Difference between revisions of "2014 AIME II Problems/Problem 14"
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==Solution 1== | ==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, | ||
Line 77: | Line 79: | ||
-Gamjawon | -Gamjawon | ||
+ | -edited by srisainandan6 to clarify and correct a small mistake | ||
− | ==Solution 2== | + | ==Solution 2== |
Here's a solution that doesn't need <math>\sin 15^\circ</math>. | Here's a solution that doesn't need <math>\sin 15^\circ</math>. | ||
Line 84: | Line 87: | ||
==Solution 3== | ==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 | + | 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{BM}{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>. | ||
− | <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 | ||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=SvJ0wDJphdU | ||
== See also == | == See also == |
Latest revision as of 20:53, 1 August 2020
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 .
Thank you.
-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
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.