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Source: 2023 AMC 12A P23

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657 posts

#1 h Nov 9, 2023, 4:46 PM

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How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation
$(1+2a)(2+2b)(2a+b) = 32ab?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$

This post has been edited 3 times. Last edited by Spectator, Nov 9, 2023, 5:03 PM

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70 posts

#2 h Nov 9, 2023, 4:47 PM • 20 Y

Y by mathleticguyyy, JingheZhang, Mashiro, Pleaseletmewin, jacoporizzo, Geometry285, Spectator, Sedro, li31415926, roribaki, tauros, lamphead, happypi31415, Zhaom, Danielzh, KnowingAnt, peelybonehead, colinwang136, centslordm, megarnie

one-liner

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1831 posts

#3 h Nov 9, 2023, 4:49 PM • 3 Y

Y by BakedPotato66, Bradygho, megarnie

Yet another excellent one. Motivated by the fact that there probably finitely many solutions (otherwise the problem is bad), we treat this as a quadratic in $b$ and note that the minimum of it must be nonpositive for there to be a solution. Find the critical points by taking the derivative wrt $b$ :
$(1+2a)(2a + 2b + 1) = 16a\implies b = \frac{12a-1-4a^2}{2(2a+1)}.$ Plug this into the original equation and rearrange to get
$(16a)^2 - (4a^2 -1)^2 = 32a(12a - 1 - 4a^2)$ $\implies 16a^4 - 128a^3 + 120a^2 - 32a + 1 =0.$ The powers of $2$ practically beg for the substitution $a = 0.5x$ , so we do that and get
$x^4 - 16x^3 + 30x^2 - 16x + 1 =0.$ Divide through by $x^2$ to get
$x^2 - 16x + 30 - \frac{16}{x} + \frac{1}{x^2} = 0.$ $\implies \left(x + \frac{1}{x}\right)^2 - 16\left(x + \frac{1}{x}\right) + 28 = 0.$ Thus $x + \tfrac{1}{x}\in \{14,2\}$ . This gives $x = 1$ or $x = 7\pm 0.5\sqrt{195}$ . The latter two solutions fail as substituting $a = \tfrac{1}{2}x$ into the equation for $b$ in terms of $a$ makes $b$ negative (check with IVT). Thus $x = 1\implies a = 1/2\implies b = 1$ . That's $\boxed{1}$ solution.

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5 posts

#4 h Nov 9, 2023, 4:55 PM

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One of the easiest 20-25 problems in recent years.

By AM-GM,
$(1+2a)(2+2b)(2a+b)\ge 2\sqrt{2a}\cdot 2\sqrt{4b}\cdot 2\sqrt{2ab} = 8\sqrt{16a^2b^2}=32ab$ .

The equality condition yields one solution, $a=\frac{1}{2}, b=1$ .

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577 posts

#5 h Nov 9, 2023, 4:57 PM

Y by

Nice problem.

If there was any difficulty here, then it was in solving the quartic. By god's grace it was symmetric, so dividing through by $x^2$ and using the $t:=x+\frac{1}{x}$ substitution it's easy to finish.

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872 posts

#6 h Nov 9, 2023, 4:59 PM • 1 Y

Y by Dansman2838

so mad about misreading this -- read as just ordered pairs, put infinitely many, proceeded to lose 6 points

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898 posts

#7 h Nov 9, 2023, 5:05 PM

Y by

AM-GM yields the LHS’s minimum value is merely 32ab

2a=1, 2b=2, 2a=b, (1/2,1) the only pair

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2982 posts

#8 h Nov 9, 2023, 5:06 PM

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pretty much immediately you can assume the answer is A or B because it's most definitely some inequality

and then find that (1/2,1) works and you're done

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2534 posts

#9 h Nov 9, 2023, 5:07 PM

Y by

why was this so trivial lol? i solved quite quickly after someone sent me this

solution

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1922 posts

#10 h Nov 9, 2023, 5:15 PM

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Let $u=2a$ and $v=b$ . Then the equation magically turns into
$(u-v)^2+u(v-1)^2+v(u-1)^2=0,$ so $u=v=1$ , so $\left(\frac12,1\right)$ is the only solution.

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12155 posts

#11 h Nov 9, 2023, 5:19 PM • 1 Y

Y by Dansman2838

The inequality solution is blatantly obvious what was MAA doing this year

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2281 posts

#12 h Nov 9, 2023, 5:35 PM

Y by

5 sec solve, idk why this was 23

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245 posts

#13 h Nov 10, 2023, 3:27 AM

Y by

I realized this manipulation works post-test (in-test I'd done something different)

Let $c = 2a$ . Then:

$\frac{(1+c)(1+b)}{4} \ge \sqrt{bc} \ge \frac{2bc}{b+c}$ with equality iff $c = b = 1$ (the first step is direct AM-GM on $1, b, c, bc$ , the second is GM-HM).

This gives only 1 solution.

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1437 posts

#14 h Nov 10, 2023, 8:20 AM

Y by

Video Solution

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280 posts

#15 h Apr 11, 2025, 10:39 PM • 2 Y

Y by fake123, KnowingAnt

I don't know AM-GM, I fear.

First, let us divide both sides by $2$ . We get
$(1+2a)(1+b)(2a+b)=16ab.$ Expanding everything, moving all terms to the LHS, and grouping in terms of $a$ , this is equivalent to
$(4b+4)a^2+(2b^2-12b+2)a+(b^2+b)=0.$ Now, by the quadratic formula, we get that
$a=\frac{-(2b^2-12b+2)\pm \sqrt{(2b^2-12b+2)^2-4(4b+4)(b^2+b)}}{2(4b+4)}.$ This has a real solution in $a$ only if the discriminant,
$(2b^2-12b+2)^2-4(4b+4)(b^2+b),$ is $\ge 0$ . Expanding, this is
$b^4-16b^3+30b^2-16b+1\ge 0.$ But this is a symmetric polynomial, so we can find its roots and use them to analyze when the polynomial yields nonnegative values! Dividing both sides by $b^2$ and grouping terms,
$\left(b^2+\frac{1}{b^2}\right)-16\left(b+\frac 1b\right)+30=0,$ and setting $c=b+\frac 1b$ ,
$(c^2-2)-16c+30=0,$ which factors into
$(c-2)(c-14)=0.$ Now, solving back for $b$ , we have
* At $c=2$ , $b+\frac 1b=2$ , yielding a double root of $b=1$ ,
* And at $c=14$ , we have $b+\frac 1b=14$ , yielding roots of $7\pm 4\sqrt 3$ .
Therefore our discriminant is
$\iff (b-1)^2(b-(7+4\sqrt3))(b-(7-4\sqrt3))\ge 0.$ If we were allowed calculators, we could graph this;

$[asy] import graph; size(300,300); Label f; f.p=fontsize(6); xaxis(-0.75,1.3,Ticks(f, 0.5)); yaxis(-1.75,1.25,Ticks(f, 0.5)); real f(real x) { return (x^4-16x^3+30x^2-16x+1); } draw(graph(f,0,1.3),RGB(160, 40, 40)+linewidth(1)); dot((1,0), RGB(160, 40, 40)); label("(1,0)", (1,0), N, RGB(160, 40, 40)); [/asy]$

However, the MAA doesn't allow calculators, so we'll also do this analytically. Since $(b-1)^2\ge 0$ always, we want $b=1$ or,
$(b-(7+4\sqrt3))(b-(7-4\sqrt3))\ge 0,$ which happens only when $b\le 7-4\sqrt 3$ , or $b\ge 7+4\sqrt 3$ . This means that the discriminant is nonnegative if and only if $\ge 1$ of the following conditions are satisfied:
* $b=1$ ,
* $b\le 7-4\sqrt3$ ,
* or $b\ge 7+4\sqrt3$ .
Now, this does not necessarily mean that $a$ is positive. We now make the following claim.

Claim. $\ge 1$ value of $a$ is positive only if $b^2-6b+1\le 0$ .

Proof.
Assume FTSOC that $b^2-6b+1$ is positive. Note that from before, we had that
$a=\frac{-(2b^2-12b+2)\pm \sqrt{(2b^2-12b+2)^2-4(4b+4)(b^2+b)}}{2(4b+4)}.$ Now, since $b^2-6b+1$ is positive, $-(2b^2-12b+2)=-2(b^2-6b+1)$ must be negative. This means that
$\frac{-(2b^2-12b+2)-\sqrt{(2b^2-12b+2)^2-4(4b+4)(b^2+b)}}{2(4b+4)},$ is negative, and cannot be a value of $a$ . Therefore if we want $\ge 1$ value of $a$ to be positive, we need the other root,
$a=\frac{-(2b^2-12b+2)+\sqrt{(2b^2-12b+2)^2-4(4b+4)(b^2+b)}}{2(4b+4)}.$ Now, since $b$ is positive,
$4(4b+4)(b^2+b),$ must also be positive, meaning that
$\left|\sqrt{(2b^2-12b+2)^2-4(4b+4)(b^2+b)}\right|<|b^2-6b+1|.$ However, this gives that
$a=\frac{-(2b^2-12b+2)+\sqrt{(2b^2-12b+2)^2-4(4b+4)(b^2+b)}}{2(4b+4)}<0,$ meaning that both possible values of $a$ are negative, a contradiction. Therefore, in order for there to exist a positive working value of $a$ , we need $b^2-6b+1\le 0$ , as desired. $\blacksquare$

Now we have an extra condition, i.e. $b^2-6b+1\le 0$ . Again, if we had the resources, we could graph this with our original one;

$[asy] import graph; size(300,300); Label f; Label g; f.p=fontsize(6); g.p=fontsize(6); xaxis(-1.75,6.5,Ticks(f, 1.0)); yaxis(-8.5,1.75,Ticks(f, 1.0)); real f(real x) { return (x^4-16x^3+30x^2-16x+1); } real g(real x) { return (x^2-6x+1); } draw(graph(f,0,1.65),RGB(255, 157, 0)+linewidth(1)); draw(graph(g,-0.125,6.125),RGB(160, 40, 40)+linewidth(1)); [/asy]$

But again, the MAA doesn't allow calculators, so we'll have to do this analytically. Taking the roots and doing sign analysis, we get that this happens if and only if
* $3-2\sqrt 2\le b\le 3+2\sqrt2$ .
Therefore we require:
* $3-2\sqrt 2\le b\le 3+2\sqrt2$ ,
* and at least one of the following:
* $b=1$ ,
* $b\le 7-4\sqrt3$ ,
* $b\ge 7+4\sqrt 3$ .

However, note that
$7-4\sqrt3<3-2\sqrt 2<3+2\sqrt 2<7+4\sqrt3,$ so the only value of $b$ that works is $b=1$ .

Plugging this back into the equation, we get that
$(1+2a)(4)(2a+1)=32a,$ which has only one root, which is $a=\frac 12$ . Therefore there is \textbf{exactly one} solution to our equation, and it is $\left(\frac 12, 1\right)$ .

This post has been edited 3 times. Last edited by peppapig_, Apr 11, 2025, 10:40 PM

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