Difference between revisions of "2003 AIME II Problems/Problem 14"
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Thus, <math>m+n = \boxed{051}</math>. | Thus, <math>m+n = \boxed{051}</math>. | ||
− | == Solution | + | == Solution 2 == |
<asy> | <asy> | ||
size(200); | size(200); | ||
Line 24: | Line 24: | ||
From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6. | From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6. | ||
− | + | <asy> | |
+ | size(200); | ||
+ | draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); | ||
+ | dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); | ||
+ | label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); | ||
+ | xaxis("$x$");yaxis("$y$"); | ||
+ | pair b=foot((10/sqrt(3),2),(0,0),(10,0)); | ||
+ | pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); | ||
+ | draw(b--(10/sqrt(3),2),dotted); | ||
+ | draw(f--(-8/sqrt(3),4),dotted); | ||
+ | label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); | ||
+ | </asy> | ||
− | + | Let the angle between the <math>x</math>-axis and segment <math>AB</math> be <math>\theta</math>, as shown above. Thus, as <math>\angle FAB=120^\circ</math>, the angle between the <math>x</math>-axis and segment <math>AF</math> is <math>60-\theta</math>, so <math>\sin{(60-\theta)}=2\sin{\theta}</math>. Expanding, we have | |
− | <center><math>\sin{60}\cos{ | + | <center><math>\sin{60}\cos{\theta}-\cos{60}\sin{\theta}=\frac{\sqrt{3}\cos{\theta}}{2}-\frac{\sin{\theta}}{2}=2\sin{\theta}</math></center> |
− | Isolating <math>\sin{ | + | Isolating <math>\sin{\theta}</math> we see that <math>\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}</math>, or <math>\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}</math>. Using the fact that <math>\sin^2{\theta}+\cos^2{\theta}=1</math>, we have <math>\frac{28}{3}\sin^2{\theta}=1</math>, or <math>\sin{\theta}=\sqrt{\frac{3}{28}}</math>. Letting the side length of the hexagon be <math>y</math>, we have <math>\frac{2}{y}=\sqrt{\frac{3}{28}}</math>. After simplification we find that that <math>y=\frac{4\sqrt{21}}{3}</math>. |
− | + | In particular, note that by the Pythagorean theorem, <math>b^2+2^2=y^2</math>, hence <math>b=10\sqrt{3}/3</math>. Also, if <math>C=(c,6)</math>, then <math>y^2=BC^2=4^2+(c-b)^2</math>, hence <math>c-b=8\sqrt{3}/3,</math> and thus <math>c=18\sqrt{3}/3</math>. Using similar methods (or symmetry), we determine that <math>D=(10\sqrt{3}/3,10)</math>, <math>E=(0,8)</math>, and <math>F=(-8\sqrt{3}/3,4)</math>. By the Shoelace theorem, | |
+ | <cmath>[ABCDEF]=\frac12\left|\begin{array}{cc} | ||
+ | 0&0\\ | ||
+ | 10\sqrt{3}/3&2\\ | ||
+ | 18\sqrt{3}/3&6\\ | ||
+ | 10\sqrt{3}/3&10\\ | ||
+ | 0&8\\ | ||
+ | -8\sqrt{3}/3&4\\ | ||
+ | 0&0\\ | ||
+ | \end{array}\right|=\frac12|60+180+80-36-60-(-64)|\sqrt{3}/3=48\sqrt{3}.</cmath> | ||
− | + | Hence the answer is <math>\boxed{51}</math>. | |
== See also == | == See also == |
Revision as of 19:53, 13 March 2015
Contents
Problem
Let and be points on the coordinate plane. Let be a convex equilateral hexagon such that and the y-coordinates of its vertices are distinct elements of the set The area of the hexagon can be written in the form where and are positive integers and n is not divisible by the square of any prime. Find
Solution
The y-coordinate of must be . All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting , and knowing that , we can use rewrite using complex numbers: . We solve for and and find that and that .
The area of the hexagon can then be found as the sum of the areas of two congruent triangles ( and , with height and base ) and a parallelogram (, with height and base ).
.
Thus, .
Solution 2
From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
Let the angle between the -axis and segment be , as shown above. Thus, as , the angle between the -axis and segment is , so . Expanding, we have
Isolating we see that , or . Using the fact that , we have , or . Letting the side length of the hexagon be , we have . After simplification we find that that .
In particular, note that by the Pythagorean theorem, , hence . Also, if , then , hence and thus . Using similar methods (or symmetry), we determine that , , and . By the Shoelace theorem,
Hence the answer is .
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.