Difference between revisions of "2011 AIME II Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
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Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>. | Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>. | ||
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Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>. | Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>. | ||
− | Now, using [[Law of Cosines]] on <math>\triangle ABP | + | Now, letting <math>x = AP</math> and using [[Law of Cosines]] on <math>\triangle ABP</math>, we have |
− | < | + | <cmath>96=144+x^{2}-24x\frac{\sqrt{2}}{2}</cmath> |
+ | <cmath>0=x^{2}-12x\sqrt{2}+48</cmath> | ||
− | + | Using the quadratic formula, we arrive at | |
− | < | + | <cmath>x = \sqrt{72} \pm \sqrt{24}</cmath> |
− | + | Taking the positive root, <math>AP=\sqrt{72}+ \sqrt{24}</math> and the answer is thus <math>\framebox[1.5\width]{096.}</math> | |
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==Solution 2== | ==Solution 2== | ||
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<math>\angle{DPC}=\angle{CPB}</math> by symmetry, and <math>\angle{APB}=\angle{DP’C}</math> because translation preserves angles. Thus <math>\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ</math>. Therefore, quadrilateral <math>CPDP’</math> is cyclic. Thus the image of <math>O_1</math> coincides with <math>O_2</math>. | <math>\angle{DPC}=\angle{CPB}</math> by symmetry, and <math>\angle{APB}=\angle{DP’C}</math> because translation preserves angles. Thus <math>\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ</math>. Therefore, quadrilateral <math>CPDP’</math> is cyclic. Thus the image of <math>O_1</math> coincides with <math>O_2</math>. | ||
− | <math>O_1P</math> is parallel to <math>O_2P’</math> so <math>\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ</math>, so <math>\angle{PDP’}=60^\circ</math> and <math>\angle{PDC}=15^\circ</math>, thus <math>\angle{ADP}=75^circ</math>. | + | <math>O_1P</math> is parallel to <math>O_2P’</math> so <math>\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ</math>, so <math>\angle{PDP’}=60^\circ</math> and <math>\angle{PDC}=15^\circ</math>, thus <math>\angle{ADP}=75^{\circ}</math>. |
Let <math>M</math> be the foot of the perpendicular from <math>D</math> to <math>AC</math>. Then <math>\triangle{AMD}</math> is a 45-45-90 triangle and <math>\triangle{DMP}</math> is a 30-60-90 triangle. Thus | Let <math>M</math> be the foot of the perpendicular from <math>D</math> to <math>AC</math>. Then <math>\triangle{AMD}</math> is a 45-45-90 triangle and <math>\triangle{DMP}</math> is a 30-60-90 triangle. Thus | ||
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<math>AM=6\sqrt{2}</math> and <math>MP=\frac{6\sqrt{2}}{\sqrt{3}}</math>. | <math>AM=6\sqrt{2}</math> and <math>MP=\frac{6\sqrt{2}}{\sqrt{3}}</math>. | ||
− | This gives us <math>AP=AM+MP=\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{ | + | This gives us <math>AP=AM+MP=\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{096}.</math> |
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==Solution 5== | ==Solution 5== | ||
− | Reflect <math>O_1</math> across <math>AP</math> to <math>O_1'</math>. By symmetry <math>O_1’</math> is the | + | Reflect <math>O_1</math> across <math>AP</math> to <math>O_1'</math>. By symmetry <math>O_1’</math> is the circumcenter of <math>\triangle{ADP}</math> |
<math>\angle{DO_1’P}</math> = <math>2*\angle{DAP} = 90^\circ</math>, so <math>\angle{O_1’PD}=45^\circ</math> | <math>\angle{DO_1’P}</math> = <math>2*\angle{DAP} = 90^\circ</math>, so <math>\angle{O_1’PD}=45^\circ</math> | ||
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By symmetry, <math>\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ</math> | By symmetry, <math>\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ</math> | ||
− | Therefore, since <math>O_1’</math> is the | + | Therefore, since <math>O_1’</math> is the circumcenter of <math>\triangle{ADP}</math>, <math>\angle{ADP}</math> = <math>0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ</math> |
Therefore <math>\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ</math> | Therefore <math>\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ</math> | ||
− | Using sine rule in <math>\triangle{ADP}</math>, <math>AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{ | + | Using sine rule in <math>\triangle{ADP}</math>, <math>AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{096}.</math> |
By Kris17 | By Kris17 | ||
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+ | ==The Simple Way of the Coordinate Plane Solution== | ||
+ | Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set <math>A=(0, 12)</math>, <math>B=(12,12)</math>, <math>C=(12, 0)</math>, <math>D=(0, 0)</math>. Let this <math>P=(a, 12-a)</math> for some <math>a</math>. | ||
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+ | We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that <math>O_{1}</math> is the intersection of <math>x=6</math> and, knowing the midpoint of <math>AP</math> to be <math>(\frac{a}{2}, \frac{12-a}{2})</math> and thus the equation to be <math>y=x+(12-a)</math>, we get <math>(6, 18-a)</math>. Likewise for <math>O_{2}</math> it's <math>(6, 6-a)</math>. Now what do we see? <math>O_{1}P=O_{2}P</math> (just look at the coordinates)! So both of those distances are <math>4\sqrt{3}</math>. Solving for <math>a</math> we get it to be <math>6+2\sqrt{3}</math>, since <math>AP>CP</math>. Multiply by <math>\sqrt{2}</math> because we are looking for <math>AP</math> to get the answer of <math>\boxed{096}</math>. | ||
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+ | ==Solution 6 (Pure Angle Chasing)== | ||
+ | Let <math>\angle APD = \theta</math>. Then <math>\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}</math>. | ||
+ | Realize that because <math>O_1</math> is a circumcenter, <math>\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}</math>. Then <math>\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta</math>. | ||
+ | Now, because <math>P</math> lies on diagonal <math>AC</math>, <math>\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}</math>. | ||
+ | To finish, we look at <math>\triangle ADP</math>. Drop a perpendicular from <math>D</math> to <math>AP</math> at <math>E</math>. Then <math>\triangle ADE</math> is a <math>45-45-90</math> and <math>\triangle PDE</math> is a <math>30-60-90</math>. Therefore, <math>DE=EA=6\sqrt{2}, EP=2\sqrt{6}</math>, so <math>AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}</math>. <math>\blacksquare</math> ~msc | ||
==See also== | ==See also== |
Revision as of 19:10, 13 October 2020
Contents
Problem
Point lies on the diagonal of square with . Let and be the circumcenters of triangles and respectively. Given that and , then , where and are positive integers. Find .
Solution 1
Denote the midpoint of be and the midpoint of be . Because they are the circumcenters, both Os lie on the perpendicular bisectors of and and these bisectors go through and .
It is given that . Because and are radii of the same circle, the have the same length. This is also true of and . Because , . Thus, and are isosceles right triangles. Using the given information above and symmetry, . Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.
Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles and have measures of 30 degrees. Thus, both triangles and are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, . Because of 45-45-90 right triangles, .
Now, letting and using Law of Cosines on , we have
Using the quadratic formula, we arrive at
Taking the positive root, and the answer is thus
Solution 2
This takes a slightly different route than Solution 1.
Solution 1 proves that and that . Construct diagonal and using the two statements above it quickly becomes clear that by isosceles triangle base angles. Let the midpoint of diagonal be , and since the diagonals are perpendicular, both triangle and triangle are 30-60-90 right triangles. Since , and . 30-60-90 triangles' sides are in the ratio , so . . Hence, .
Solution 3
Use vectors. In an plane, let be , be , be , be , and be P, where . It remains to find .
The line is the perpendicular bisector of and , so and lies on the line. Now compute the perpendicular bisector of . The center has coordinate , and the segment is part of the -axis, so the perpendicular bisector has equation . Since is the circumcenter of triangle , it lies on the perpendicular bisector of both and , so Similarly, The relation can now be written using dot product as Computation of both sides yields Solve for gives , so . The answer is 72+24
Solution 4
Translate so that the image of coincides . Let the image of be .
by symmetry, and because translation preserves angles. Thus . Therefore, quadrilateral is cyclic. Thus the image of coincides with .
is parallel to so , so and , thus .
Let be the foot of the perpendicular from to . Then is a 45-45-90 triangle and is a 30-60-90 triangle. Thus
and .
This gives us , and the answer is
Solution 5
Reflect across to . By symmetry is the circumcenter of
= , so
similarly = , so
Thus, , so that
By symmetry,
Therefore, since is the circumcenter of , =
Therefore
Using sine rule in , , and the answer is
By Kris17
The Simple Way of the Coordinate Plane Solution
Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set , , , . Let this for some .
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that is the intersection of and, knowing the midpoint of to be and thus the equation to be , we get . Likewise for it's . Now what do we see? (just look at the coordinates)! So both of those distances are . Solving for we get it to be , since . Multiply by because we are looking for to get the answer of .
Solution 6 (Pure Angle Chasing)
Let . Then . Realize that because is a circumcenter, . Then . Now, because lies on diagonal , . To finish, we look at . Drop a perpendicular from to at . Then is a and is a . Therefore, , so . ~msc
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.