Difference between revisions of "1994 AIME Problems/Problem 8"
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− | == Problem == | + | ==Problem== |
The points <math>(0,0)\,</math>, <math>(a,11)\,</math>, and <math>(b,37)\,</math> are the vertices of an equilateral triangle. Find the value of <math>ab\,</math>. | The points <math>(0,0)\,</math>, <math>(a,11)\,</math>, and <math>(b,37)\,</math> are the vertices of an equilateral triangle. Find the value of <math>ab\,</math>. | ||
− | == Solution == | + | ==Solution== |
+ | ===Solution 1=== | ||
Consider the points on the [[complex plane]]. The point <math>b+37i</math> is then a rotation of <math>60</math> degrees of <math>a+11i</math> about the origin, so: | Consider the points on the [[complex plane]]. The point <math>b+37i</math> is then a rotation of <math>60</math> degrees of <math>a+11i</math> about the origin, so: | ||
− | <cmath>(a+11i)\left(\ | + | <cmath>(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.</cmath> |
Equating the real and imaginary parts, we have: | Equating the real and imaginary parts, we have: | ||
Line 13: | Line 14: | ||
Solving this system, we find that <math>a=21\sqrt{3}, b=5\sqrt{3}</math>. Thus, the answer is <math>\boxed{315}</math>. | Solving this system, we find that <math>a=21\sqrt{3}, b=5\sqrt{3}</math>. Thus, the answer is <math>\boxed{315}</math>. | ||
− | '''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math> | + | '''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>. However, this triangle is just a reflection of the first triangle by the <math>y</math>-axis, and the signs of <math>a</math> and <math>b</math> are flipped. The product <math>ab</math> is unchanged. |
− | == See | + | ===Solution 2=== |
+ | Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? <math>\sqrt{3}</math> and perpendiculars inspires this solution: | ||
+ | |||
+ | First, drop a perpendicular from <math>O</math> to <math>AB</math>. Call this midpoint of <math>AB M</math>. Thus, <math>M=\left(\frac{a+b}{2}, 24\right)</math>. The vector from <math>O</math> to <math>M</math> is <math>\left[\frac{a+b}{2}, 24\right]</math>. Meanwhile from point <math>M</math> we can use a vector with <math>\frac{\sqrt{3}}{3}</math> the distance; we have to switch the <math>x</math> and <math>y</math> and our displacement is <math>\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]</math>. (Do you see why we switched <math>x</math> and <math>y</math> due to the rotation of 90 degrees?) | ||
+ | |||
+ | |||
+ | We see this displacement from <math>M</math> to <math>A</math> is <math>\left[\frac{a-b}{2}, 13\right]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>. | ||
+ | |||
+ | '''Note''': This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems". | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Plot this equilateral triangle on the complex plane. | ||
+ | Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives <math>\left(\frac{a+b}{3}, 16i\right)</math>. The new coordinates of the equilateral triangle are <math>\left(-\frac{a+b}{3}-16i\right), \left(a-\frac{a+b}{3}-5i\right), \left(b-\frac{a+b}{3}+21i\right)</math>. These three vertices are solutions of a cubic polynomial of form <math>x^3 + C</math>. By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots <math>r_1, r_2,</math> and <math>r_3,</math><math>\, r_1r_2 + r_2r_3 + r_3r_1 = 0</math>.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation <math>5a = 21b.</math> | ||
+ | Now use the equation with only real parts. This should give you a quadratic <math>a^2 - ab + b^2 = 1083</math>. Use your previously obtained equation to plug in for <math>a</math> and solve for <math>b</math>, which should yield <math>5\sqrt{3}</math>. <math>a</math> is then <math>\frac{21}{5}\sqrt{3}</math>. Multiplying <math>a</math> and <math>b</math> yields <math>\boxed{315}</math>. | ||
+ | |||
+ | ===Solution 4 (gigantic numbers)=== | ||
+ | Just using the Pythagorean Theorem, we get that <math>a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2</math>. | ||
+ | |||
+ | <math>a^2 + 121 = b^2 + 1369 ==> a^2 = b^2 + 1248</math>. Expanding the second and subtracting the first equation from it we get <math>b^2 = 2ab - 555</math> | ||
+ | . | ||
+ | <math>b^2 = 2ab - 555 ==> a^2 = 2ab + 693</math>. | ||
+ | |||
+ | We have <math>b^2 + 1248 = 2b\sqrt{b^2+1248} + 693</math>. | ||
+ | |||
+ | Moving the square root to one side and non square roots to the other we eventually get <math>b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2</math>. | ||
+ | |||
+ | <math>3b^4 + 3882b^2 - 308025 = 0</math>. | ||
+ | |||
+ | This factors to <math>(3b^2-225)(b^2+1369)</math>, so <math>3b^2 = 225, b = 5\sqrt 3</math>. | ||
+ | |||
+ | Plugging it back in, we find that a = <math>\sqrt{1323}</math> which is <math>21\sqrt3</math>, so the product <math>ab</math> is <math>\boxed {315}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=1994|num-b=7|num-a=9}} | {{AIME box|year=1994|num-b=7|num-a=9}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 02:09, 28 June 2024
Contents
Problem
The points ,
, and
are the vertices of an equilateral triangle. Find the value of
.
Solution
Solution 1
Consider the points on the complex plane. The point is then a rotation of
degrees of
about the origin, so:
Equating the real and imaginary parts, we have:
Solving this system, we find that . Thus, the answer is
.
Note: There is another solution where the point is a rotation of
degrees of
. However, this triangle is just a reflection of the first triangle by the
-axis, and the signs of
and
are flipped. The product
is unchanged.
Solution 2
Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? and perpendiculars inspires this solution:
First, drop a perpendicular from to
. Call this midpoint of
. Thus,
. The vector from
to
is
. Meanwhile from point
we can use a vector with
the distance; we have to switch the
and
and our displacement is
. (Do you see why we switched
and
due to the rotation of 90 degrees?)
We see this displacement from to
is
as well. Equating the two vectors, we get
and
. Therefore,
and
. And the answer is
.
Note: This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".
Solution 3
Plot this equilateral triangle on the complex plane.
Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives . The new coordinates of the equilateral triangle are
. These three vertices are solutions of a cubic polynomial of form
. By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots
and
.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation
Now use the equation with only real parts. This should give you a quadratic
. Use your previously obtained equation to plug in for
and solve for
, which should yield
.
is then
. Multiplying
and
yields
.
Solution 4 (gigantic numbers)
Just using the Pythagorean Theorem, we get that .
. Expanding the second and subtracting the first equation from it we get
.
.
We have .
Moving the square root to one side and non square roots to the other we eventually get .
.
This factors to , so
.
Plugging it back in, we find that a = which is
, so the product
is
.
~Arcticturn
See Also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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