Difference between revisions of "2005 AIME II Problems/Problem 12"
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=== Solution 2 (synthetic) === | === Solution 2 (synthetic) === | ||
<center><asy> | <center><asy> | ||
+ | size(200); | ||
defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(1,0));label("\(J\)",J,(1,0));label("\(O\)",O,(1,-1)); label("\(x\)",(B+F)/2,(0,1)); label("\(400\)",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1)); | defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(1,0));label("\(J\)",J,(1,0));label("\(O\)",O,(1,-1)); label("\(x\)",(B+F)/2,(0,1)); label("\(400\)",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1)); | ||
</asy></center> | </asy></center> |
Revision as of 22:06, 19 November 2016
Problem
Square has center and are on with and between and and Given that where and are positive integers and is not divisible by the square of any prime, find
Contents
Solutions
Solution 1 (trigonometry)
Let be the foot of the perpendicular from to . Denote and , and (since and ). Then , and .
By the tangent addition rule , we see that Since , this simplifies to . We know that , so we can substitute this to find that .
Substituting again, we know have . This is a quadratic with roots . Since , use the smaller root, .
Now, . The answer is .
Solution 2 (synthetic)
Label , so . Rotate about until lies on . Now we know that therefore also since is the center of the square. Label the new triangle that we created . Now we know that rotation preserves angles and side lengths, so and . Draw and . Notice that since rotations preserve the same angles so too. By SAS we know that so . Now we have a right with legs and and hypotenuse . By the Pythagorean Theorem,
and applying the quadratic formula we get that . Since we take the positive root, and our answer is .
Solution 3
Let the midpoint of be and let , so then and . Drawing , we have , so By the Pythagorean Theorem on , Setting these two expressions for equal and solving for (it is helpful to scale the problem down by a factor of 50 first), we get . Since , we want the value , and the answer is .
See also
2005 AIME II (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.