Difference between revisions of "2014 AIME II Problems/Problem 11"
5849206328x (talk | contribs) (Created page with "==Problem 11== In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>\abs{RD}=1</math>. Let <math>M</mat...") |
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==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== | + | ==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 = (CD)/2</math>. | + | 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>MP = x.</math> Meanwhile, because <math>\triangle RPM</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> | ||
+ | |||
+ | Finally, what is <math>RE</math>? It comes out to <math>\frac{\sqrt{6}}{2}</math>. | ||
+ | |||
+ | We got the three sides. Now all that is left is using the Law of Cosines. There we can equate <math>x</math> and solve for it. | ||
+ | |||
+ | Taking <math>\triangle AER</math> and using <math>\angle AER</math>, of course, we find out (after some calculation) that <math>AE = \frac{7 - \sqrt{27}}{22}</math>. The step before? <math>x=\frac{\sqrt{3}-1}{4\sqrt{3}+2}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2014|n=II|num-b=10|num-a=12}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 17:28, 29 August 2021
Contents
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, because 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
Finally, what is ? It comes out to .
We got the three sides. Now all that is left is using the Law of Cosines. There we can equate and solve for it.
Taking and using , of course, we find out (after some calculation) that . The step before? .
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 |
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