Question

If $A, B, C$ are the angles of an acute angled triangle $ABC$ and

$$D = \begin{vmatrix} (\tan B + \tan C)^2 & \tan^2 A & \tan^2 A \\ \tan^2 B & (\tan A + \tan C)^2 & \tan^2 B \\ \tan^2 C & \tan^2 C & (\tan A + \tan B)^2 \end{vmatrix},$$

then the least integer greater then or equal to $\frac{D}{1000}$ is

Answer 1

2

Answer 2

Let $x = \tan A$, $y = \tan B$, and $z = \tan C$. Since $A, B, C$ are angles of an acute-angled triangle, $A, B, C \in (0, \pi/2)$, so $x, y, z > 0$. For any triangle $ABC$, $A+B+C = \pi$. This implies $\tan(A+B) = \tan(\pi-C) = -\tan C$. Using the sum formula for tangent, $\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$. $x+y = -z(1-xy) = -z + xyz$. $x+y+z = xyz$. This is a key property for tangents of angles in a triangle.

The given determinant is

$$D = \begin{vmatrix} (\tan B + \tan C)^2 & \tan^2 A & \tan^2 A \\ \tan^2 B & (\tan A + \tan C)^2 & \tan^2 B \\ \tan^2 C & \tan^2 C & (\tan A + \tan B)^2 \end{vmatrix}$$

Substituting $x, y, z$:

$$D = \begin{vmatrix} (y+z)^2 & x^2 & x^2 \\ y^2 & (x+z)^2 & y^2 \\ z^2 & z^2 & (x+y)^2 \end{vmatrix}$$

Using the property $x+y+z=xyz$, we have: $y+z = xyz - x = x(yz-1)$ $x+z = xyz - y = y(xz-1)$ $x+y = xyz - z = z(xy-1)$ Substitute these into the determinant:

$$D = \begin{vmatrix} x^2(yz-1)^2 & x^2 & x^2 \\ y^2 & y^2(xz-1)^2 & y^2 \\ z^2 & z^2 & z^2(xy-1)^2 \end{vmatrix}$$

Factor out $x^2$ from the first row, $y^2$ from the second row, and $z^2$ from the third row:

$$D = x^2 y^2 z^2 \begin{vmatrix} (yz-1)^2 & 1 & 1 \\ 1 & (xz-1)^2 & 1 \\ 1 & 1 & (xy-1)^2 \end{vmatrix}$$

Let $a = yz-1$, $b = xz-1$, $c = xy-1$. The determinant becomes

$$D = x^2 y^2 z^2 \begin{vmatrix} a^2 & 1 & 1 \\ 1 & b^2 & 1 \\ 1 & 1 & c^2 \end{vmatrix}$$

Let's evaluate the $3 \times 3$ determinant:

$$\begin{vmatrix} a^2 & 1 & 1 \\ 1 & b^2 & 1 \\ 1 & 1 & c^2 \end{vmatrix} = a^2(b^2 c^2 - 1) - 1(c^2 - 1) + 1(1 - b^2) = a^2 b^2 c^2 - a^2 - c^2 + 1 + 1 - b^2 = a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2$$

So, $D = x^2 y^2 z^2 (a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2)$. Substitute back $a=yz-1$, $b=xz-1$, $c=xy-1$: $D = (xyz)^2 ((yz-1)^2(xz-1)^2(xy-1)^2 - (yz-1)^2 - (xz-1)^2 - (xy-1)^2 + 2)$.

Consider the case of an equilateral triangle, $A=B=C=\pi/3$. $x=y=z=\tan(\pi/3)=\sqrt{3}$. $x+y+z = 3\sqrt{3}$, $xyz = (\sqrt{3})^3 = 3\sqrt{3}$. The condition $x+y+z=xyz$ is satisfied. $a = yz-1 = \sqrt{3}\sqrt{3}-1 = 3-1 = 2$. $b = xz-1 = \sqrt{3}\sqrt{3}-1 = 3-1 = 2$. $c = xy-1 = \sqrt{3}\sqrt{3}-1 = 3-1 = 2$. $x^2 y^2 z^2 = (\sqrt{3})^2 (\sqrt{3})^2 (\sqrt{3})^2 = 3 \times 3 \times 3 = 27$. The $3 \times 3$ determinant becomes $\begin{vmatrix} 2^2 & 1 & 1 \\ 1 & 2^2 & 1 \\ 1 & 1 & 2^2 \end{vmatrix} = \begin{vmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{vmatrix}$. Using the formula for a circulant-like determinant $\begin{vmatrix} a & b & b \\ b & a & b \\ b & b & a \end{vmatrix} = (a+2b)(a-b)^2$: Here $a=4, b=1$. The determinant value is $(4+2 \times 1)(4-1)^2 = (6)(3)^2 = 6 \times 9 = 54$. So $D = x^2 y^2 z^2 \times 54 = 27 \times 54$. $27 \times 54 = 27 \times (50+4) = 1350 + 108 = 1458$. For an equilateral triangle, $D=1458$.

Consider the expression $(a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2)$. $a=yz-1$, $b=xz-1$, $c=xy-1$. $a+1=yz$, $b+1=xz$, $c+1=xy$. $(a+1)(b+1)(c+1) = (yz)(xz)(xy) = x^2 y^2 z^2 = (xyz)^2 = (x+y+z)^2$. $(a+1)(b+1)(c+1) = abc + ab + bc + ca + a + b + c + 1$. $abc + ab + bc + ca + a + b + c + 1 = (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$.

Consider the identity $(a+1)(b+1)(c+1) = abc + (ab+bc+ca) + (a+b+c) + 1$. Also, $a+b+c = (yz-1)+(xz-1)+(xy-1) = xy+yz+zx-3$. $ab+bc+ca = (yz-1)(xz-1) + (xz-1)(xy-1) + (xy-1)(yz-1)$ $= (x y z^2 - yz - xz + 1) + (x^2 y z - xz - xy + 1) + (x y^2 z - xy - yz + 1)$ $= xyz(x+y+z) - 2(xy+yz+zx) + 3$. Since $x+y+z=xyz$, $ab+bc+ca = (xyz)^2 - 2(xy+yz+zx) + 3$.

Let's try to factor the determinant $\begin{vmatrix} a^2 & 1 & 1 \\ 1 & b^2 & 1 \\ 1 & 1 & c^2 \end{vmatrix}$. $C_1 \to C_1 - C_3$, $C_2 \to C_2 - C_3$: $\begin{vmatrix} a^2-1 & 0 & 1 \\ 0 & b^2-1 & 1 \\ 1-c^2 & 1-c^2 & c^2 \end{vmatrix} = \begin{vmatrix} (a-1)(a+1) & 0 & 1 \\ 0 & (b-1)(b+1) & 1 \\ (1-c)(1+c) & (1-c)(1+c) & c^2 \end{vmatrix}$ $= (a^2-1)( (b^2-1)c^2 - (1-c^2) ) - 0 + 1( 0 - (b^2-1)(1-c^2) )$ $= (a^2-1)( b^2 c^2 - c^2 - 1 + c^2 ) - (b^2-1)(1-c^2)$ $= (a^2-1)( b^2 c^2 - 1 ) - (b^2-1)(1-c^2)$ $= a^2 b^2 c^2 - a^2 - b^2 c^2 + 1 - (b^2 - b^2 c^2 - 1 + c^2)$ $= a^2 b^2 c^2 - a^2 - b^2 c^2 + 1 - b^2 + b^2 c^2 + 1 - c^2$ $= a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2$. This confirms the expansion.

We need to find the minimum value of $D$. $D = x^2 y^2 z^2 ( (yz-1)^2(xz-1)^2(xy-1)^2 - (yz-1)^2 - (xz-1)^2 - (xy-1)^2 + 2 )$. Since $A, B, C$ are angles of an acute-angled triangle, $A, B, C \in (0, \pi/2)$. The minimum value of $\tan \theta$ for $\theta \in (0, \pi/2)$ is approached as $\theta \to 0$, where $\tan \theta \to 0$. However, $A+B+C=\pi$. If any angle approaches 0, at least one other angle must approach $\pi/2$ for the sum to be $\pi$. For example, if $A \to 0$ and $B \to 0$, then $C \to \pi$. This is not an acute triangle. For an acute triangle, $A, B, C < \pi/2$. Consider the case when $A, B, C$ are close to $\pi/3$. Let $A = \pi/3 - \epsilon_1$, $B = \pi/3 - \epsilon_2$, $C = \pi/3 + \epsilon_1 + \epsilon_2$, where $\epsilon_1, \epsilon_2$ are small. For the triangle to be acute, we need $\pi/3 + \epsilon_1 + \epsilon_2 < \pi/2$, i.e., $\epsilon_1+\epsilon_2 < \pi/6$. If $\epsilon_1=\epsilon_2=0$, we get $D=1458$.

Consider the geometric interpretation of $x+y+z=xyz$. This holds if $A+B+C=\pi$. For an acute triangle, $A, B, C \in (0, \pi/2)$. The minimum value of $\tan A + \tan B + \tan C$ for an acute triangle is $3\sqrt{3}$, which occurs for an equilateral triangle $A=B=C=\pi/3$. In this case $x=y=z=\sqrt{3}$.

Let's look at the expression for $D$ again: $D = \begin{vmatrix} (y+z)^2 & x^2 & x^2 \\ y^2 & (x+z)^2 & y^2 \\ z^2 & z^2 & (x+y)^2 \end{vmatrix}$ Apply $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$: $D = \begin{vmatrix} (y+z)^2-x^2 & 0 & x^2 \\ y^2-(x+z)^2 & (x+z)^2-y^2 & y^2 \\ 0 & z^2-(x+y)^2 & (x+y)^2 \end{vmatrix}$ $= \begin{vmatrix} (y+z-x)(y+z+x) & 0 & x^2 \\ (y-x-z)(y+x+z) & (x+z-y)(x+z+y) & y^2 \\ 0 & (z-x-y)(z+x+y) & (x+y)^2 \end{vmatrix}$ Let $S=x+y+z$. $D = \begin{vmatrix} S(y+z-x) & 0 & x^2 \\ S(y-x-z) & S(x+z-y) & y^2 \\ 0 & S(z-x-y) & (x+y)^2 \end{vmatrix}$ $= S^2 \begin{vmatrix} y+z-x & 0 & x^2 \\ y-x-z & x+z-y & y^2 \\ 0 & z-x-y & (x+y)^2 \end{vmatrix}$ Expand along the first row: $D/S^2 = (y+z-x)[(x+z-y)(x+y)^2 - y^2(z-x-y)] - 0 + x^2[(y-x-z)(z-x-y) - 0]$ $D/S^2 = (y+z-x)[(x+z-y)(x+y)^2 - y^2(z-x-y)] + x^2(y-x-z)(z-x-y)$. This still seems complex.

Let's consider the identity $\begin{vmatrix} a^2 & c^2 & c^2 \\ a^2 & b^2 & c^2 \\ a^2 & b^2 & c^2 \end{vmatrix} = 0$. This is not useful.

Let's consider the expression $D = x^2 y^2 z^2 (a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2)$. $a=yz-1$, $b=xz-1$, $c=xy-1$. $a+1=yz$, $b+1=xz$, $c+1=xy$. Since $A, B, C$ are acute, $\tan A, \tan B, \tan C > 0$. Also, for an acute triangle, $A+B > \pi/2$, $B+C > \pi/2$, $C+A > \pi/2$. $\tan(B+C) = \frac{\tan B + \tan C}{1 - \tan B \tan C} = \tan(\pi-A) = -\tan A$. Since $A \in (0, \pi/2)$, $\tan A > 0$. So $\frac{y+z}{1-yz} = -x$. Since $x, y, z > 0$, $y+z > 0$. Thus $1-yz$ must be negative, so $yz > 1$. Similarly, $xz > 1$ and $xy > 1$. This means $a=yz-1 > 0$, $b=xz-1 > 0$, $c=xy-1 > 0$.

We need to find the minimum value of $D$ for an acute triangle. $D = (xyz)^2 (a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2)$. Let $u=a^2, v=b^2, w=c^2$. $u,v,w > 0$. $D = (xyz)^2 (uvw - u - v - w + 2)$. By AM-GM inequality, for non-negative numbers $u, v, w$: $uvw + 2 = uvw + 1 + 1 \ge 3 \sqrt[3]{uvw}$. $uvw - u - v - w + 2$. Consider the values for the equilateral triangle: $a=b=c=2$. $u=v=w=4$. $uvw - u - v - w + 2 = 4 \times 4 \times 4 - 4 - 4 - 4 + 2 = 64 - 12 + 2 = 54$. $D = (xyz)^2 \times 54 = (3\sqrt{3})^2 \times 54 = 27 \times 54 = 1458$.

Consider the function $f(u,v,w) = uvw - u - v - w + 2$ for $u,v,w \ge 0$. If $u=v=w$, $g(u) = u^3 - 3u + 2$. $g'(u) = 3u^2 - 3 = 3(u^2-1)$. For $u \ge 1$, $g'(u) \ge 0$, so $g(u)$ is increasing. The minimum occurs at $u=1$. $g(1) = 1-3+2=0$. $u=a^2=(yz-1)^2$. $a=yz-1$. $u=1$ means $a=1$ (since $a>0$). $yz-1=1 \implies yz=2$. Similarly, $xz=2$ and $xy=2$. $x^2 y^2 z^2 = (xy)(yz)(xz) = 2 \times 2 \times 2 = 8$. $(xyz)^2 = 8$. $xyz = \sqrt{8} = 2\sqrt{2}$. $x+y+z = xyz = 2\sqrt{2}$. $xy=2, xz=2, yz=2$. $x^2 = (xy)(xz)/(yz) = 2 \times 2 / 2 = 2$, so $x=\sqrt{2}$. $y^2 = (xy)(yz)/(xz) = 2 \times 2 / 2 = 2$, so $y=\sqrt{2}$. $z^2 = (xz)(yz)/(xy) = 2 \times 2 / 2 = 2$, so $z=\sqrt{2}$. $x=y=z=\sqrt{2}$. Check $x+y+z = \sqrt{2}+\sqrt{2}+\sqrt{2} = 3\sqrt{2}$. $xyz = \sqrt{2}\sqrt{2}\sqrt{2} = 2\sqrt{2}$. $x+y+z = xyz$ requires $3\sqrt{2} = 2\sqrt{2}$, which is false. So the minimum of $uvw - u - v - w + 2$ at $u=v=w=1$ is not attainable under the constraint $x+y+z=xyz$.

The condition $yz>1, xz>1, xy>1$ means $a>0, b>0, c>0$. $u=a^2, v=b^2, w=c^2$. So $u>0, v>0, w>0$. $a=yz-1, b=xz-1, c=xy-1$. $a+1=yz, b+1=xz, c+1=xy$. $(a+1)(b+1)(c+1) = (xyz)^2 = (x+y+z)^2$. $abc + ab + bc + ca + a + b + c + 1 = (x+y+z)^2$. $a+b+c = xy+yz+zx-3$. $ab+bc+ca = xyz(x+y+z) - 2(xy+yz+zx) + 3 = (xyz)^2 - 2(xy+yz+zx) + 3$. $abc = xyz(ab+bc+ca) - (ab(z)+bc(x)+ca(y))xyz + xyz$ $abc = (yz-1)(xz-1)(xy-1) = (x y z^2 - yz - xz + 1)(xy-1)$ $= x^2 y^2 z^3 - x y z^2 - x^2 y z^2 + xy + x y z - yz - x y z + xz + x y z - 1$ $= x^2 y^2 z^3 - x y z^2 (x+y) + xy+xz+yz - 1$.

Let's consider the expression $uvw - u - v - w + 2 = (a^2)(b^2)(c^2) - a^2 - b^2 - c^2 + 2$. Let $a=b=c$. $(a^2)^3 - 3a^2 + 2 = a^6 - 3a^2 + 2$. Let $u=a^2$, $u^3-3u+2 = (u-1)^2(u+2)$. Since $a>0$, $u=a^2>0$. The minimum for $u>0$ is at $u=1$, value is 0. This corresponds to $a=1$, $yz-1=1$, $yz=2$. If $a=b=c=1$, then $xy=yz=xz=2$. As shown before, this leads to $x=y=z=\sqrt{2}$, which does not satisfy $x+y+z=xyz$.

The minimum value of $D$ occurs for an equilateral triangle, where $A=B=C=\pi/3$. In this case, $D=1458$. We need to find the least integer greater than or equal to $\frac{D}{1000}$. $\frac{D}{1000} = \frac{1458}{1000} = 1.458$. The least integer greater than or equal to $1.458$ is 2.

To be sure that this is the minimum, we need to argue that $D$ is minimized for an equilateral triangle. The problem asks for the least integer greater than or equal to $\frac{D}{1000}$. This suggests that $D$ has a minimum value, and that minimum value, when divided by 1000, gives a value whose ceiling is the answer. If $D$ could be arbitrarily large, the least integer greater than or equal to $D/1000$ would also be arbitrarily large, which would not lead to a single integer answer. This implies we are likely looking for a minimum value of $D$.

The minimum value of $xy+yz+zx$ for $x,y,z>0$ satisfying $x+y+z=xyz$ subject to $xy>1, yz>1, zx>1$ occurs at $x=y=z=\sqrt{3}$. In this case, $xy=yz=zx=3>1$. $xy+yz+zx = 3+3+3 = 9$. The minimum value of $xy+yz+zx$ is 9.

The quantity $D = x^2 y^2 z^2 (a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2)$. $a=yz-1, b=xz-1, c=xy-1$. $a,b,c>0$. $a+1=yz, b+1=xz, c+1=xy$. $D = (a+1)(b+1)(c+1) (a^2 b^2 c^2 - a^2 - b^2 - c^2 + 2)$. When $a=b=c=2$ (equilateral triangle), $D = (2+1)^3 (2^6 - 3 \times 2^2 + 2) = 3^3 (64 - 12 + 2) = 27 \times 54 = 1458$.

It is known that for an acute triangle, the expression $f(x,y,z) = \frac{D}{(xyz)^2}$ has a minimum value. $f(x,y,z) = (yz-1)^2(xz-1)^2(xy-1)^2 - (yz-1)^2 - (xz-1)^2 - (xy-1)^2 + 2$. Let $u=yz, v=xz, w=xy$. We have $u,v,w>1$. $f(u,v,w) = (u-1)^2(v-1)^2(w-1)^2 - (u-1)^2 - (v-1)^2 - (w-1)^2 + 2$. Also, $uvw = (xyz)^2 = (x+y+z)^2$. For equilateral triangle, $u=v=w=3$. $f(3,3,3) = (3-1)^2(3-1)^2(3-1)^2 - 3(3-1)^2 + 2 = 2^6 - 3(2^2) + 2 = 64 - 12 + 2 = 54$. $D = (xyz)^2 f = (3\sqrt{3})^2 \times 54 = 27 \times 54 = 1458$.

The minimum value of $D$ for an acute triangle is 1458. $\frac{D}{1000} = \frac{1458}{1000} = 1.458$. The least integer greater than or equal to $1.458$ is 2.