This was not a USAMO problem. USAJMO #2 was the USAMO problem.

Thanks, edited!

BarbieRocks wrote:

I don't want to double-post, but answer this please! I'll delete this post once its answered.

I didn't get that number; I only proceeded with the coordinates of the center relative to the original points...

PHEW I thought that when i got the line y=-1/4 I was crazy or something (cause my drawings didn't look like it AT ALL). But anyways, I think it extends both ways infinitely, cause after getting the equation of the locus of points, taking the limits on both sides gives infinity. (I was expressing everything in $\tan \theta$). Though now that I think of it, that doesn't really prove it for sure does it?

I did get $y=-1/4$ but didn't prove that $x=2a_1+\frac{16a_1}{1-12a_1^2}$ could be any value. I checked afterward on Wolfram Alpha and it turns out $x$ can be any value. http://www.wolframalpha.com/input/?i=x%2B16x%2F%281-12x^2%29 Aaah the link's not working you should copy and paste the url. How many points do you think I could get if I didn't prove it?

I basically analytic geometried the whole thing. You can derive a formula for angle between two lines based upon their slopes, and from there do a lot of algebra to get the x co-ordinate as a rational function based on $a_1$, which has two vertical asymptotes and so the range of the function goes from negative infinity to infinity. y co-ordinate is -1/4 so yeah. (sorry for lack of latex, typing this on my phone)

y=-1/4 is the directrix of the parabola so it made sense to me. Anyone showed that all points on the line could be achieved? What i did was rather bashy. First I let the points be (2a, 4a^2) etc. And int he end got that ab+ac+bc=-3/16, and used that to show that the y-coordiante of the centroid of the equilateral triangle was 1/4. Then i just said that given the value of ab+ac+bc, there are infinite possible of values for a+b+c, the variable in the x-coordinate of the centroid. (I kinda made up a proof if it too)But I didn't show they worked.

I think it was actually easier to let the points be $\frac{tan \theta}{2},\,\frac{tan ^2 theta}{4}$. I ended up with $(-tan \theta,\,-1)$ unfortunately though...

If the problem says just "find the locus", how many points would i get if i wrote "y= -1/4"

You would get zero points. All USA(J)MO solutions require a proof.

Following up on what I put earlier, you can find the formula for the angle between two lines $\theta$ with slope $m$ and $n$ (a bit of technicalities due to choosing which is higher, but it doesn't really matter in the context) is $tan(\theta)=\frac{m-n}{1+mn}$ pretty easy to prove if you draw a picture. So then you know that the angle between the line at $a_1$ and the one at $a_2$ is $60$ degrees, and the other $120$. Explicit formulas for $a_2$ and $a_3$ can be formed for $a_1$, and as it is easy to find the centroid is $(x,y)=(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_1a_3+a_2a_3}{3})$, you just algebra bash it out to the rational function given above for the x co-ordinate, and the y co-ordinate is obviously $\frac{-1}{4}$. Then you argue that the rational function has a range of all real numbers due to the behavior at the asymptotes, and you're done.

You could also take the centroid's x-coordinate and express it solely in terms of a_1. Then let the whole thing equal n. You then have a cubic in a_1, and a cubic always has at least one real root. Thus, for any value of n, you can obtain a solution for a_1, so all points on the line are in the locus.

fml, I simplified algebra wrong in like the second to last step on the x co-ordinate. Hopefully a 6 though.

We can prove a more general result: Let the tangents to the parabola at $P_1$, $P_2$, and $P_3$ determine a triangle $\Delta$. Then the orthocenter of $\Delta$ lies on the directrix of the parabola. Using coordinates, we can compute that the orthocenter of the triangle is \[H = \left( \frac{a_1 + a_2 + a_3 + 4a_1 a_2 a_3}{2}, -\frac{1}{4} \right).\] We can also prove the result synthetically, as follows: Lemma 1. Let $F$ and $V$ be the focus and vertex of the parabola, respectively. For any point $P$ on the parabola, let $\ell(P)$ denote the tangent to the parabola at $P$. Then for any point $P$ on the parabola, the projection of $F$ onto $\ell(P)$ lies on $\ell(V)$. (We can prove this, for example, using the reflective property of the parabola.) Lemma 2. Let $H$ and $\omega$ denote the orthocenter and circumcircle of triangle $ABC$. Then for any point $P$ on $\omega$, the Simson line of $P$ with respect to $\omega$ passes through the midpoint of $PH$. Let $Q_1 = \ell(P_2) \cap \ell(P_3)$, $Q_2 = \ell(P_3) \cap \ell(P_1)$, and $Q_3 = \ell(P_1) \cap \ell(P_2)$. Let $R_i$ denote the projection of $F$ onto $\ell(P_i)$. By Lemma 1, $R_1$, $R_2$, and $R_3$ all lie on $\ell(V)$. Hence, $\ell(V)$ is the Simson line of $F$ with respect to the circumcircle of triangle $R_1 R_2 R_3$, which means that $F$ also lies on this circumcircle. Let $H$ be the orthocenter of triangle $Q_1 Q_2 Q_3$. Then by Lemma 2, the midpoint of $FH$ lies on $\ell(V)$, which means that $H$ lies on the directrix of the parabola. [asy][asy] import graph; import geometry; real f(real x){return x^2;} unitsize(2.5 cm); pair F, H; pair[] P, Q, R; real[] a, m; //m[1] = 2.6; //m[2] = (m[1] + sqrt(3))/(1 - m[1]*sqrt(3)); //m[3] = (m[2] + sqrt(3))/(1 - m[2]*sqrt(3)); m[1] = 2.6; m[2] = -1.2; m[3] = 0.6; for(int i = 1; i <= 3; ++i) { a[i] = m[i]/2; P[i] = (a[i],a[i]^2); } F = (0,1/4); Q[1] = extension(P[2], P[2] + (1,m[2]), P[3], P[3] + (1,m[3])); Q[2] = extension(P[3], P[3] + (1,m[3]), P[1], P[1] + (1,m[1])); Q[3] = extension(P[1], P[1] + (1,m[1]), P[2], P[2] + (1,m[2])); H = orthocentercenter(Q[1],Q[2],Q[3]); for(int i = 1; i <= 3; ++i) { R[i] = (F + reflect(P[i], P[i] + (1,m[i]))*(F))/2; } draw((-1.5,0)--(1.5,0),gray(0.5)); draw(graph(f,-1.5,1.5)); draw((-1.5,-1/4)--(1.5,-1/4)); draw(P[1]--Q[3]); draw(P[2]--Q[3]); draw(Q[1]--Q[2]); draw(circumcircle(Q[1],Q[2],Q[3])); draw(F--R[1],dashed); draw(F--R[2],dashed); draw(F--R[3],dashed); draw(F--H); dot("$F$", F, N); dot("$H$", H, NE); dot("$P_1$", P[1], E); dot("$P_2$", P[2], SW); dot("$P_3$", P[3], SE); dot("$Q_1$", Q[1], W); dot("$Q_2$", Q[2], E); dot("$Q_3$", Q[3], S); dot(R[1]); dot(R[2]); dot(R[3]); dot((F + H)/2); label("$\ell(V)$", (1.5,0), E); label("directrix", (1.5,-1/4), E); [/asy][/asy]

After I found the line, I took the limit of a3 as a1 approached $-(\frac {\sqrt{3}}{6})$ and I did the same thing for the other side.

pi37 wrote: I did get $y=-1/4$ but didn't prove that $x=2a_1+\frac{16a_1}{1-12a_1^2}$ could be any value. I checked afterward on Wolfram Alpha and it turns out $x$ can be any value. http://www.wolframalpha.com/input/?i=x%2B16x%2F%281-12x^2%29 Aaah the link's not working you should copy and paste the url. How many points do you think I could get if I didn't prove it? just decompose into partial fractions, and note the upper and lower bounds of a_1

Abe27342 wrote: Following up on what I put earlier, you can find the formula for the angle between two lines $\theta$ with slope $m$ and $n$ (a bit of technicalities due to choosing which is higher, but it doesn't really matter in the context) is $tan(\theta)=\frac{m-n}{1+mn}$ pretty easy to prove if you draw a picture. So then you know that the angle between the line at $a_1$ and the one at $a_2$ is $60$ degrees, and the other $120$. Explicit formulas for $a_2$ and $a_3$ can be formed for $a_1$, and as it is easy to find the centroid is $(x,y)=(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_1a_3+a_2a_3}{3})$, you just algebra bash it out to the rational function given above for the x co-ordinate, and the y co-ordinate is obviously $\frac{-1}{4}$. Then you argue that the rational function has a range of all real numbers due to the behavior at the asymptotes, and you're done. Yay that's what I got. However...I somehow got the y coordinate as -1...how many points will I lose from that?

There's an important condition that you haven't used yet - the triangle is equilateral.

nsato wrote: Lemma 2. Let $H$ and $\omega$ denote the orthocenter and circumcircle of triangle $ABC$. Then for any point $P$ on $\omega$, the Simson line of $P$ with respect to $\omega$ passes through the midpoint of $PH$. Can someone give me a hint on how to prove this lemma?

^Let the Simson line intersect $AH$ at $M$ and $BC$ at $N$; show that $MHNP$ is a parallelogram.

For ease of viewing, let $a_1,a_2,a_3=a,b,c$ respectively. It is not hard to see that if $P=(a,a^2)$, then $l(P)\rightarrow y=2ax-a^2$. To find the intersections, we have that $2ax-a^2=2bx-b^2\rightarrow x=\frac{a+b}{2} \text{ and } y=ab$. Hence we let the vertices of our triangle be $$A=\bigg(\frac{a+b}{2},ab\bigg) B=\bigg(\frac{b+c}{2},bc\bigg) C=\bigg(\frac{c+a}{2},ca\bigg)$$Finding the distances, we have $\overline{AB}=(a-c)\sqrt{b^2+\frac{1}{4}}$. Setting them equal, subtracting, and simplifying, we obtain that $4(ab+bc+ac)=-3\rightarrow ab+bc+ac=-\frac{3}{4}$. The centroid being the average of the vertices, we find that $$G=\bigg(\frac{a+b+c}{3},\frac{-\frac{3}{4}}{3}\bigg)=\bigg(\frac{a+b+c}{3},-\frac{1}{4}\bigg)$$Since there are no bounds on $a+b+c$, we have that the locus is $y=-\frac{1}{4}$. $\square$

^ Actually, this is not the only condition we get. We get two conditions: $$ab+bc+ca = -3/4$$and $$a+b+c=-12abc$$ We want to show that $a+b+c$ can take all values, that is, we want to show that the cubic $$f(t)=t^3+12kt^2-\frac{3}{4}t-k$$has three distinct real roots. $$f'(t)=3t^2+24kt-\frac{3}{4}$$Let $p,q$ be the roots of $f'(t)=0$. It's not hard to see that $p,q$ are real. In fact, we can assume $p<0<q$. Now it is easy to calculate that $$f(p)= - \left ( 32k^2+\frac{1}{2} \right )p >0$$and similarly $f(q)<0$, which ends the proof.

Outline(?) for another not-as-nice synthetic solution. WLoG let $a_2<a_1<a_3$. Define $A\equiv \ell(P_2)\cap\ell(P_3)$, $B\equiv \ell(P_3)\cap\ell(P_1)$, $C\equiv \ell(P_1)\cap\ell(P_2)$, and $O$ as the center of $\triangle ABC$—our main claim is that the locus of $O$ is the directrix of the parabola $f(x)=x^2$. First, because $\tfrac{d}{dx}(x^2)=2x$, $\ell(P_n)$ is tangent to the parabola $f(x)=x^2$ at $x=a_n$. Exploiting the reflection property and doing some angle chasing, we obtain a few important results: the focus of our parabola, denoted by $F$ from here on out, is the Fermat point of $\triangle P_1P_2P_3$; the points $A$, $B$, $C$, and $F$ are concylic; and the line $\overline{P_1F}$ is the interior bisector of $\angle BFC$. To finish, let $F'$ be the reflection of $F$ over $\overline{BC}$. Now, one can see that $B$, $O$, $C$, and $F'$ must be concyclic, which, in conjunction with $OB=OC$, implies that $\overline{OF'}$ is the exterior bisector of $\angle BF'C$. However, because $P$ lies on $\overline{BC}$, $\overline{P_1F}$ bisecting $\angle BFC$ implies that $\overline{P_1F'}$ must also bisect $\angle BF'C$; hence, $\overline{P_1F'}\perp\overline{OF'}$. Applying the reflection property once more yields that $F'$ is the foot of $P_1$ on the directrix of our parabola, so $\overline{OF'}$ is, itself, the directrix of our parabola. Showing that every point on the directrix can be obtained is trivial and left as an exercise for the reader (meaning that I'm too lazy to do it).

Let $A=(a,a^2), B=(b,b^2), C=(c,c^2)$ be three points such that $l(A),l(B),l(C)$ determine an equilateral triangle. Note that $l(A)\equiv y - a^2=2a(x-a)$, and similarly for the other points. Finding the pairwise intersections of these lines, we find that the vertices of our equilateral triangle are $$\left(\frac{a+b}{2},ab\right),\left(\frac{b+c}{2},bc\right),\left(\frac{c+a}{2},ca\right).$$Throw this onto the complex plane, and dilate from the origin (scale everything up) by a factor of $2$ to get rid of fractions. We know that $\triangle XYZ$ is equilateral iff $x^2+y^2+z^2=xy+yz+zx$, which means, after some algebra and equating real/complex parts, $$a^2+b^2+c^2-4(a^2b^2+b^2c^2+c^2a^2)=ab+bc+ca-4abc(a+b+c)$$$$a^2b+a^2c+b^2c+b^2a+c^2a+c^2b=6abc.$$Letting $(P, Q, R)=(a+b+c,ab+bc+ca,abc)$, we can write this as $$P^2-3Q-4Q^2+12PR=0$$$$PQ=9R$$Substituting for $R$ in our first equation, we have $$P^2-3Q-4Q^2+\frac{12P^2Q}{9}=0$$$$\implies 3P^2-9Q-12Q^2+4P^2Q=0$$$$\implies (4Q+3)(P^2-3Q)=0.$$We now have two cases: either $Q=-\frac{3}{4}$ or $P^2=3Q$. For the sake of contradiction, assume that $P^2=3Q$ $$\implies (a+b+c)^2\ge 3(ab+bc+ca)$$$$\implies a^2+b^2+c^2=ab+bc+ca.$$By the Cauchy-Schwarz Inequality, we know that $$(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2\implies a^2+b^2+c^2\ge ab+bc+ca,$$with equality iff $a=b=c$. Thus, if $P^2=3Q$, then $a=b=c$, and all three vertices of the equilateral triangle are the same point, which doesn't actually form an equilateral triangle. Thus, we can conclude that $P^2\neq 3Q$, so $Q=-\frac{3}{4}$. Since the center of our dilated triangle is $\left(\frac{2a+2b+2c}{3},\frac{2ab+2bc+2ca}{3}\right)$, by dilating back the actual center of our triangle is $\left(\frac{a+b+c}{3},\frac{ab+bc+ca}{3}\right)=\left(\frac{P}{3},\frac{Q}{3}\right)=\left(\frac{P}{3},-\frac{1}{4}\right)$. So far, we know that all equilateral triangles have a center with $y$-coordinate $-\frac{1}{4}$. Now, we will show that all points on the line $y=-\frac{1}{4}$ can be the center of the triangle. This is equivalent to showing that if $ab+bc+ca=-\frac{3}{4}$ and $a+b+c=-12abc$ (this is from our earlier conclusion that $PQ=9R$), then $a+b+c$ can take any real value. Let $abc=p$, so that $a+b+c=-12p$. Then, $a, b, c$ are the roots of $f(z)=z^3+12pz^2-\frac{3z}{4}-p$. We want to show that, no matter what $p$ is equal to, $f(z)$ has 3 distinct roots (which are $a, b, c$). Case 1: $p>0$. Note that $f(0)=-p<0$, $f\left(-\frac{1}{2}\right)=\frac{1}{4}+2p>0$, and $f(1)=\frac{1}{4}+11p>0$. Therefore, since the cubic is continuous and by observing the end behavior, we must have a root in each of the intervals $\left(-\infty,-\frac{1}{2}\right), \left(-\frac{1}{2},0\right),(0,1)$. Case 2: $p<0$. We have $f(0)=-p<0$, $f\left(\frac{1}{2}\right)=-\frac{1}{4}+2p<0$, and $f(-1)=-\frac{1}{4}+11p<0$. Therefore, since the cubic is continuous and by observing the end behavior, we must have a root in each of the intervals $\left(-\infty,-1\right), \left(1,0\right),\left(0,\frac{1}{2}\right)$. Case 3: $p=0$ This is trivial and obviously has three distinct roots, namely $0, \frac{\sqrt{3}}{2},-\frac{\sqrt3}{2}$. This shows that $f(z)$ has three distinct roots regardless of $p$, so $p=abc$ can be any real value and thus $-12p=a+b+c$ also can be any real value. The locus is therefore the line $y=-\frac{1}{4}$.

Take two arbitrary lines $\ell(B)$ and $\ell(C)$, passing through $(b,b^2)$ and $(c,c^2)$, respectively. The equation of $\ell(B)$ is $y= 2b \cdot x -b^2$ and the equation of $\ell(C)$ is $y=2c \cdot x -c^2$. The intersection of where the graphs meet is when $2bx-b^2=2cx-c^2$ or $2x(b-c)=(b-c)(b+c)$ or $x=\frac{b+c}{2}$. If follows that $y=bc$. We now note that the points $(\frac{a_1+a_2}{2},a_1a_1), (\frac{a_2+a_3}{2},a_2a_3),$ and $(\frac{a_3+a_1}{2},a_3a_1)$ must form an equilateral triangle $\Delta$. Now let $a_1+a_2+a_3=x$, $a_1a_2+a_2a_3+a_3a_1=y$, and $a_1a_2a_3=z$. By the distance formula, $$\sqrt{(\frac{a_3-a_1}{2})^2+a_2^2(a_3-a_1)^2}=\sqrt{(\frac{a_3-a_2}{2})^2+a_1^2(a_3-a_2)^2}=\sqrt{(\frac{a_2-a_1}{2})^2+a_3^2(a_2-a_1)^2}$$Let's focus on the first two expressions. Squaring both sides and multiplying be $4$, we get $$(a_3-a_1)^2+(2a_2a_3-2a_1a_2)^2=(a_3-a_2)^2+(2a_1a_3-2a_1a_2)^2$$Rearranging and using difference of squares, it follows that $$(2a_3-a_2-a_1)(a_2-a_1)=(2a_1a_3+2a_2a_3-4a_1a_2)(2a_1a_3-2a_2a_3)=$$$$(2a_1a_3+2a_2a_3-4a_1a_2)2a_3(a_1-a_2)$$Dividing by $a_2-a_1$ and then using our earlier substitutions, we get $$x-3a_3=(2y-6a_1a_2)2a_3=4a_3 \cdot y -12z$$Similar computations give the following system. $$x-3a_1=4a_1 \cdot y -12z$$$$x-3a_2=4a_2 \cdot y -12z$$$$x-3a_3=4a_3 \cdot y -12z$$Subtractiong the first equation from the second, we get $$3a_2-3a_1=(4a_1-4a_2) \cdot y$$It follows that $y=-\frac{3}{4}$. Then we get that $x=12z$ and $y=-\frac{3}{4}$. Note that the center of $\Delta$ is the average of the coordinates. It follows that that is $(\frac{x}{3},\frac{y}{3})=(\frac{x}{3},-\frac{1}{4})$. Since $x$ can be anything, the answer is $y=-\frac{1}{4}$. Note: How many points will I lose for not showing $x$ can be anything?

OK here is my write-up. I will say that I think showing every point is achievable is the bulk of problem. (In particular, I feel most solutions on this thread are incomplete.) The answer is the line $y = -1/4$. We piggy-back off the claim in nsato's post. Claim: [Naoki Sato] In general, the orthocenter of $\Delta$ lies on the directrix $y = -1/4$ of the parabola (even if the triangle $\Delta$ is not equilateral). Proof. By writing out the equation $y = 2a_i x - a_i^2$ for $\ell(P_i)$, we find the vertices of the triangle are located at \[ \left( \frac{a_1+a_2}{2}, a_1a_2 \right); \quad \left( \frac{a_2+a_3}{2}, a_2a_3 \right); \quad \left( \frac{a_3+a_1}{2}, a_3a_1 \right). \]The coordinates of the orthocenter can be checked explicitly to be \[ H = \left( \frac{a_1 + a_2 + a_3 + 4 a_1 a_2 a_3}{2}, - \frac 14 \right). \qquad \blacksquare \] This claim already shows that every point lies on $y = -1/4$. We now turn to showing that, even when restricted to equilateral triangles, we can achieve every point on $y = -1/4$. In what follows $a = a_1$, $b = a_2$, $c = a_3$ for legibility. Claim: Lines $\ell(a)$, $\ell(b)$, $\ell(c)$ form an equilateral triangle if and only if \begin{align*} a+b+c &= -12abc \\ ab+bc+ca &= -\frac34. \end{align*}Moreover, the $x$-coordinate of the equilateral triangle is $\frac13(a+b+c)$. Proof. The triangle is equilateral if and only if the centroid and orthocenter coincide, i.e. \[ \left( \frac{a+b+c}{3}, \frac{ab+bc+ca}{3} \right) = G = H = \left( \frac{a+b+c+4abc}{2}, -\frac14 \right). \]Setting the $x$ and $y$ coordinates equal, we derive the claimed equations. $\blacksquare$ Let $\lambda$ be any real number. We are tasked to show that \[ P(X) = X^3 - 3\lambda \cdot X^2 - \frac34 X + \frac{\lambda}{4} \]has three real roots (with multiplicity); then taking those roots as $(a,b,c)$ yields a valid equilateral-triangle triple whose $x$-coordinate is exactly $\lambda$, be the previous claim. To prove that, pick the values \begin{align*} P(-\sqrt{3}/2) &= -2\lambda \\ P(0) &= \tfrac14 \lambda \\ P(\sqrt{3}/2) &= -2\lambda. \end{align*}The intermediate value theorem (at least for $\lambda \neq 0$) implies that $P$ should have at least two real roots now, and since $P$ has degree $3$, it has all real roots. That's all.

v_Enhance wrote: Claim: Lines $\ell(a)$, $\ell(b)$, $\ell(c)$ form an equilateral triangle if and only if \begin{align*} a+b+c &= -12abc \\ ab+bc+ca &= -12. \end{align*}Moreover, the $x$-coordinate of the equaliteral triangle is $\frac13(a+b+c)$. You mean $ab+bc+ca=-\dfrac{3}{4}$?

myh2910 wrote: You mean $ab+bc+ca=-\dfrac{3}{4}$? Yes! Thanks for the correction. Looks like I'll have to edit the end of the proof too.

I believe you can coordinate bash this by finding the three equations of the lines in terms of a_1, x, and y, though I'm not sure how computationally feasible it is...

This is my first USAJMO 3 so I am probably wrong

Since $l(p)$ passes $P$ with a slope $2a$ so we can define $l(p)$ as $y=2a_1x-a_1^2 $. Therefore the 3 sides of the equilateral triangle are $$y=2a_1-a_1^2$$$$y=2a_2-a_2^2$$$$y=2a_3-a_3^2$$and if we find the intercepts of these tree lines we get $(\frac{a_1+a_2}{2}, a_1a_2)$, $(\frac{a_2+a_3}{2}, a_2a_3)$, $(\frac{a_3+a_1}{2}, a_3a_1)$. The centroid of this triangle is $(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_1a_3}{3})$. sine this triangle is an equilateral triangle, $(a_1a_2-a_2a_3)^2+(\frac{a_1-a_3}{2})^2=(a_1a_3-a_2a_3)^2+(\frac{a_1-a_2}{2})^2$ satisfies and also after expanding this equation you can get $4a_1(a_1a_2+a_1a_3-2a_2a_3)=a_2+a_3-2a_1$. If you find a different form of this you get $4a_2(a_2a_1+a_2a_3-2a_1a_3)=a_1+a_3-2a_2$ and $4a_3(a_3a_2+a_3a_1-2a_2a_1)=a_2+a_1-2a_3$ and by subtracting each other you can finally get $a_1a_2+a_1a_3+a_2a_3=-3/4$ therefore at the centroid $y=\frac{a_1a_2+a_2a_3+a_1a_3}{3}=-1/4$