Difference between revisions of "2013 AIME I Problems/Problem 12"
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− | == Problem | + | == Problem == |
Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^\circ</math> and <math>\angle Q = 60^\circ</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^\circ</math> and <math>\angle Q = 60^\circ</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | ||
− | == Solution == | + | == Solution 1 == |
First, find that <math>\angle R = 45^\circ</math>. | First, find that <math>\angle R = 45^\circ</math>. | ||
− | Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = | + | Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}</math> |
+ | |||
+ | ==Solution 2 (Cartesian Variation)== | ||
+ | Use coordinates. Call <math>Q</math> the origin and <math>QP</math> be on the x-axis. It is easy to see that <math>F</math> is the vertex on <math>RP</math>. After labeling coordinates (noting additionally that <math>QBC</math> is an equilateral triangle), we see that the area is <math>QP</math> times <math>0.5</math> times the coordinate of <math>R</math>. Draw a perpendicular of <math>F</math>, call it <math>H</math>, and note that <math>QP = 1 + \sqrt{3}</math> after using the trig functions for <math>75</math> degrees. | ||
+ | |||
+ | Now, get the lines for <math>QR</math> and <math>RP</math>: <math>y=\sqrt{3}x</math> and <math>y=-(2+\sqrt{3})x + (5+\sqrt{3})</math>, whereupon we get the ordinate of <math>R</math> to be <math>\frac{3+2\sqrt{3}}{2}</math>, and the area is <math>\frac{5\sqrt{3} + 9}{4}</math>, so our answer is <math>\boxed{021}</math>. | ||
+ | |||
+ | == Solution 3 (Trig) == | ||
+ | |||
+ | Angle chasing yields that both triangles <math>PAF</math> and <math>PQR</math> are <math>75</math>-<math>60</math>-<math>45</math> triangles. First look at triangle <math>PAF</math>. Using Law of Sines, we find: | ||
+ | |||
+ | <math>\frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{1} = \frac{\frac{\sqrt{2}}{2}}{PA}</math> | ||
+ | |||
+ | Simplifying, we find <math>PA = \sqrt{3} - 1</math>. | ||
+ | Since <math>\angle{Q} = 60^\circ</math>, WLOG assume triangle <math>BQC</math> is equilateral, so <math>BQ = 1</math>. So <math>PQ = \sqrt{3} + 1</math>. | ||
+ | |||
+ | Apply Law of Sines again, | ||
+ | |||
+ | <math>\frac{\frac{\sqrt{2}}{2}}{\sqrt{3} + 1} = \frac{\frac{\sqrt{3}}{2}}{PR}</math> | ||
+ | |||
+ | Simplifying, we find <math>PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})</math>. | ||
+ | |||
+ | <math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot \sin 75^\circ</math>. | ||
+ | |||
+ | Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | -by What do Humanitarians Eat? | ||
+ | |||
+ | ==Solution 5 (Trig)== | ||
+ | |||
+ | [[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=\fbox{021}</math>. | ||
+ | |||
+ | ==Solution 6 (Elementary Geo)== | ||
+ | |||
+ | We can find that <math>AF || CD || QR</math>. This means that the perpendicular from <math>P</math> to <math>QR</math> is perpendicular to <math>AF</math> as well, so let that perpendicular intersect <math>AF</math> at <math>G</math>, and the perpendicular intersect <math>QR</math> at <math>H</math>. Set <math>AP=x</math>. Note that <math>\angle {PAG} = 60^\circ</math>, so <math>AG=\frac{x}{2}</math> and <math>PG = GF = \frac{x\sqrt3}{2}</math>. Also, <math>1=AF=AG+GF=\frac{x}{2} + \frac{x\sqrt{3}}{2}</math>, so <math>x=\sqrt{3} - 1</math>. It's easy to calculate the area now, because the perpendicular from <math>P</math> to <math>QR</math> splits <math>\triangle{PQR}</math> into a <math>30-60-90</math> (PHQ) and a <math>45-45-90</math> (PHR). From these triangles' ratios, it should follow that <math>QH=\frac{\sqrt{3} + 1}{2}, PH=HR=\frac{\sqrt{3}+3}{2}</math>, so the area is <math>\frac{1}{2} * PH * QR = \frac{1}{2} * PH * (QH + HR) = \frac{1}{2} * \frac{\sqrt{3} + 3}{2} * \frac{2\sqrt{3}+4}{2} = \boxed{\frac{9+5\sqrt{3}}{4}}</math>. <math>9+5+3+4=021</math>. | ||
+ | By Mathscienceclass | ||
+ | |||
+ | ==Solution 7 (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}} |
Revision as of 02:25, 2 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 .
-by What do Humanitarians Eat?
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 6 (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 7 (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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