Difference between revisions of "2015 AIME I Problems/Problem 4"
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==Problem== | ==Problem== | ||
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="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST) | ||
+ | |||
+ | ==Solution== | ||
+ | Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>. | ||
+ | ==Solution 2== | ||
+ | Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle <math>BMN</math> to solve for its area. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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. | ||
+ | |||
+ | From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>. | ||
+ | This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral. | ||
+ | |||
+ | Using the Law of Cosines on <math>\triangle ABE</math>, | ||
+ | <cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)</cmath> | ||
+ | <cmath>AE = 4\sqrt{21}</cmath> | ||
+ | Thus, <math>AM=ME=2\sqrt{21}</math>. | ||
+ | |||
+ | Using Stewart's Theorem on <math>\triangle ABE</math>, | ||
+ | <cmath>AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath> | ||
+ | <cmath>BM = 2\sqrt{13}</cmath> | ||
+ | |||
+ | Calculating the area of <math>\triangle BMN</math>, | ||
+ | <cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath> | ||
+ | <cmath>[BMN] = 13\sqrt{3}</cmath> | ||
+ | Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>. | ||
+ | |||
+ | Admittedly, this is much more tedious than the coordinate solutions. | ||
+ | |||
+ | I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral: | ||
+ | |||
+ | One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>. | ||
+ | |||
+ | 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. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=3|num-a=5}} | {{AIME box|year=2015|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] |
Revision as of 19:52, 15 February 2021
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) 18:52, 15 February 2021 (EST)
Solution
Let point be at . Then, is at , and is at . Due to symmetry, it is allowed to assume and are in quadrant 1. By equilateral triangle calculations, Point is at , and Point is at . By Midpoint Formula, is at , and is at . The distance formula shows that . Therefore, by equilateral triangle area formula, , so is .
Solution 2
Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle to solve for its area.
Solution 3
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.