Difference between revisions of "2015 AIME I Problems/Problem 4"
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Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>. | Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>. | ||
− | == | + | ==Diagram== |
− | + | <asy> | |
+ | pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3)); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(B--C--EE--cycle); | ||
+ | draw(A--EE); | ||
+ | draw(C--D); | ||
+ | draw(B--M--NN--cycle); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(EE); | ||
+ | dot(M); | ||
+ | dot(NN); | ||
+ | label("A", A, SW); | ||
+ | label("B", B, S); | ||
+ | label("C", C, SE); | ||
+ | label("D", D, N); | ||
+ | label("E", EE, N); | ||
+ | label("M", M, NW); | ||
+ | label("N", NN, NE); | ||
+ | </asy> | ||
+ | |||
+ | Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FFFFFF">talk</font>]]) | ||
+ | |||
+ | ==Solution 1== | ||
+ | Let <math>A</math> be the origin, so <math>B=(16,0)</math> and <math>C=(20,0).</math> Using equilateral triangle properties tells us that <math>D=(8,8\sqrt3)</math> and <math>E=(18,2\sqrt3)</math> as well. Therefore, <math>M=(9,\sqrt3)</math> and <math>N=(14,4\sqrt3).</math> Applying the Shoelace Theorem to triangle <math>BMN</math> gives | ||
+ | |||
+ | <cmath>x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt3,</cmath> | ||
+ | |||
+ | so <math>x^2=\boxed{507}.</math> | ||
+ | |||
==Solution 2== | ==Solution 2== | ||
− | |||
− | |||
− | |||
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS. | Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS. | ||
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The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral. | The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral. | ||
+ | |||
+ | |||
+ | Note: A much easier way to go about finding <math>BM</math> without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that <math>AY = 18 \implies AX = 9 \implies BX = 16-9 = 7</math>. Additionally, <math>MX = \frac{1}{2} (2\sqrt{3}) = \sqrt{3}</math> from similar triangles meaning we can now just do pythagorean theorem on right triangle <math>MXB</math> to get <math>MB = \sqrt{52}</math> - SuperJJ | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | [[File:2015 AIME I 4.png|430px|right]] | ||
+ | |||
+ | <math> AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE \cong \triangle DBC</math> | ||
+ | |||
+ | Medians are equal, so <math>MB = MN, \angle ABM = \angle DBN \implies</math> | ||
+ | <math>\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies </math> | ||
+ | |||
+ | <math>\triangle MNB</math> is equilateral triangle. | ||
+ | |||
+ | The height of <math>\triangle BCE</math> is <math>2 \sqrt{3},</math> distance from <math>A</math> to midpoint <math>BC</math> is <math>16 + 2 = 18 \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} = 81 + 3 = 84.</math> | ||
+ | |||
+ | <math>BM</math> is the median of <math>\triangle ABE \implies</math> | ||
+ | <math>BM^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 - 84 = 52.</math> | ||
+ | |||
+ | The area of <math>\triangle BMN</math> | ||
+ | |||
+ | <cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2 =13 \sqrt{3} \implies \boxed{\textbf{507}}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==See Also== | ==See Also== |
Latest revision as of 23:12, 7 May 2024
Problem
Point lies on line segment with and . Points and lie on the same side of line forming equilateral triangles and . Let be the midpoint of , and be the midpoint of . The area of is . Find .
Diagram
Diagram by RedFireTruck (talk)
Solution 1
Let be the origin, so and Using equilateral triangle properties tells us that and as well. Therefore, and Applying the Shoelace Theorem to triangle gives
so
Solution 2
Note that and . Also, . Thus, by SAS.
From this, it is clear that a rotation about will map to . This rotation also maps to . Thus, and . Thus, is equilateral.
Using the Law of Cosines on , Thus, .
Using Stewart's Theorem on ,
Calculating the area of , Thus, , so . Our final answer is .
Admittedly, this is much more tedious than the coordinate solutions.
I also noticed that there are two more ways of showing that is equilateral:
One way is to show that , , and are related by a spiral similarity centered at .
The other way is to use the Mean Geometry Theorem. Note that and are similar and have the same orientation. Note that is the weighted average of and , is the weighted average of and , and is the weighted average of and . The weights are the same for all three averages. (The weights are actually just and , so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, is similar to both and , which means that is equilateral.
Note: A much easier way to go about finding without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that . Additionally, from similar triangles meaning we can now just do pythagorean theorem on right triangle to get - SuperJJ
Solution 3
Medians are equal, so
is equilateral triangle.
The height of is distance from to midpoint is
is the median of
The area of
vladimir.shelomovskii@gmail.com, vvsss
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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