{3}+1</math> | ||
+ | |||
+ | |||
+ | To get <math>\overline{RQ}</math> and <math>\overline{Pb}</math>, we first work bit by bit. | ||
+ | |||
+ | <math>\overline{Qb}=\frac{\overline{PQ}}{2}=\frac{\sqrt{3}+1}{2}</math> (30-60-90 Right Triangle) | ||
+ | |||
+ | <math>\overline{Pb}=\sqrt{3}\overline{Qb}=\frac{\sqrt{3}+3}{2}</math> (same 30-60-90 Right Triangle) | ||
+ | |||
+ | Since <math>\overline{Pb}=\overline{Rb}</math> because of 45-45-90 right triangles, | ||
+ | |||
+ | <math>\overline{Rb}=\frac{\sqrt{3}+3}{2}</math> too. | ||
+ | |||
+ | Now, we can finally calculate <math>\overline{RQ}</math>, and it is <math>\overline{Rb}+\overline{Qb}=\frac{\sqrt{3}+3}{2}+\frac{\sqrt{3}+1}{2}=\sqrt{3}+2</math>. | ||
+ | |||
+ | Finally, the area of <math>\triangle PRQ</math> can be calculated by <math>\frac{1}{2}\cdot\overline{RQ}\cdot\overline{Pb}</math>, which is equal to <math>[\triangle PRQ]=\frac{1}{2} \cdot (\sqrt{3}+2) \cdot \frac{\sqrt{3}+3}{2} =\frac{9+5\sqrt{3}}{4}</math>. So the final answer is <math>9+5+3+4=\fbox{021}</math>. | ||
+ | ==Solution 5 (Trig)== | ||
[[File:2013_AIME_I_Problem_12.png]] | [[File:2013_AIME_I_Problem_12.png]] | ||
− | With some simple angle chasing we can show that <math>\triangle OJL</math> and <math>\triangle MPL</math> are congruent. This means we have a large equilateral triangle with side length <math>3</math> and quadrilateral <math>OJQN</math>. We know that <math>[OJQN] = [\triangle NQL] - [\triangle OJL]</math>. Using Law of Sines and the fact that <math>\angle N = 45^{\circ}</math> we know that <math>\overline{NL} = \sqrt{6}</math> and the height to that side is <math>\frac{\sqrt{3} -1}{\sqrt{2}}</math> so <math>[\triangle NQL] = \frac{3-\sqrt{3}}{2}</math>. Using an extremely similar process we can show that <math>\overline{OJ} = 2-\sqrt{3}</math> which means the height to <math>\overline{LJ}</math> is <math>\frac{2\sqrt{3}-3}{2}</math>. So the area of <math>\triangle OJL = \frac{2\sqrt{3}-3}{4}</math>. This means the area of quadrilateral <math>OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}</math>. So the area of our larger triangle is <math>\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}</math>. Therefore <math>9+5+3+4=021</math>. | + | With some simple angle chasing we can show that <math>\triangle OJL</math> and <math>\triangle MPL</math> are congruent. This means we have a large equilateral triangle with side length <math>3</math> and quadrilateral <math>OJQN</math>. We know that <math>[OJQN] = [\triangle NQL] - [\triangle OJL]</math>. Using Law of Sines and the fact that <math>\angle N = 45^{\circ}</math> we know that <math>\overline{NL} = \sqrt{6}</math> and the height to that side is <math>\frac{\sqrt{3} -1}{\sqrt{2}}</math> so <math>[\triangle NQL] = \frac{3-\sqrt{3}}{2}</math>. Using an extremely similar process we can show that <math>\overline{OJ} = 2-\sqrt{3}</math> which means the height to <math>\overline{LJ}</math> is <math>\frac{2\sqrt{3}-3}{2}</math>. So the area of <math>\triangle OJL = \frac{2\sqrt{3}-3}{4}</math>. This means the area of quadrilateral <math>OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}</math>. So the area of our larger triangle is <math>\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}</math>. Therefore <math>9+5+3+4=\fbox{021}</math>. |
==Solution 5 (Elementary Geo)== | ==Solution 5 (Elementary Geo)== |
Revision as of 02:21, 1 May 2024
Contents
Problem
Let be a triangle with and . A regular hexagon with side length 1 is drawn inside so that side lies on , side lies on , and one of the remaining vertices lies on . There are positive integers and such that the area of can be expressed in the form , where and are relatively prime, and c is not divisible by the square of any prime. Find .
Solution 1
First, find that . Draw . Now draw around such that is adjacent to and . The height of is , so the length of base is . Let the equation of be . Then, the equation of is . Solving the two equations gives . The area of is .
Solution 2 (Cartesian Variation)
Use coordinates. Call the origin and be on the x-axis. It is easy to see that is the vertex on . After labeling coordinates (noting additionally that is an equilateral triangle), we see that the area is times times the coordinate of . Draw a perpendicular of , call it , and note that after using the trig functions for degrees.
Now, get the lines for and : and , whereupon we get the ordinate of to be , and the area is , so our answer is .
Solution 3 (Trig)
Angle chasing yields that both triangles and are -- triangles. First look at triangle . Using Law of Sines, we find:
Simplifying, we find . Since , WLOG assume triangle is equilateral, so . So .
Apply Law of Sines again,
Simplifying, we find .
.
Evaluating and reducing, we get thus the answer is
Solution 4 (Special Triangles)
As we can see, the angle of can be split into a angle and a angle. This allows us to drop an altitude from point for which intersects at point and at point . The main idea of our solution is to obtain enough sides of that we are able to directly figure out its area (specifically by figuring out side and .
We first begin by figuring out the length of . This can be easily done, since is simply (given in the problem) and because is an equilateral after some simple angle calculations. Now we need to find . This is when we bring in some simple algebra.
PREPARATION: (45-45-90 Right Triangle)
(30-60-90 Right Triangle)
SOLVING:
so
Finally,
Now, we can finally get the length of by adding up , which is simply
To get and , we first work bit by bit.
(30-60-90 Right Triangle)
(same 30-60-90 Right Triangle)
Since because of 45-45-90 right triangles,
too.
Now, we can finally calculate , and it is .
Finally, the area of can be calculated by , which is equal to . So the final answer is .
Solution 5 (Trig)
With some simple angle chasing we can show that and are congruent. This means we have a large equilateral triangle with side length and quadrilateral . We know that . Using Law of Sines and the fact that we know that and the height to that side is so . Using an extremely similar process we can show that which means the height to is . So the area of . This means the area of quadrilateral . So the area of our larger triangle is . Therefore .
Solution 5 (Elementary Geo)
We can find that . This means that the perpendicular from to is perpendicular to as well, so let that perpendicular intersect at , and the perpendicular intersect at . Set . Note that , so and . Also, , so . It's easy to calculate the area now, because the perpendicular from to splits into a (PHQ) and a (PHR). From these triangles' ratios, it should follow that , so the area is . . By Mathscienceclass
Solution 6 (Combination of 1 & 2)
We can observe that (because & are both ). Thus we know that is equivalent to the height of the hexagon, which is . Now we look at triangle and apply the Law of Sines to it. . From here we can solve for and get that . Now we use the Sine formula for the area of a triangle with sides , , and to get the answer. Setting and we get the expression which is . Thus our final answer is . By AwesomeLife_Math
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.