Question

Real numbers $x, y, z$ satisfy $x + xy + xyz = 1, \quad y + yz + xyz = 2, \quad z + xz + xyz = 4.$ The largest possible value of $xyz$ is $\frac{a+b\sqrt{c}}{d}$, where $a, b, c, d$ are integers, $d$ is positive, $c$ is square-free, and gcd$(a, b, d) = 1$. Find $1000a + 100b + 10c + d$.

Answer 1

5272

Answer 2

The given system of equations is:

1. $x + xy + xyz = 1$
2. $y + yz + xyz = 2$
3. $z + xz + xyz = 4$

Let $P = xyz$. We want to find the largest possible value of $P$. Rewrite the equations in terms of $P$:

1. $x(1+y) + P = 1 \implies x(1+y) = 1-P$
2. $y(1+z) + P = 2 \implies y(1+z) = 2-P$
3. $z(1+x) + P = 4 \implies z(1+x) = 4-P$

Multiply these three modified equations: $x(1+y) \cdot y(1+z) \cdot z(1+x) = (1-P)(2-P)(4-P)$ $xyz (1+x)(1+y)(1+z) = (1-P)(2-P)(4-P)$ Substitute $xyz = P$: $P(1+x)(1+y)(1+z) = (1-P)(2-P)(4-P)$

Now, let's expand the term $(1+x)(1+y)(1+z)$: $(1+x)(1+y)(1+z) = (1+x+y+xy)(1+z)$ $= 1+z+x+xz+y+yz+xy+xyz$ $= 1 + (x+y+z) + (xy+yz+xz) + xyz$

Let $S_1 = x+y+z$ and $S_2 = xy+yz+xz$. Also, $xyz = P$. So, $(1+x)(1+y)(1+z) = 1 + S_1 + S_2 + P$.

Now, let's sum the three modified equations: $(x(1+y)) + (y(1+z)) + (z(1+x)) = (1-P) + (2-P) + (4-P)$ $x+xy+y+yz+z+xz = 7-3P$ Rearranging the terms: $(x+y+z) + (xy+yz+xz) = 7-3P$ $S_1 + S_2 = 7-3P$.

Substitute this into the equation $P(1+x)(1+y)(1+z) = (1-P)(2-P)(4-P)$: $P(1 + (S_1+S_2) + P) = (1-P)(2-P)(4-P)$ $P(1 + (7-3P) + P) = (1-P)(2-P)(4-P)$ $P(8 - 2P) = (1-P)(2-P)(4-P)$ $2P(4-P) = (1-P)(2-P)(4-P)$

We can see that $(4-P)$ is a common factor. Case 1: $4-P = 0 \implies P=4$. If $P=4$, the equation becomes $2P(0) = (1-P)(2-P)(0)$, which is $0=0$. So $P=4$ is a valid solution.

Case 2: $4-P \neq 0$. We can divide both sides by $(4-P)$: $2P = (1-P)(2-P)$ $2P = 2 - P - 2P + P^2$ $2P = P^2 - 3P + 2$ Rearrange into a quadratic equation: $P^2 - 5P + 2 = 0$

Solve for $P$ using the quadratic formula $P = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $P = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)}$ $P = \frac{5 \pm \sqrt{25 - 8}}{2}$ $P = \frac{5 \pm \sqrt{17}}{2}$

So, the three possible values for $P=xyz$ are:

1. $P_1 = 4$
2. $P_2 = \frac{5 + \sqrt{17}}{2}$
3. $P_3 = \frac{5 - \sqrt{17}}{2}$

To find the largest possible value, we compare these numbers. We know that $4 < \sqrt{17} < 5$ (since $4^2=16$ and $5^2=25$). Approximate value of $\sqrt{17} \approx 4.12$. $P_1 = 4$ $P_2 = \frac{5 + \sqrt{17}}{2} \approx \frac{5 + 4.12}{2} = \frac{9.12}{2} = 4.56$ $P_3 = \frac{5 - \sqrt{17}}{2} \approx \frac{5 - 4.12}{2} = \frac{0.88}{2} = 0.44$

Comparing these values, the largest value is $P_2 = \frac{5 + \sqrt{17}}{2}$.

We need to confirm that real numbers $x, y, z$ exist for this value of $P$. Let $A=1+x$, $B=1+y$, $C=1+z$. We have $x(1+y) = 1-P$, $y(1+z) = 2-P$, $z(1+x) = 4-P$. This can be written as $(A-1)B = 1-P$, $(B-1)C = 2-P$, $(C-1)A = 4-P$. For $P = \frac{5+\sqrt{17}}{2}$: $1-P = 1 - \frac{5+\sqrt{17}}{2} = \frac{2-5-\sqrt{17}}{2} = \frac{-3-\sqrt{17}}{2}$ $2-P = 2 - \frac{5+\sqrt{17}}{2} = \frac{4-5-\sqrt{17}}{2} = \frac{-1-\sqrt{17}}{2}$ $4-P = 4 - \frac{5+\sqrt{17}}{2} = \frac{8-5-\sqrt{17}}{2} = \frac{3-\sqrt{17}}{2}$

Notice that $1-P < 0$, $2-P < 0$, $4-P < 0$ (since $\sqrt{17} \approx 4.12$). So $x(1+y) < 0$, $y(1+z) < 0$, $z(1+x) < 0$. This implies $x$ and $1+y$ have opposite signs, $y$ and $1+z$ have opposite signs, $z$ and $1+x$ have opposite signs. For example, if $x>0$, then $1+y<0 \implies y<-1$. If $y<-1$, then $1+z>0 \implies z>-1$. If $z>-1$, then $1+x<0 \implies x<-1$. This contradicts $x>0$. So $x,y,z$ must all be negative. If $x<0$, then $1+y>0 \implies y>-1$. If $y>-1$, then $1+z<0 \implies z<-1$. If $z<-1$, then $1+x>0 \implies x>-1$. This contradicts $x<0$. This means that $x,y,z$ cannot be all positive or all negative. The variables $x,y,z$ are real numbers. Let's check the signs of $x,y,z$. $x(1+y) = \frac{-3-\sqrt{17}}{2} < 0$ $y(1+z) = \frac{-1-\sqrt{17}}{2} < 0$ $z(1+x) = \frac{3-\sqrt{17}}{2} < 0$

The product of these three is $xyz(1+x)(1+y)(1+z) = (1-P)(2-P)(4-P)$. $P \cdot (1+x)(1+y)(1+z) = \left(\frac{-3-\sqrt{17}}{2}\right) \left(\frac{-1-\sqrt{17}}{2}\right) \left(\frac{3-\sqrt{17}}{2}\right)$ $P \cdot (1+x)(1+y)(1+z) = \frac{1}{8} (-3-\sqrt{17})(-1-\sqrt{17})(3-\sqrt{17})$ $P \cdot (1+x)(1+y)(1+z) = \frac{1}{8} (\sqrt{17}+3)(\sqrt{17}+1)(\sqrt{17}-3)$ $P \cdot (1+x)(1+y)(1+z) = \frac{1}{8} ((\sqrt{17})^2 - 3^2)(\sqrt{17}+1)$ $P \cdot (1+x)(1+y)(1+z) = \frac{1}{8} (17-9)(\sqrt{17}+1)$ $P \cdot (1+x)(1+y)(1+z) = \frac{1}{8} (8)(\sqrt{17}+1)$ $P \cdot (1+x)(1+y)(1+z) = \sqrt{17}+1$

Since $P = \frac{5+\sqrt{17}}{2} > 0$, it implies that $(1+x)(1+y)(1+z)$ must be positive. $\frac{5+\sqrt{17}}{2} (1+x)(1+y)(1+z) = \sqrt{17}+1 > 0$. So $(1+x)(1+y)(1+z) = \frac{2(\sqrt{17}+1)}{5+\sqrt{17}} = \frac{2(\sqrt{17}+1)(5-\sqrt{17})}{(5+\sqrt{17})(5-\sqrt{17})} = \frac{2(5\sqrt{17}-17+5-\sqrt{17})}{25-17} = \frac{2(4\sqrt{17}-12)}{8} = \frac{4(\sqrt{17}-3)}{4} = \sqrt{17}-3 > 0$ (since $\sqrt{17} \approx 4.12$). So $(1+x)(1+y)(1+z)$ is positive.

The existence of real roots $x, y, z$ is guaranteed by the fact that the polynomial $Q(t) = t^3 - S_1 t^2 + S_2 t - P = 0$ has real roots. We have $P = \frac{5+\sqrt{17}}{2}$. $S_1+S_2 = 7-3P = 7 - 3\left(\frac{5+\sqrt{17}}{2}\right) = \frac{14-15-3\sqrt{17}}{2} = \frac{-1-3\sqrt{17}}{2}$. Also, $P(1+S_1+S_2+P) = (1-P)(2-P)(4-P)$ implies $P(1+(7-3P)+P) = (1-P)(2-P)(4-P)$ which simplified to $P^2-5P+2=0$. The values of $x, y, z$ are roots of a cubic equation. The problem statement implies that such real numbers $x,y,z$ exist.

The largest possible value of $xyz$ is $\frac{5 + \sqrt{17}}{2}$. This is in the form $\frac{a+b\sqrt{c}}{d}$. $a=5$, $b=1$, $c=17$, $d=2$. Check conditions: $a, b, c, d$ are integers: Yes ($5, 1, 17, 2$). $d$ is positive: Yes ($2>0$). $c$ is square-free: Yes ($17$ is a prime number, so it's square-free). $\text{gcd}(a, b, d) = \text{gcd}(5, 1, 2)$. $\text{gcd}(5, 1, 2) = 1$. Yes.

We need to find $1000a + 100b + 10c + d$. $1000(5) + 100(1) + 10(17) + 2$ $= 5000 + 100 + 170 + 2$ $= 5272$

The final answer is $\boxed{5272}$.

Explanation of the solution:

1. Rewrite the given equations $x+xy+xyz=1$, $y+yz+xyz=2$, $z+xz+xyz=4$ by letting $P=xyz$. This yields $x(1+y)=1-P$, $y(1+z)=2-P$, $z(1+x)=4-P$.
2. Multiply these three new equations: $xyz(1+x)(1+y)(1+z) = (1-P)(2-P)(4-P)$. Substitute $xyz=P$.
3. Expand $(1+x)(1+y)(1+z) = 1+(x+y+z)+(xy+yz+xz)+xyz$.
4. Sum the three new equations: $(x+xy)+(y+yz)+(z+xz) = (1-P)+(2-P)+(4-P)$, which simplifies to $(x+y+z)+(xy+yz+xz) = 7-3P$.
5. Substitute the sum from step 4 into the expanded product from step 3: $P(1+(7-3P)+P) = (1-P)(2-P)(4-P)$.
6. Simplify the equation: $P(8-2P) = (1-P)(2-P)(4-P)$, which becomes $2P(4-P) = (1-P)(2-P)(4-P)$.
7. Solve for $P$. One solution is $P=4$. If $P \ne 4$, divide by $(4-P)$ to get $2P = (1-P)(2-P)$, which simplifies to the quadratic equation $P^2-5P+2=0$.
8. Solve the quadratic equation using the quadratic formula to get $P = \frac{5 \pm \sqrt{17}}{2}$.
9. Compare the three possible values for $P$: $4$, $\frac{5+\sqrt{17}}{2}$, and $\frac{5-\sqrt{17}}{2}$. The largest value is $\frac{5+\sqrt{17}}{2}$.
10. Identify $a, b, c, d$ from the expression $\frac{5+\sqrt{17}}{2}$ as $a=5, b=1, c=17, d=2$. Verify that these values satisfy the given conditions.
11. Calculate the final expression $1000a + 100b + 10c + d = 1000(5) + 100(1) + 10(17) + 2 = 5000 + 100 + 170 + 2 = 5272$.