Difference between revisions of "2011 USAJMO Problems/Problem 3"
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===Solution 1=== | ===Solution 1=== | ||
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+ | Note that the lines <math>l(P_1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of intersection are <cmath>\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).</cmath> The slopes of each side of this equilateral triangle are <cmath>2a_1,2a_2,2a_3,</cmath> and we want to find the locus of <cmath>\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).</cmath> Define the three complex numbers <math>w_n = 1+2a_ni</math> for <math>n=1,2,3</math>. Then note that the slope - that is, the imaginary part divided by the real part - of all <math>w_n^3</math> is constant, say it is <math>k</math>. Then for <math>n=1,2,3</math>, | ||
+ | |||
+ | \begin{align*} | ||
+ | \frac{\mathbb{Im}(w_n^3)}{\mathbb{Re}(w_n^3)} | ||
+ | &= \frac{\mathbb{Im}((1+2a_ni)^3)}{\mathbb{Re}((1+2a_ni)^3)}\\ | ||
+ | &= \frac{\mathbb{Im}(1+6a_ni-12a_n^2-8a_n^3i)}{\mathbb{Re}(1+6a_ni-12a_n^2-8a_n^3i)}\\ | ||
+ | &= \frac{6a_n-8a_n^3}{1-12a_n^2}\\ | ||
+ | &= k. | ||
+ | \end{align*} | ||
+ | |||
+ | Rearranging, we get that<cmath>8a_n^3 -12ka_n^2-6a_n+k=0,</cmath>or<cmath>a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.</cmath>Note that this is a cubic, and the roots are <math>a_1,a_2</math> and <math>a_3</math> which are all distinct, and so there are no other roots. Using Vieta's, we get that | ||
+ | <cmath>a_1+a_2+a_3=\frac{3k}2,</cmath>and<cmath>a_1a_2+a_2a_3+a_3a_1=\frac34.</cmath> | ||
+ | Obviously all values of <math>k</math> are possible, and so our answer is the line <cmath>\boxed{y=-\frac{1}{4}.}</cmath> <math>\blacksquare</math> | ||
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+ | ===Solution 2=== | ||
Note that all the points <math>P=(a,a^2)</math> belong to the parabola <math>y=x^2</math> which we will denote <math>p</math>. This parabola has a focus <math>F=\left(0,\frac{1}{4}\right)</math> and directrix <math>y=-\frac{1}{4}</math> which we will denote <math>d</math>. We will prove that the desired locus is <math>d</math>. | Note that all the points <math>P=(a,a^2)</math> belong to the parabola <math>y=x^2</math> which we will denote <math>p</math>. This parabola has a focus <math>F=\left(0,\frac{1}{4}\right)</math> and directrix <math>y=-\frac{1}{4}</math> which we will denote <math>d</math>. We will prove that the desired locus is <math>d</math>. | ||
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--Killbilledtoucan | --Killbilledtoucan | ||
− | ===Solution | + | ===Solution 3=== |
Note that the lines <math>l(P_1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of intersection are <cmath>\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).</cmath> The slopes of each side of this equilateral triangle are <cmath>2a_1,2a_2,2a_3,</cmath> and we want to find the locus of <cmath>\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).</cmath> We know that <cmath>2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)</cmath> for some <math>\theta.</math> Therefore, we can use the tangent addition formula to deduce <cmath>\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}</cmath> and <cmath>\begin{align*} | Note that the lines <math>l(P_1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of intersection are <cmath>\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).</cmath> The slopes of each side of this equilateral triangle are <cmath>2a_1,2a_2,2a_3,</cmath> and we want to find the locus of <cmath>\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).</cmath> We know that <cmath>2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)</cmath> for some <math>\theta.</math> Therefore, we can use the tangent addition formula to deduce <cmath>\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}</cmath> and <cmath>\begin{align*} | ||
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&=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\ | &=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\ | ||
&=-\frac{1}{4}.\end{align*}</cmath> Now we show that <math>\frac{a_1+a_2+a_3}{3}</math> can be any real number. Let's say <cmath>\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k</cmath> for some real number <math>k.</math> Multiplying both sides by <math>2-\tan^2\theta</math> and rearranging yields a cubic in <math>\tan\theta.</math> Clearly this cubic has at least one real solution. As <math>\tan \theta</math> can take on any real number, all values of <math>k</math> are possible, and our answer is the line <cmath>\boxed{y=-\frac{1}{4}.}</cmath> Of course, as the denominator could equal 0, we must check <math>\tan \theta=\pm \frac{1}{\sqrt{3}}.</math> <cmath>3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).</cmath> The left side is nonzero, while the right side is zero, so these values of <math>\theta</math> do not contribute to any values of <math>k.</math> So, our answer remains the same. <math>\blacksquare</math> ~ Benq | &=-\frac{1}{4}.\end{align*}</cmath> Now we show that <math>\frac{a_1+a_2+a_3}{3}</math> can be any real number. Let's say <cmath>\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k</cmath> for some real number <math>k.</math> Multiplying both sides by <math>2-\tan^2\theta</math> and rearranging yields a cubic in <math>\tan\theta.</math> Clearly this cubic has at least one real solution. As <math>\tan \theta</math> can take on any real number, all values of <math>k</math> are possible, and our answer is the line <cmath>\boxed{y=-\frac{1}{4}.}</cmath> Of course, as the denominator could equal 0, we must check <math>\tan \theta=\pm \frac{1}{\sqrt{3}}.</math> <cmath>3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).</cmath> The left side is nonzero, while the right side is zero, so these values of <math>\theta</math> do not contribute to any values of <math>k.</math> So, our answer remains the same. <math>\blacksquare</math> ~ Benq | ||
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== See also == | == See also == | ||
{{USAJMO newbox|year=2011|num-b=2|num-a=4}} | {{USAJMO newbox|year=2011|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:58, 15 June 2020
Problem
For a point in the coordinate plane, let denote the line passing through with slope . Consider the set of triangles with vertices of the form , , , such that the intersections of the lines , , form an equilateral triangle . Find the locus of the center of as ranges over all such triangles.
Solutions
Solution 1
Note that the lines are respectively. It is easy to deduce that the three points of intersection are The slopes of each side of this equilateral triangle are and we want to find the locus of Define the three complex numbers for . Then note that the slope - that is, the imaginary part divided by the real part - of all is constant, say it is . Then for ,
\begin{align*} \frac{\mathbb{Im}(w_n^3)}{\mathbb{Re}(w_n^3)} &= \frac{\mathbb{Im}((1+2a_ni)^3)}{\mathbb{Re}((1+2a_ni)^3)}\\ &= \frac{\mathbb{Im}(1+6a_ni-12a_n^2-8a_n^3i)}{\mathbb{Re}(1+6a_ni-12a_n^2-8a_n^3i)}\\ &= \frac{6a_n-8a_n^3}{1-12a_n^2}\\ &= k. \end{align*}
Rearranging, we get thatorNote that this is a cubic, and the roots are and which are all distinct, and so there are no other roots. Using Vieta's, we get that and Obviously all values of are possible, and so our answer is the line
Solution 2
Note that all the points belong to the parabola which we will denote . This parabola has a focus and directrix which we will denote . We will prove that the desired locus is .
First note that for any point on , the line is the tangent line to at . This is because contains and because . If you don't like calculus, you can also verify that has equation and does not intersect at any point besides . Now for any point on let be the foot of the perpendicular from onto . Then by the definition of parabolas, . Let be the perpendicular bisector of . Since , passes through . Suppose is any other point on and let be the foot of the perpendicular from to . Then in right , is a leg and so . Therefore cannot be on . This implies that is exactly the tangent line to at , that is . So we have proved Lemma 1: If is a point on then is the perpendicular bisector of .
We need another lemma before we proceed. Lemma 2: If is on the circumcircle of with orthocenter , then the reflections of across , , and are collinear with .
Proof of Lemma 2: Say the reflections of and across are and , and the reflections of and across are and . Then we angle chase where is the measure of minor arc on the circumcircle of . This implies that is on the circumcircle of , and similarly is on the circumcircle of . Therefore , and . So . Since , , and are collinear it follows that , and are collinear. Similarly, the reflection of over also lies on this line, and so the claim is proved.
Now suppose , , and are three points of and let , , and . Also let , , and be the midpoints of , , and respectively. Then since and , it follows that , , and are collinear. By Lemma 1, we know that , , and are the feet of the altitudes from to , , and . Therefore by the Simson Line Theorem, is on the circumcircle of . If is the orthocenter of , then by Lemma 2, it follows that is on . It follows that the locus described in the problem is a subset of .
Since we claim that the locus described in the problem is , we still need to show that for any choice of on there exists an equilateral triangle with center such that the lines containing the sides of the triangle are tangent to . So suppose is any point on and let the circle centered at through be . Then suppose is one of the intersections of with . Let , and construct the ray through on the same halfplane of as that makes an angle of with . Say this ray intersects in a point besides , and let be the perpendicular bisector of . Since and , we have . By the inscribed angles theorem, it follows that . Also since and are both radii, is isosceles and . Let be the reflection of across . Then , and so . It follows that is on , which means is the perpendicular bisector of .
Let intersect in points and and let be the point diametrically opposite to on . Also let intersect at . Then . Therefore is a right triangle and so . So and by the inscribed angles theorem, . Since it follows that is and equilateral triangle with center .
By Lemma 2, it follows that the reflections of across and , call them and , lie on . Let the intersection of and the perpendicular to through be , the intersection of and the perpendicular to through be , and the intersection of and the perpendicular to through be . Then by the definitions of , , and it follows that for and so , , and are on . By lemma 1, , , and . Therefore the intersections of , , and form an equilateral triangle with center , which finishes the proof. --Killbilledtoucan
Solution 3
Note that the lines are respectively. It is easy to deduce that the three points of intersection are The slopes of each side of this equilateral triangle are and we want to find the locus of We know that for some Therefore, we can use the tangent addition formula to deduce and Now we show that can be any real number. Let's say for some real number Multiplying both sides by and rearranging yields a cubic in Clearly this cubic has at least one real solution. As can take on any real number, all values of are possible, and our answer is the line Of course, as the denominator could equal 0, we must check The left side is nonzero, while the right side is zero, so these values of do not contribute to any values of So, our answer remains the same. ~ Benq
See also
2011 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.