Difference between revisions of "2005 AIME II Problems/Problem 12"
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Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = \boxed{307}</math>. | Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = \boxed{307}</math>. | ||
− | === Solution 2 === | + | === Solution 2 (synthetic) === |
<center><asy> | <center><asy> | ||
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)); | ||
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Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. Now we know that <math>\angle{EOF}=45^\circ</math> therefore <math>\angle BOF+\angle AOE=45^\circ</math> also since <math>O</math> is the center of the square. Label the new triangle that we created <math>\triangle OGJ</math>. Now we know that rotation preserves angles and side lengths, so <math>BG=500-x</math> and <math>JC=x</math>. Draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle BOG =\angle OAE</math> since rotations preserve the same angles so | Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. Now we know that <math>\angle{EOF}=45^\circ</math> therefore <math>\angle BOF+\angle AOE=45^\circ</math> also since <math>O</math> is the center of the square. Label the new triangle that we created <math>\triangle OGJ</math>. Now we know that rotation preserves angles and side lengths, so <math>BG=500-x</math> and <math>JC=x</math>. Draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle BOG =\angle OAE</math> since rotations preserve the same angles so | ||
− | <math>\angle{FOG}=45^\circ</math> too | + | <math>\angle{FOG}=45^\circ</math> too. By SAS we know that <math>\triangle FOE\cong \triangle FOG,</math> so <math>FG=400</math>. Now we have a right <math>\triangle BFG</math> with legs <math>x</math> and <math>500-x</math> and hypotenuse <math>400</math>. By the [[Pythagorean Theorem]], |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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and applying the [[quadratic formula]] we get that | and applying the [[quadratic formula]] we get that | ||
− | <math>x=250\pm 50\sqrt{7}</math>. Since <math>BF > AE</math> we take the positive | + | <math>x=250\pm 50\sqrt{7}</math>. Since <math>BF > AE,</math> we take the positive root, and our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>. |
== See also == | == See also == |
Revision as of 20:53, 9 March 2014
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
Solution
Solution 1
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 .
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.