Difference between revisions of "2014 AIME II Problems/Problem 14"
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Let us just drop the perpendicular from <math>B</math> to <math>AC</math> and label the point of intersection <math>O</math>. We will use this point later in the problem. | Let us just drop the perpendicular from <math>B</math> to <math>AC</math> and label the point of intersection <math>O</math>. We will use this point later in the problem. | ||
As we can see, | As we can see, | ||
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<math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math> | <math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math> | ||
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<math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>. | <math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>. | ||
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<math>AHD</math> is <math>30-60-90</math> triangle. | <math>AHD</math> is <math>30-60-90</math> triangle. | ||
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<math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also. | <math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also. | ||
− | + | Then if we use those informations we get <math>AD=2HD</math> and <math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math>. | |
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− | Then if we use those informations we get <math>AD=2HD</math> and | ||
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− | <math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> | ||
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Now we know that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find. | Now we know that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find. | ||
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We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>, | We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>, | ||
− | + | We can chase those lengths and we would get <math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math> | |
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− | We can chase those lengths and we would get | ||
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− | <math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math> | ||
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We can also use Law of Sines: | We can also use Law of Sines: | ||
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<cmath>\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}</cmath> | <cmath>\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}</cmath> | ||
<cmath>\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}</cmath> | <cmath>\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}</cmath> | ||
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Then using right triangle <math>AHB</math>, we have <math>HB=10 \sin 15^\circ</math> | Then using right triangle <math>AHB</math>, we have <math>HB=10 \sin 15^\circ</math> | ||
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So <math>HB=10 \sin 15^\circ=\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>. | So <math>HB=10 \sin 15^\circ=\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>. | ||
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And we know that <math>AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}</math>. | And we know that <math>AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}</math>. | ||
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Finally if we calculate <math>(AP)^2</math>. | Finally if we calculate <math>(AP)^2</math>. | ||
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<math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>. | <math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>. | ||
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<math>m+n=\boxed{077}</math> | <math>m+n=\boxed{077}</math> | ||
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-Gamjawon | -Gamjawon |
Latest revision as of 19:40, 24 October 2024
Contents
Problem
In , and . Let and be points on the line such that , , and . Point is the midpoint of the segment , and point is on ray such that . Then , where and are relatively prime positive integers. Find .
Diagram
Solution 1
Let us just drop the perpendicular from to and label the point of intersection . We will use this point later in the problem. As we can see, is the midpoint of and is the midpoint of is a triangle, so . is triangle.
and are parallel lines so is triangle also. Then if we use those informations we get and and or . Now we know that , we can find for which is simpler to find. We can use point to split it up as , We can chase those lengths and we would get , so , so , so We can also use Law of Sines: Then using right triangle , we have So . And we know that . Finally if we calculate . . So our final answer is .
-Gamjawon -edited by srisainandan6 to clarify and correct a small mistake
Solution 2
Here's a solution that doesn't need .
As above, get to . As in the figure, let be the foot of the perpendicular from to . Then is a 45-45-90 triangle, and is a 30-60-90 triangle. So and ; also, , , and . But and are parallel, both being orthogonal to . Therefore , or , and we're done.
Solution 3
Break our diagram into 2 special right triangle by dropping an altitude from to we then get that Since is a 45-45-90,
We know that and are 30-60-90. Thus,
. So our final answer is .
Solution 4
Draw the . Now, take the perpendicular bisector of to intersect the circumcircle of and at as shown, and denote to be the circumcenter of . It is not difficult to see by angle chasing that is cyclic, namely with diameter . Then, by symmetry, and as are both subtended by equal arcs they are equal. Hence, . Now, draw line and intersect it at at point in the diagram. It is not hard to use angle chase to arrive at a parallelogram, and from our length condition derived earlier, . From here, it is clear that ; that is, is just the intersection of the perpendicular from down to and ! After this point, note that . It is easily derived that the circumradius of is . Now, is a triangle, and from here it is easy to arrive at the final answer of . ~awang11's sol
Solution 5
Let
Let
points are collinear.
In
In vladimir.shelomovskii@gmail.com, vvsss
Video solution
https://www.youtube.com/watch?v=SvJ0wDJphdU
See also
2014 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.