Question

Let $ABCD$ be a regular tetrahedron. Suppose point $X, Y$ and $Z$ lie on rays $AB, AC$ and $AD$ respectively such that $XY = YZ = 7$ and $XZ = 5$. The length of $AX, AY$ and $AZ$ are all distinct. If the volume of tetrahedron $AXYZ$ is $\sqrt{\lambda}$, then $\lambda$ equals

Answer 1

122

Answer 2

Let $A$ be the origin. Let $\vec{AB}, \vec{AC}, \vec{AD}$ be vectors representing the edges of the regular tetrahedron $ABCD$ originating from $A$. Let the side length of the tetrahedron be $a$. Then $|\vec{AB}| = |\vec{AC}| = |\vec{AD}| = a$. Since it is a regular tetrahedron, the angle between any two of these vectors is $60^\circ$. So, $\vec{AB} \cdot \vec{AC} = \vec{AC} \cdot \vec{AD} = \vec{AD} \cdot \vec{AB} = a^2 \cos(60^\circ) = a^2/2$.

Point $X$ lies on ray $AB$, so $\vec{AX} = \frac{AX}{a} \vec{AB}$. Let $x = AX$. Then $\vec{AX} = \frac{x}{a} \vec{AB}$. Point $Y$ lies on ray $AC$, so $\vec{AY} = \frac{AY}{a} \vec{AC}$. Let $y = AY$. Then $\vec{AY} = \frac{y}{a} \vec{AC}$. Point $Z$ lies on ray $AD$, so $\vec{AZ} = \frac{AZ}{a} \vec{AD}$. Let $z = AZ$. Then $\vec{AZ} = \frac{z}{a} \vec{AD}$. We are given that $x, y, z$ are distinct positive numbers.

We are given the lengths $XY=7$, $YZ=7$, $XZ=5$. $XY^2 = |\vec{XY}|^2 = |\vec{AY} - \vec{AX}|^2 = |\frac{y}{a}\vec{AC} - \frac{x}{a}\vec{AB}|^2 = \frac{y^2}{a^2}|\vec{AC}|^2 - \frac{2xy}{a^2}(\vec{AC} \cdot \vec{AB}) + \frac{x^2}{a^2}|\vec{AB}|^2$ $XY^2 = \frac{y^2}{a^2}a^2 - \frac{2xy}{a^2}\frac{a^2}{2} + \frac{x^2}{a^2}a^2 = y^2 - xy + x^2$. So, $x^2 - xy + y^2 = 7^2 = 49$. (1)

$YZ^2 = |\vec{YZ}|^2 = |\vec{AZ} - \vec{AY}|^2 = |\frac{z}{a}\vec{AD} - \frac{y}{a}\vec{AC}|^2 = \frac{z^2}{a^2}|\vec{AD}|^2 - \frac{2zy}{a^2}(\vec{AD} \cdot \vec{AC}) + \frac{y^2}{a^2}|\vec{AC}|^2$ $YZ^2 = \frac{z^2}{a^2}a^2 - \frac{2zy}{a^2}\frac{a^2}{2} + \frac{y^2}{a^2}a^2 = z^2 - zy + y^2$. So, $y^2 - yz + z^2 = 7^2 = 49$. (2)

$XZ^2 = |\vec{XZ}|^2 = |\vec{AZ} - \vec{AX}|^2 = |\frac{z}{a}\vec{AD} - \frac{x}{a}\vec{AB}|^2 = \frac{z^2}{a^2}|\vec{AD}|^2 - \frac{2zx}{a^2}(\vec{AD} \cdot \vec{AB}) + \frac{x^2}{a^2}|\vec{AB}|^2$ $XZ^2 = \frac{z^2}{a^2}a^2 - \frac{2zx}{a^2}\frac{a^2}{2} + \frac{x^2}{a^2}a^2 = z^2 - zx + x^2$. So, $x^2 - xz + z^2 = 5^2 = 25$. (3)

From (1) and (2): $x^2 - xy + y^2 = y^2 - yz + z^2 \implies x^2 - xy = z^2 - yz \implies x^2 - z^2 = xy - yz \implies (x-z)(x+z) = y(x-z)$. Since $x, y, z$ are distinct, $x \neq z$, so we can divide by $(x-z)$. $x+z = y$.

Substitute $y = x+z$ into (1) and (3). From (1): $x^2 - x(x+z) + (x+z)^2 = 49 \implies x^2 - x^2 - xz + x^2 + 2xz + z^2 = 49 \implies x^2 + xz + z^2 = 49$. (4) From (3): $x^2 - xz + z^2 = 25$. (5)

We have a system of equations for $x$ and $z$: (4) $x^2 + xz + z^2 = 49$ (5) $x^2 - xz + z^2 = 25$ Adding (4) and (5): $2x^2 + 2z^2 = 74 \implies x^2 + z^2 = 37$. (6) Subtracting (5) from (4): $2xz = 24 \implies xz = 12$. (7)

We can find $y^2 = (x+z)^2 = x^2 + 2xz + z^2 = (x^2+z^2) + 2xz = 37 + 2(12) = 37 + 24 = 61$. So $y = \sqrt{61}$ (since $y>0$). We can find $(x-z)^2 = x^2 - 2xz + z^2 = (x^2+z^2) - 2xz = 37 - 2(12) = 37 - 24 = 13$. So $x-z = \pm \sqrt{13}$. We have $x+z = \sqrt{61}$ and $x-z = \pm \sqrt{13}$. If $x-z = \sqrt{13}$, then $2x = \sqrt{61}+\sqrt{13}$ and $2z = \sqrt{61}-\sqrt{13}$. $x = \frac{\sqrt{61}+\sqrt{13}}{2}, z = \frac{\sqrt{61}-\sqrt{13}}{2}$. If $x-z = -\sqrt{13}$, then $2x = \sqrt{61}-\sqrt{13}$ and $2z = \sqrt{61}+\sqrt{13}$. $x = \frac{\sqrt{61}-\sqrt{13}}{2}, z = \frac{\sqrt{61}+\sqrt{13}}{2}$. In either case, the set of lengths $\{x, z\}$ is $\{\frac{\sqrt{61}-\sqrt{13}}{2}, \frac{\sqrt{61}+\sqrt{13}}{2}\}$. The three lengths are $x, y, z$. The set of lengths is $\{\frac{\sqrt{61}-\sqrt{13}}{2}, \sqrt{61}, \frac{\sqrt{61}+\sqrt{13}}{2}\}$. These three values are distinct since $\sqrt{13} \neq 0$.

The volume of the tetrahedron $AXYZ$ is given by $V = \frac{1}{6} |\vec{AX} \cdot (\vec{AY} \times \vec{AZ})|$. $V = \frac{1}{6} |(\frac{x}{a}\vec{AB}) \cdot ((\frac{y}{a}\vec{AC}) \times (\frac{z}{a}\vec{AD}))| = \frac{xyz}{6a^3} |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|$. The scalar triple product $\vec{AB} \cdot (\vec{AC} \times \vec{AD})$ is related to the volume of the tetrahedron $ABCD$. The volume of tetrahedron $ABCD$ is $V_{ABCD} = \frac{1}{6} |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|$. For a regular tetrahedron with side length $a$, the volume is $V_{ABCD} = \frac{a^3}{6\sqrt{2}}$. So, $|\vec{AB} \cdot (\vec{AC} \times \vec{AD})| = 6 \cdot \frac{a^3}{6\sqrt{2}} = \frac{a^3}{\sqrt{2}}$. $V_{AXYZ} = \frac{xyz}{6a^3} \frac{a^3}{\sqrt{2}} = \frac{xyz}{6\sqrt{2}}$.

We need the product $xyz$. $y = \sqrt{61}$. $xz = 12$. $xyz = \sqrt{61} \cdot 12 = 12\sqrt{61}$.

$V_{AXYZ} = \frac{12\sqrt{61}}{6\sqrt{2}} = \frac{2\sqrt{61}}{\sqrt{2}} = \frac{2\sqrt{61}\sqrt{2}}{2} = \sqrt{122}$. The volume is given as $\sqrt{\lambda}$. So, $\sqrt{\lambda} = \sqrt{122}$. $\lambda = 122$.

The final answer is $\boxed{122}$.