Difference between revisions of "2014 AIME II Problems/Problem 11"
(→Solution) |
m (→Solution 2) |
||
(17 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
==Problem 11== | ==Problem 11== | ||
− | In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math> | + | In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>RD=1</math>. Let <math>M</math> be the midpoint of segment <math>\overline{RD}</math>. Point <math>C</math> lies on side <math>\overline{ED}</math> such that <math>\overline{RC}\perp\overline{EM}</math>. Extend segment <math>\overline{DE}</math> through <math>E</math> to point <math>A</math> such that <math>CA=AR</math>. Then <math>AE=\frac{a-\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer. Find <math>a+b+c</math>. |
+ | |||
+ | ==Solution 1== | ||
+ | Let <math>P</math> be the foot of the perpendicular from <math>A</math> to <math>\overline{CR}</math>, so <math>\overline{AP}\parallel\overline{EM}</math>. Since triangle <math>ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and <math>AE = PM = \frac{CD}{2}</math>. We can then use coordinates. Let <math>O</math> be the foot of altitude <math>RO</math> and set <math>O</math> as the origin. Now we notice special right triangles! In particular, <math>DO = \frac{1}{2}</math> and <math>EO = RO = \frac{\sqrt{3}}{2}</math>, so <math>D\left(\frac{1}{2}, 0\right)</math>, <math>E\left(-\frac{\sqrt{3}}{2}, 0\right)</math>, and <math>R\left(0, \frac{\sqrt{3}}{2}\right).</math> <math>M =</math> midpoint<math>(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)</math> and the slope of <math>ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}</math>, so the slope of <math>RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.</math> Instead of finding the equation of the line, we use the definition of slope: for every <math>CO = x</math> to the left, we go <math>\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}</math> up. Thus, <math>x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.</math> <math>DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}</math>, and <math>AE = \frac{7 - \sqrt{27}}{22}</math>, so the answer is <math>\boxed{056}</math>. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(8cm); | ||
+ | pair a, o, d, r, e, m, cm, c,p; | ||
+ | o =(0,0); | ||
+ | d = (0.5, 0); | ||
+ | r = (0,sqrt(3)/2); | ||
+ | e = (-sqrt(3)/2,0); | ||
+ | |||
+ | m = midpoint(d--r); | ||
+ | draw(e--m); | ||
+ | cm = foot(r, e, m); | ||
+ | draw(L(r, cm,1, 1)); | ||
+ | c = IP(L(r, cm, 1, 1), e--d); | ||
+ | clip(r--d--e--cycle); | ||
+ | draw(r--d--e--cycle); | ||
+ | draw(rightanglemark(e, cm, c, 1.5)); | ||
+ | a = -(4sqrt(3)+9)/11+0.5; | ||
+ | dot(a); | ||
+ | draw(a--r, dashed); | ||
+ | draw(a--c, dashed); | ||
+ | pair[] PPAP = {a, o, d, r, e, m, c}; | ||
+ | for(int i = 0; i<7; ++i) { | ||
+ | dot(PPAP[i]); | ||
+ | } | ||
+ | label("$A$", a, W); | ||
+ | label("$E$", e, SW); | ||
+ | label("$C$", c, S); | ||
+ | label("$O$", o, S); | ||
+ | label("$D$", d, SE); | ||
+ | label("$M$", m, NE); | ||
+ | label("$R$", r, N); | ||
+ | p = foot(a, r, c); | ||
+ | label("$P$", p, NE); | ||
+ | draw(p--m, dashed); | ||
+ | draw(a--p, dashed); | ||
+ | dot(p); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>MN = x.</math> Meanwhile, since <math>\triangle R PM</math> is similar to <math>\triangle RCD</math> (angle, side, and side- <math>RP</math> and <math>RC</math> ratio), <math>CD</math> must be 2<math>x</math>. Now, notice that <math>AE</math> is <math>x</math>, because of the parallel segments <math>\overline A\overline E</math> and <math>\overline P\overline M</math>. | ||
+ | |||
+ | Now we just have to calculate <math>ED</math>. Using the Law of Sines, or perhaps using altitude <math>\overline R\overline O</math>, we get <math>ED = \frac{\sqrt{3}+1}{2}</math>. <math>CA=RA</math>, which equals <math>ED - x</math> | ||
+ | |||
+ | Using Law of Sine in <math>\triangle RED</math>, we find <math>RE</math> = <math>\frac{\sqrt{6}}{2}</math>. | ||
+ | |||
+ | We got the three sides of <math>\triangle AER</math>. Now using the Law of Cosines on <math>\angle AER</math>. There we can equate <math>x</math> and solve for it. We got <math>AE=x=\frac{\sqrt{3}-1}{4\sqrt{3}+2}</math>. Then rationalize the denominator, we get <math>AE = \frac{7 - \sqrt{27}}{22}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let <math>P</math> be the foot of the perpendicular from <math>A</math> to <math>\overline{CR}</math>, so <math>\overline{AP}\parallel\overline{EM}</math>. Since <math>\triangle ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and by midpoint theorem <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and therefore <math>AE = PM = \tfrac 12 CD</math>. | ||
+ | <asy> unitsize(8cm); pair a, d, r, e, m, cm, c,p; | ||
+ | d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m); | ||
+ | draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed); | ||
+ | pair[] PPAP = {a, d, r, e, m, c, p}; | ||
+ | for(int i = 0; i<7; ++i) { dot(PPAP[i]); } | ||
+ | label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW); | ||
+ | MA("60^\circ",black,c,d,m,0.07, black); | ||
+ | </asy> | ||
+ | We can now use coordinates with <math>D(0,0)</math> as origin and <math>DE</math> along the <math>x</math>-axis. | ||
+ | |||
+ | Let <math>RD=4</math> instead of <math>1</math> (in the end we will scale down by <math>4</math>). Since <math>\angle D = 60^\circ</math>, we get <math>R(2,2\sqrt{3})</math>, and therefore <math>M(1, \sqrt{3})</math>. | ||
+ | |||
+ | We use sine-law in <math>\triangle RED</math> to find the coordinates <math>E(2+2\sqrt{3}, 0)</math>:<cmath>DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}. </cmath> | ||
+ | Since slope<math>(ME)= -\sqrt{3}/(1+2\sqrt{3})</math>, and <math>RC\perp ME</math>, it follows that slope<math>(RC)=(1+2\sqrt{3})/\sqrt{3}</math>. If <math>C(c,0)</math> then we have<cmath>\frac{2\sqrt{3}}{2-c}=\frac{1+2\sqrt{3}}{\sqrt{3}}\qquad \Longrightarrow\qquad c=\frac{4\sqrt{3}-4}{1+2\sqrt{3}}</cmath> | ||
+ | Now <math>\tfrac 12 CD = \tfrac 12c =(2\sqrt{3}-2)/(1+2\sqrt{3})= \tfrac 1{11}(14-6\sqrt{3})</math>. | ||
+ | |||
+ | Scaling down by <math>4</math>, we get <math>AE=\tfrac 1{22}(7-3\sqrt{3})</math>, so our answer is <math>7+27+22=056</math>. | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
− | |||
− | |||
== See also == | == See also == |
Revision as of 17:24, 20 January 2024
Problem 11
In , and . . Let be the midpoint of segment . Point lies on side such that . Extend segment through to point such that . Then , where and are relatively prime positive integers, and is a positive integer. Find .
Solution 1
Let be the foot of the perpendicular from to , so . Since triangle is isosceles, is the midpoint of , and . Thus, is a parallelogram and . We can then use coordinates. Let be the foot of altitude and set as the origin. Now we notice special right triangles! In particular, and , so , , and midpoint and the slope of , so the slope of Instead of finding the equation of the line, we use the definition of slope: for every to the left, we go up. Thus, , and , so the answer is .
Solution 2
Let Meanwhile, since is similar to (angle, side, and side- and ratio), must be 2. Now, notice that is , because of the parallel segments and .
Now we just have to calculate . Using the Law of Sines, or perhaps using altitude , we get . , which equals
Using Law of Sine in , we find = .
We got the three sides of . Now using the Law of Cosines on . There we can equate and solve for it. We got . Then rationalize the denominator, we get .
Solution 3
Let be the foot of the perpendicular from to , so . Since is isosceles, is the midpoint of , and by midpoint theorem . Thus, is a parallelogram and therefore . We can now use coordinates with as origin and along the -axis.
Let instead of (in the end we will scale down by ). Since , we get , and therefore .
We use sine-law in to find the coordinates : Since slope, and , it follows that slope. If then we have Now .
Scaling down by , we get , so our answer is .
Video Solution
https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv
~MathProblemSolvingSkills.com
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.