Difference between revisions of "2014 AIME II Problems/Problem 14"
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− | + | ==Problem== | |
+ | In <math>\triangle{ABC}, AB=10, \angle{A}=30^\circ</math> , and <math>\angle{C=45^\circ}</math>. Let <math>H, D,</math> and <math>M</math> be points on the line <math>BC</math> such that <math>AH\perp{BC}</math>, <math>\angle{BAD}=\angle{CAD}</math>, and <math>BM=CM</math>. Point <math>N</math> is the midpoint of the segment <math>HM</math>, and point <math>P</math> is on ray <math>AD</math> such that <math>PN\perp{BC}</math>. Then <math>AP^2=\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | + | ==Diagram== | |
− | + | <asy> | |
+ | unitsize(20); | ||
+ | pair A = MP("A",(-5sqrt(3),0)), B = MP("B",(0,5),N), C = MP("C",(5,0)), M = D(MP("M",0.5(B+C),NE)), D = MP("D",IP(L(A,incenter(A,B,C),0,2),B--C),N), H = MP("H",foot(A,B,C),N), N = MP("N",0.5(H+M),NE), P = MP("P",IP(A--D,L(N,N-(1,1),0,10))); | ||
+ | D(A--B--C--cycle); | ||
+ | D(B--H--A,blue+dashed); | ||
+ | D(A--D); | ||
+ | D(P--N); | ||
+ | markscalefactor = 0.05; | ||
+ | D(rightanglemark(A,H,B)); | ||
+ | D(rightanglemark(P,N,D)); | ||
+ | MP("10",0.5(A+B)-(-0.1,0.1),NW); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | 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, | ||
− | |||
<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> | ||
− | |||
<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>. | ||
− | |||
<math>AHD</math> is <math>30-60-90</math> triangle. | <math>AHD</math> is <math>30-60-90</math> triangle. | ||
<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>. | ||
+ | Now we know that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find. | ||
+ | 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> | ||
+ | We can also use Law of Sines: | ||
+ | <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> | ||
+ | Then using right triangle <math>AHB</math>, we have <math>HB=10 \sin 15^\circ</math> | ||
+ | So <math>HB=10 \sin 15^\circ=\dfrac{5(\sqrt{6}-\sqrt{2})}{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>. | ||
+ | Finally if we calculate <math>(AP)^2</math>. | ||
+ | <math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>. | ||
+ | <math>m+n=\boxed{077}</math> | ||
− | + | -Gamjawon | |
+ | -edited by srisainandan6 to clarify and correct a small mistake | ||
− | + | ==Solution 2== | |
+ | Here's a solution that doesn't need <math>\sin 15^\circ</math>. | ||
− | + | As above, get to <math>AP=HM</math>. As in the figure, let <math>O</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>. Then <math>BCO</math> is a 45-45-90 triangle, and <math>ABO</math> is a 30-60-90 triangle. So <math>BO=5</math> and <math>AO=5\sqrt{3}</math>; also, <math>CO=5</math>, <math>BC=5\sqrt2</math>, and <math>MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}</math>. But <math>MO</math> and <math>AH</math> are parallel, both being orthogonal to <math>BC</math>. Therefore <math>MH:AO=MC:CO</math>, or <math>MH=\dfrac{5\sqrt3}{\sqrt2}</math>, and we're done. | |
+ | |||
+ | ==Solution 3== | ||
+ | Break our diagram into 2 special right triangle by dropping an altitude from <math>B</math> to <math>AC</math> we then get that <cmath>AC=5+5\sqrt{3}, BC=5\sqrt{2}.</cmath> | ||
+ | Since <math>\triangle{HCA}</math> is a 45-45-90, | ||
+ | |||
+ | <cmath>HC=\frac{5\sqrt2+5\sqrt6}{2}</cmath> | ||
+ | <math>MC=\frac{BC}{2},</math> | ||
+ | <cmath>HM=\frac{5\sqrt6}{2}</cmath> | ||
+ | <cmath>HN=\frac{5\sqrt6}{4}</cmath> | ||
+ | We know that <math>\triangle{AHD}\simeq \triangle{PND}</math> and are 30-60-90. | ||
+ | Thus, <cmath>AP=2 \cdot HN=\frac{5\sqrt6}{2}.</cmath> | ||
+ | |||
+ | <math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=\boxed{077}</math>. | ||
− | + | ==Solution 4== | |
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | import graph; size(15cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */ | ||
+ | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); | ||
− | + | draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq); | |
+ | /* draw figures */ | ||
+ | draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr); | ||
+ | draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr); | ||
+ | draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr); | ||
+ | draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr); | ||
+ | draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr); | ||
+ | draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr); | ||
+ | draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq); | ||
+ | draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq); | ||
+ | draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq); | ||
+ | draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq); | ||
+ | draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq); | ||
+ | /* dots and labels */ | ||
+ | dot((-1.4934334172297545,2.6953043701763835),dotstyle); | ||
+ | label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor); | ||
+ | dot((1.1286284157632023,-6.954814372303504),dotstyle); | ||
+ | label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor); | ||
+ | dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle); | ||
+ | label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor); | ||
+ | dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle); | ||
+ | label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor); | ||
+ | dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle); | ||
+ | label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor); | ||
+ | dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle); | ||
+ | label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor); | ||
+ | dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle); | ||
+ | label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor); | ||
+ | dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle); | ||
+ | label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor); | ||
+ | dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle); | ||
+ | label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor); | ||
+ | dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle); | ||
+ | label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor); | ||
+ | dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle); | ||
+ | label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor); | ||
+ | dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle); | ||
+ | label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | Draw the <math>45-45-90 \triangle AHC</math>. Now, take the perpendicular bisector of <math>BC</math> to intersect the circumcircle of <math>\triangle ABC</math> and <math>AC</math> at <math>F, L, G</math> as shown, and denote <math>O</math> to be the circumcenter of <math>\triangle ABC</math>. It is not difficult to see by angle chasing that <math>AHBGO</math> is cyclic, namely with diameter <math>AB</math>. Then, by symmetry, <math>EH = HB</math> and as <math>HB, OG</math> are both subtended by equal arcs they are equal. Hence, <math>EH = GO</math>. Now, draw line <math>HL</math> and intersect it at <math>AC</math> at point <math>K</math> in the diagram. It is not hard to use angle chase to arrive at <math>AEOL</math> a parallelogram, and from our length condition derived earlier, <math>AL \parallel HG</math>. From here, it is clear that <math>AK = KG</math>; that is, <math>P</math> is just the intersection of the perpendicular from <math>K</math> down to <math>BC</math> and <math>AD</math>! After this point, note that <math>AP = PF</math>. It is easily derived that the circumradius of <math>\triangle ABC</math> is <math>\frac{10}{\sqrt{2}}</math>. Now, <math>APO</math> is a <math>30-60-90</math> triangle, and from here it is easy to arrive at the final answer of <math>\boxed{077}</math>. ~awang11's sol | ||
− | + | ==Solution 5== | |
+ | [[File:2014 AIME II 14.png|500px|right]] | ||
+ | Let <math>BO \perp AC, O \in AC.</math> | ||
+ | |||
+ | Let <math>ME \perp BC, E \in AD.</math> | ||
− | + | <math>MB = MC, \angle C = 45^\circ \implies</math> points <math>M, E, O</math> are collinear. | |
− | + | <math>HN = NM, AH||NP||ME \implies AP = PE.</math> | |
− | + | In <math>\triangle ABO \hspace{10mm} \angle A = 30^\circ \implies AO = AB \cos 60^\circ = 5 \sqrt{3}.</math> | |
+ | |||
+ | In <math>\triangle AEO \hspace{10mm} \angle A = 15^\circ, \angle O = 90^\circ + 45^\circ = 135^\circ \implies</math> | ||
+ | <cmath>\angle AEO = 30^\circ \implies</cmath> | ||
+ | <cmath>AE = AO \frac {\sin 135^\circ}{\sin 30^\circ} = 5 \sqrt{3} \cdot \sqrt{2} = 5 \sqrt{6} \implies</cmath> | ||
+ | <cmath>AP = 5 \sqrt {\frac {3}{2}} \implies AP^2 = \frac {75}{2} \implies \boxed{\textbf{077}}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | |||
− | + | ==Video solution== | |
− | + | https://www.youtube.com/watch?v=SvJ0wDJphdU | |
− | + | == See also == | |
+ | {{AIME box|year=2014|n=II|num-b=13|num-a=15}} | ||
− | + | [[Category:Intermediate Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 19:40, 24 October 2024
Contents
[hide]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.