You want...an EASIER way?

Ya I knew there was something easy like that...

AIME15 wrote: You want...an EASIER way?

DUDE THATS WHAT I GOT like down to the EXACT vertices even the 0 case is the same

Basically what I did, except slightly differently:

a_0=1 b_0=1 a_n=a_n-1 ^2 + 2b_n-1 ^2 b_n=2a_n-1b_n-1 (0,0),(1,1),(a_n,a_n^2) However that is not what I came up with during the test. What I basically did is (0,0),(1,1),(2,4); (0,0),(1,1),(9,81); (0,0),(1,1),(289,289^2) gives one for n=0,1,2. Multiply all x-coords by 4 to multiply area by 64 WIN

I only found the lame inductive solution.. I figured the inductive step out like instantly but took over 1/2 hour to find one for the case n=1... =/

For me n=0,1 were easy but it took me an hour to find a solution for n=2, using my definition of a "nice parabolic triangle"

AIME15 wrote: You want...an EASIER way?

Hmm, this is what I got. Yay.

This actually makes the problem really nice: the area of a triangle with vertices (a,a^2), (b,b^2), and (c,c^2) where a<b<c is 1/2*(b-a)(c-a)(c-b) (U can either use shoelace and factor or use the cross-product of the two vectors) It's a slightly stronger result

I did the same thing as numbersense. From there, I substituted that $c-b = 2^{2n+1} +1$ and $b-a=2^{2n+1}(2^{2n+1}+1)$. This reduces the area to the following. $A = \frac{1}{2}(2^{2n+1})(2^{2n+1}+1)^{4} = [2^{n}(2^{2n+1}+1)^{2}]^{2}$

Mewto55555 wrote: a_0=1 b_0=1 a_n=a_n-1 ^2 + 2b_n-1 ^2 b_n=2a_n-1b_n-1 (0,0),(1,1),(a_n,a_n^2) However that is not what I came up with during the test. What I basically did is (0,0),(1,1),(2,4); (0,0),(1,1),(9,81); (0,0),(1,1),(289,289^2) gives one for n=0,1,2. Multiply all x-coords by 4 to multiply area by 64 WIN I did the exact same thing except I wasn't able to find initial value when n=2 How did you find (0,1,289)?? When I set one variable for 0 or 1, I kept disproved the problem

Then set b=0, because you the factorized version makes it invariant. That's what I did. My solution: We factor like number.sense. Because it is invariant when we decrease each variable by any real number, we decrease it by b, to get: $\dfrac{ac(a-c)}{2}=2^{(2n)}m^2$. The coordinates of b are (0,0). To get it, let $a=a_1^2, c=c_1^2$, and let $a_1=2^{2n-1}+1$, and $c_1=2^{2n-1}-1$. Then $a_1^2c_1^2(a_1^2-c^2_1)=(2^{4n-2}-1)^2*2^{2n+1}$. (After bashing) This is always of that form, so it works. (Except for n=0, which is done by the case (9,9) (6,6) (0,0)) This problem was okay, pretty good.

AIME15 wrote: You want...an EASIER way?

Yay that was my solution too, except I used $4^n$ instead of $2^{2n}$ because it looks nicer

Does mine work? (Several posts above; I think/hope it does)

Just a note: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=206892 Altheman's mock ARML #8 immediately told me how to do the problem.

OneShotSniper wrote: Mewto55555 wrote: a_0=1 b_0=1 a_n=a_n-1 ^2 + 2b_n-1 ^2 b_n=2a_n-1b_n-1 (0,0),(1,1),(a_n,a_n^2) However that is not what I came up with during the test. What I basically did is (0,0),(1,1),(2,4); (0,0),(1,1),(9,81); (0,0),(1,1),(289,289^2) gives one for n=0,1,2. Multiply all x-coords by 4 to multiply area by 64 WIN I did the exact same thing except I wasn't able to find initial value when n=2 How did you find (0,1,289)?? When I set one variable for 0 or 1, I kept disproved the problem basically the area is a(a-1)/2 if the last point is a,a^2 you want this to be a perfect square, so a is just the square of the constant term of the expansion of (rt2-1)^n. and you try them until you get something divisible by 4 but not 8

mgao wrote: AIME15 wrote: You want...an EASIER way?

DUDE THATS WHAT I GOT like down to the EXACT vertices even the 0 case is the same isn't it funny how we all did it the same way...

thinker123 wrote: This problem is basically just finding an example. I don't know why it was accepted to be in the 6 problems.

(Because of this, I believe that the problem is actually quite instructive--it took me a nontrivial but reasonable amount of time to play around with the problem and obtain this construction.)

Consider the triangle with vertices $(-a, a^2), (b, b^2), (a, a^2)$. Assume that for convenience $b > a$. Then the area of triangle (using determinants or the shoelace formula) is $$\frac{1}{2}|(-ab^2 + ba^2 + a^3 + a^3 -ba^2 -ab^2)| = |a^3 - ab^2| = a(b^2 - a^2).$$We want to prove the existence of solutions to $a(b^2 - a^2) = 2^{2n}m^2$. Make the substitutions $a = 2^{2n}, b^2 - a^2 = m^2$. Then using the pythagorean triples generator $(k^2 - \ell^2, 2k \ell, k^2 + \ell^2)$, set $k = 2^{2n - 1}, \ell = 1$. Then $b = 4^{2n - 1} + 1$. Now evaluating $a(b^2 - a^2)$ yields; \begin{align*} a(b^2 - a^2) &= 2^{2n}(4^{4n - 2} + 2 \cdot 2^{4n - 2} + 1 - 2^{4n}) \\ &= 2^{2n}(4^{4n - 2} - 2^{4n - 1} + 1) \\ &= 2^{2n}(2^{8n - 4} - 2 \cdot 2^{4n - 2} + 1) \\ &= 2^{2n}(2^{4n - 2} - 1)^2. \end{align*} Now as $2^{4n - 2} - 1$ is always odd, we have the desired solution for any $n$; namely take $\boxed{(a, b) = (2^{2n}, 2^{4n - 2} + 1)}$, and the vertices of the triangle as $\boxed{(-a, a^2), (b, b^2), (a, a^2)}$. $\blacksquare$

I revisted this problem and found a more "natural" solution, though I am suprised that not many people did it this way. For $n=0$, take the triangle $(1,1),(-1,1),(0,0)$ which as area 1. For $n=1$, take $(5,25),(-4,16),(4,16)$. Assume $n\geq2$ from now on. Otherwise, let the vertices of the triangle be $(a,a^2),(b,b^2),(c,c^2).$ By Shoelace, the area is $$|ab^2+bc^2+ca^2-(a^2b+b^2c+c^2a)|.$$When $a=0$, this is just $$|bc^2-b^2c|.$$WLOG $c>b$, so that this is just $$bc(c-b).$$ We want this to be a perfect square. To do this, we will try to make $b,c,b-c$ all perfect squares, so we need a Pythagorean triple. Amusingly, we can do the following: Consider a Pythagorean triple $(x,y,z)$ with $x^2+y^2=z^2.$ Then, assign $c-b$ as $x^2$, $b$ as $y^2$, so $c$ is $z^2,$ so the area would be $(xyz)^2$ so $xyz=2^nm.$ Therefore, we just need to show that for any positive integer $n\geq2$ there exists a Pythagorean triple whose product of side lengths has $v_2$ of $n$. This can be achieved by taking a power of 2, $2^n$ as one of the sides, and making the other sides $(2^n/2)^2\pm 1,$ which will both be odd for $n\geq2$, so the product has a $v_2$ of $n$ and we are done.

AIME15 wrote: You want...an EASIER way?

How did you get the idea of taking these co-ordinates?

Am I the only one who did mod 3 for n and got if the triangle vertices are at $(a,a^2),(b,b^2),(c,c^2)$ then If $n \equiv 0 (\mod 3)$ then $a=0, b=2^{2k}, c=-2^{2k}$ If $n \equiv 1(\mod 3)$ then $a=0, b=7^2 \cdot 2^{2k}, c=7\cdot 8\cdot 2^{2k}$ If $n\equiv 2(\mod 3)$ then $a=0, b=31^2\cdot 2^{2k}, c=31\cdot 32\cdot 2^{2k}$ Where $k=\lfloor \dfrac{n}{3} \rfloor$? Got this by using Shoelace formula ignoring sign and setting $a=0$ for no reason Full proof here: https://infinityintegral.substack.com/p/usajmo-2010-contest-review

It turns out that we can always have $(0,0)$ as a vertex of the triangle. By Shoelace Theorem, the area of the triangle with vertices $(0,0)$, $(a,a^2)$, and $(b,b^2)$ is $\frac12 ab|a-b|$. Therefore, it suffices to specify a pair $(a,b)$ for each $n$. For $n=0$, use $(2,1)$. For $n=1$, use $(9,8)$. For $n=2$, use $(81,32)$. For $n=3k+r$, use the construction for $n=r$ but multiply both $a$ and $b$ by $2^k$. $\blacksquare$

Solved with Falcon311 We take points $O$, $(0, 0)$, $A$, $(m^2, m)$ and $B$, $(n^2, n)$. The area of the triangle formed by these three points is absolute value of $\frac{mn(m-n)}{2}$. Then, taking $m = -n$, we get $n^3$, taking $m = 4n$, we get $6n^3$ and $m = 8n$ we get $28n^3$. Then, taking $n = 2^{2k}$ in the first case, $n = 3\cdot 2^{2k-1}$ in the second case, and $n = 7\cdot 2^{2k}$ in the last case, we are done. (Whether we are given $n = 3k-2$, $3k-1$ or $3k$.

wao :O Letting the vertices be $(-a, a^2)$, $(a, a^2)$, and $(b, b^2)$ with $0<a<b$, we claim that $(a, b, m)$ can be \[(4^n, 4^{2n-1}+1, 4^{2n-1}-1).\][for $n\ge 1$, if $n=0$ take $(0, 0)$, $(1, 1)$, $(-1, 1)$ as the vertices] We may obtain that \[a(b^2-a^2)=4^n\cdot m^2\]by Shoelace (actually this isn't necessary at all lol just use normal base x height). Let $a=2^{2n}$ for “convenience”. So $b^2-2^{4n}=m^2$, or $m^2+2^{4n}=b^2$. The general form of Pythagorean triples is $(x^2-y^2, 2xy, x^2+y^2)$. So, choosing $x=2^{2n-1}$ and $y=1$, $m=4^{2n-1}-1$, and $b=4^{2n-1}+1$. Checking, it’s easy to see that $a(b^2-a^2)=2^{2n}\cdot m^2$ holds, so we are done. $\square$

Triangle with vertices $(0, 0), (-x, x^2), (x, x^2)$ has area $x^3$. And triangles with vertices $(x, x^2), (2x, 4x^2), (-2x, 4x^2)$ and $(x, x^2), (4x, 16x^2), (-4x, 16x^2)$ have areas $6x^3$ and $60x^3$, respectively. Note that $\{\nu_2(x^3), \nu_2(6x^3), \nu_2(60x^3)\}$ is a complete residue class modulo 3, hence, for given $n$, we can adjust $x$ such that $ax^3$ is a perfect square and $\nu_2(ax^3) = n$, where $a \in \{1, 6, 60\}$. Thus we're done. $\blacksquare$

I am not fully sure if this is right- any feedback would be appreciated. Let's define the 3 points of the triangle to be (a,a^2) (-a,a^2) and (b,b^2) and the variable a in this case is odd due to the problem statment. The base of the triangle is simply 2a and the height of the triangle is (a^2-b^2) where a is greater than b. We know (a^2-b^2) is less than or equal to a^2 since we know a square of a real number is always non-negative. So, using base times height/2, we find the area of the triangle is a(a^2-b^2). At this point, we can use inequalites to finish off the problem. The area the problem that is requested is (a^2)(2^2n). So, a(a^2-b^2) ≤ a^3=(a^2)(2^2n) and we can divide by a^2 ( for the latter half of the inequality), leaving us with a≤(2^2n). Taking log base 2 on both sides and dividing by 2 on both sides, we find that log₂a/2≤n. We know that the lowest possible value of a=1 will leave us with n=0, which satisfies the problem condition. As a increases, n will increase due to the nature of the log graph( as when the x increases on the log graph, the y also increases due to the nature of the log graph).

14 years of discussion... yet this one I believe hasn't been posted yet. EDIT: I could be wrong, so I'm looking for some feedback as well....

I thought this was harder than it actually was However, it is a very cute problem!! We look for a triangle with vertices $X(-a, a^2)$, $Y(a, a^2)$, $Z(b, b^2)$. Using the shoelace formula, we have: $[XYZ]= \lvert 2a^3- 2ab^2\rvert$ Take $Z$ “above” $XY$ so that the area is $[XYZ]=2ab^2-2a^3$ We want $$a(b-a)(b+a)=2^{2n-1}\cdot m^2$$Set $a=2^{n-1}$ and calculating, we get $b=2^{4n-4}-1$. This is a solution for $n\geq 2$. If $n=1$ we may pick: $X(-1,1)$, $Y(1,1)$, $Z(0,0)$.

We split $n$ into cases and construct a working $m$ and parabolic triangle. $\textbf{Case 1: } n \equiv 0 \pmod{3}$ Let $n = 3k$, then $m = 1$ and the parabolic triangle with vertices $(0, 0), (2^{2k}, 2^{4k}), (2^{2k + 1}, 2^{4k + 2})$ has area $\dfrac{1}{2}|2^{6k + 2} - 2^{6k + 1}| = 2^{6k} = (2^{3k} \cdot 1)^2$ which works. $\textbf{Case 2: } n \equiv 1 \pmod{3}$ Let $n = 3k + 1$, then $m = 49$ and the parabolic triangle with vertices $(0, 0), (7 \cdot 2^{2k}, 7^2 \cdot 2^{4k}), (7 \cdot 2^{2k + 3}, 7^2 \cdot 2^{4k + 6})$ has area $\dfrac{1}{2}|7^3 \cdot 2^{6k + 6} - 7^3 \cdot 2^{6k + 3}| = 7^4 \cdot 2^{6k + 2} = (2^{3k} \cdot 49)^2$ which works. $\textbf{Case 3: } n \equiv 2 \pmod{3}$ Let $n = 3k + 2$, then $m = 9$ and the parabolic triangle with vertices $(0, 0), (3 \cdot 2^{2k + 1}, 3 \cdot 2^{4k + 2}), (3 \cdot 2^{2k + 3}, 3^2 \cdot 2^{4k + 6})$ has area $\dfrac{1}{2}|3^3 \cdot 2^{6k + 7} - 3^3 \cdot 2^{6k + 5}| = 3^4 \cdot 2^{6k + 4} = (2^{3k + 2} \cdot 9)^2$ which works. We have exhausted all cases and have provided a valid construction. It is done.

Let the coordinates be $(a,a^2),(b,b^2),(c,c^2)$ Then by the shoelace formula, the area of the triangle will be :- \[Area = \dfrac{1}{2} \left|(ab^2 + bc^2 + ca^2) - (ba^2 + cb^2 + ac^2) \right|\]which is equal to :- $\frac{(a-b)(b-c)(c-a)}{2}$ Now we will apply a trick here, If we change the values of $a_2 = a_1 + k$ and $b_2 = b_1 +k$ then $a_2-b_2$ = $a_1-b_1$ This thing is pretty obvious, but here this matters a lot. And if we set $a = 0$, this will not matter, as then we can subtract the value a from b, c. Now, This sets the value of the area to $\frac{bc(b-c)}{2}$ \newline \newline Now, observe that if we set $b=4 b'$ and $c=4 c'$ Let the area be $A$ Now here if $A'$ = $\frac{b'c'(b'-c)}{2}$ Then $A = 64A'$ therefore, all that's left is to prove that for $n={0,1,2}$ Hence, proved.

The (signed) area of a parabolic triangle with vertices at $(x_1, x_1^2)$ and so on is given by the determinant \[\frac 12 \begin{vmatrix} x_1 & x_1^2 & 1 \\ x_2 & x_2^2 & 1 \\ x_3 & x_3^2 & 1 \end{vmatrix} = \frac 12(x_1-x_2)(x_2-x_3)(x_3-x_1).\]Now take $x_1-x_2 = \left(\frac{m^2-2}2\right)^2$ and $x_2-x_3 = 2m^2$, such that \[\frac 12 (x_1-x_2)(x_2-x_3)(x_3-x_1) = m^2\left(\frac{m^2-2}2\right)^2 \left(\frac{m^2+2}2\right)^2.\]By varying $\nu_2(m)$, we solve the problem for all $n \geq 1$. For the $n=0$ case, just use $(0, 0)$, $(1, 1)$, $(-1, 1)$, lol.