Difference between revisions of "2013 AIME I Problems/Problem 12"
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− | 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 = | + | 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>. |
==Solution 4 (Elementary Geo)== | ==Solution 4 (Elementary Geo)== | ||
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By Mathscienceclass | By Mathscienceclass | ||
+ | ==Solution 5 (Combination of 1 & 2)== | ||
+ | We can observe that <math>RD=DF</math> (because <math>\angle R</math> & <math>\angle RFD</math> are both <math>45^\circ</math>). Thus we know that <math>RD</math> is equivalent to the height of the hexagon, which is <math>\sqrt3</math>. Now we look at triangle <math>\triangle AFP</math> and apply the Law of Sines to it. <math>\frac{1}{\sin{75}}=\frac{AP}{\sin{45}}</math>. From here we can solve for <math>AP</math> and get that <math>AP=\sqrt{3}-1</math>. Now we use the Sine formula for the area of a triangle with sides <math>RQ</math>, <math>PQ</math>, and <math>\angle {RQP}</math> to get the answer. Setting <math>PQ=\sqrt{3}+1</math> and <math>QR=\sqrt{3}+2</math> we get the expression <math>\frac{(\sqrt{3}+1)(\sqrt{3}+2)(\frac{\sqrt{3}}{2})}{2}</math> which is <math>\frac{9 + 5\sqrt{3}}{4}</math>. Thus our final answer is <math>9+5+3+4=\fbox{021}</math>. | ||
+ | By AwesomeLife_Math | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=11|num-a=13}} | {{AIME box|year=2013|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:33, 15 March 2020
Contents
Problem 12
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 .
Cartesian Variation Solution
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 ordinate 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 2 (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 3(Trig with Diagram)
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 4 (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 5 (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 |
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