Question

There exists a unique triple $(a,b,c)$ of positive real numbers that satisfies the equations

$2(a^2+1) = 3(b^2+1) = 4(c^2+1)$ and $ab+bc+ca = 1$.

Compute $a+b+c$.

Answer 1

$\frac{9\sqrt{23}}{23}$

Answer 2

Let the common value of the first equation be $K$. $2(a^2+1) = 3(b^2+1) = 4(c^2+1) = K$

From this, we can express $a^2, b^2, c^2$ in terms of $K$: $2a^2+2 = K \Rightarrow a^2 = \frac{K-2}{2}$ $3b^2+3 = K \Rightarrow b^2 = \frac{K-3}{3}$ $4c^2+4 = K \Rightarrow c^2 = \frac{K-4}{4}$

Since $a,b,c$ are positive real numbers, we must have $a^2 > 0, b^2 > 0, c^2 > 0$. This implies: $K-2 > 0 \Rightarrow K > 2$ $K-3 > 0 \Rightarrow K > 3$ $K-4 > 0 \Rightarrow K > 4$ Combining these, we must have $K > 4$.

Now, we can write $a,b,c$ as: $a = \sqrt{\frac{K-2}{2}}$ $b = \sqrt{\frac{K-3}{3}}$ $c = \sqrt{\frac{K-4}{4}}$

To simplify the second equation $ab+bc+ca=1$, we use a trigonometric substitution. Let $a = \tan x$, $b = \tan y$, $c = \tan z$. Since $a,b,c > 0$, we can choose $x,y,z \in (0, \pi/2)$. Then $a^2+1 = \tan^2 x + 1 = \sec^2 x = \frac{1}{\cos^2 x}$. $b^2+1 = \sec^2 y = \frac{1}{\cos^2 y}$. $c^2+1 = \sec^2 z = \frac{1}{\cos^2 z}$.

Substitute these into the first equation: $\frac{2}{\cos^2 x} = \frac{3}{\cos^2 y} = \frac{4}{\cos^2 z} = K$ This gives: $\cos^2 x = \frac{2}{K}$ $\cos^2 y = \frac{3}{K}$ $\cos^2 z = \frac{4}{K}$

Since $x,y,z \in (0, \pi/2)$, $\cos x, \cos y, \cos z$ are positive. $\cos x = \sqrt{\frac{2}{K}}$ $\cos y = \sqrt{\frac{3}{K}}$ $\cos z = \sqrt{\frac{4}{K}}$

Now we find $\sin x, \sin y, \sin z$: $\sin^2 x = 1 - \cos^2 x = 1 - \frac{2}{K} = \frac{K-2}{K} \Rightarrow \sin x = \sqrt{\frac{K-2}{K}}$ $\sin^2 y = 1 - \cos^2 y = 1 - \frac{3}{K} = \frac{K-3}{K} \Rightarrow \sin y = \sqrt{\frac{K-3}{K}}$ $\sin^2 z = 1 - \cos^2 z = 1 - \frac{4}{K} = \frac{K-4}{K} \Rightarrow \sin z = \sqrt{\frac{K-4}{K}}$

The second equation is $ab+bc+ca = 1$. In terms of $x,y,z$: $\tan x \tan y + \tan y \tan z + \tan z \tan x = 1$. This is a known trigonometric identity that holds if $x+y+z = \frac{\pi}{2}$ (since $x,y,z \in (0, \pi/2)$, their sum must be in $(0, 3\pi/2)$, and $\frac{\pi}{2}$ is the only possibility for this identity to hold true).

So, we have $x+y+z = \frac{\pi}{2}$. This implies $z = \frac{\pi}{2} - (x+y)$. Taking cosine on both sides: $\cos z = \cos\left(\frac{\pi}{2} - (x+y)\right) = \sin(x+y)$ Using the sine addition formula: $\sin(x+y) = \sin x \cos y + \cos x \sin y$. So, $\cos z = \sin x \cos y + \cos x \sin y$.

Substitute the expressions in terms of $K$: $\sqrt{\frac{4}{K}} = \sqrt{\frac{K-2}{K}}\sqrt{\frac{3}{K}} + \sqrt{\frac{2}{K}}\sqrt{\frac{K-3}{K}}$ $\frac{2}{\sqrt{K}} = \frac{\sqrt{3(K-2)}}{\sqrt{K}\sqrt{K}} + \frac{\sqrt{2(K-3)}}{\sqrt{K}\sqrt{K}}$ Multiply by $K$: $2\sqrt{K} = \sqrt{3(K-2)} + \sqrt{2(K-3)}$ Square both sides: $(2\sqrt{K})^2 = (\sqrt{3K-6} + \sqrt{2K-6})^2$ $4K = (3K-6) + (2K-6) + 2\sqrt{(3K-6)(2K-6)}$ $4K = 5K - 12 + 2\sqrt{6K^2 - 18K - 12K + 36}$ $4K = 5K - 12 + 2\sqrt{6K^2 - 30K + 36}$ Rearrange the terms to isolate the square root: $12 - K = 2\sqrt{6K^2 - 30K + 36}$

For the left side to be non-negative, $12-K \ge 0 \Rightarrow K \le 12$. Square both sides again: $(12-K)^2 = 4(6K^2 - 30K + 36)$ $144 - 24K + K^2 = 24K^2 - 120K + 144$ $0 = 24K^2 - K^2 - 120K + 24K + 144 - 144$ $0 = 23K^2 - 96K$ $K(23K - 96) = 0$

Since $K > 4$, $K \ne 0$. So, $23K - 96 = 0 \Rightarrow K = \frac{96}{23}$.

Let's check if this value of $K$ satisfies the conditions $K > 4$ and $K \le 12$: $K = \frac{96}{23} \approx 4.17$. $4 < 4.17 \le 12$. Both conditions are satisfied.

Now, we can compute $a, b, c$: $a = \sqrt{\frac{K-2}{2}} = \sqrt{\frac{96/23 - 2}{2}} = \sqrt{\frac{(96-46)/23}{2}} = \sqrt{\frac{50/23}{2}} = \sqrt{\frac{25}{23}} = \frac{5}{\sqrt{23}}$ $b = \sqrt{\frac{K-3}{3}} = \sqrt{\frac{96/23 - 3}{3}} = \sqrt{\frac{(96-69)/23}{3}} = \sqrt{\frac{27/23}{3}} = \sqrt{\frac{9}{23}} = \frac{3}{\sqrt{23}}$ $c = \sqrt{\frac{K-4}{4}} = \sqrt{\frac{96/23 - 4}{4}} = \sqrt{\frac{(96-92)/23}{4}} = \sqrt{\frac{4/23}{4}} = \sqrt{\frac{1}{23}} = \frac{1}{\sqrt{23}}$

Finally, compute $a+b+c$: $a+b+c = \frac{5}{\sqrt{23}} + \frac{3}{\sqrt{23}} + \frac{1}{\sqrt{23}} = \frac{5+3+1}{\sqrt{23}} = \frac{9}{\sqrt{23}}$ Rationalizing the denominator: $a+b+c = \frac{9\sqrt{23}}{23}$.