Triangle and Parallelogram Questions in English

(C) Let the vertices be $A(1,2,3), B(3,-1,5),$ and $C(4,0,-3)$.
First,we calculate the direction ratios of the sides:
$\overrightarrow{AB} = (3-1, -1-2, 5-3) = (2, -3, 2)$
$\overrightarrow{BC} = (4-3, 0+1, -3-5) = (1, 1, -8)$
$\overrightarrow{AC} = (4-1, 0-2, -3-3) = (3, -2, -6)$
Now,check for orthogonality:
$\overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(3) + (-3)(-2) + (2)(-6) = 6 + 6 - 12 = 0$.
Since the dot product is $0$,$\overrightarrow{AB} \perp \overrightarrow{AC}$,which means $\angle A = 90^{\circ}$.
In a right-angled triangle,the circumcenter is the midpoint of the hypotenuse $BC$.
Circumcenter $(\alpha, \beta, \gamma) = \left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2}\right) = \left(\frac{7}{2}, -\frac{1}{2}, 1\right)$.
Thus,$\alpha = \frac{7}{2}, \beta = -\frac{1}{2}, \gamma = 1$.
Then,$|\alpha| + |\beta| = |\frac{7}{2}| + |-\frac{1}{2}| = \frac{7}{2} + \frac{1}{2} = 4$.
Since $|\gamma| = |1| = 1$,we have $4 = 4|\gamma|$.

(A) Given points are $A(1, 3, 5)$,$B(2, 4, 6)$,and $C(4, 5, k)$.
First,calculate the squared distances between the points:
$AB^2 = (2-1)^2 + (4-3)^2 + (6-5)^2 = 1^2 + 1^2 + 1^2 = 3$
$BC^2 = (4-2)^2 + (5-4)^2 + (k-6)^2 = 4 + 1 + (k-6)^2 = k^2 - 12k + 41$
$AC^2 = (4-1)^2 + (5-3)^2 + (k-5)^2 = 9 + 4 + (k-5)^2 = k^2 - 10k + 38$
Case $1$: Right-angled at $A$ $(AB^2 + AC^2 = BC^2)$
$3 + k^2 - 10k + 38 = k^2 - 12k + 41$
$41 - 10k = 41 - 12k$
$2k = 0 \Rightarrow k = 0$
Case $2$: Right-angled at $B$ $(AB^2 + BC^2 = AC^2)$
$3 + k^2 - 12k + 41 = k^2 - 10k + 38$
$44 - 12k = 38 - 10k$
$6 = 2k \Rightarrow k = 3$
Case $3$: Right-angled at $C$ $(AC^2 + BC^2 = AB^2)$
$k^2 - 10k + 38 + k^2 - 12k + 41 = 3$
$2k^2 - 22k + 76 = 0$
$k^2 - 11k + 38 = 0$
The discriminant $D = (-11)^2 - 4(1)(38) = 121 - 152 = -31 < 0$. No real values for $k$.
Thus,the possible values for $k$ are $0$ and $3$. The number of possible values is $2$.

(A) Let the vertices of $\triangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given that $D, E, F$ are midpoints of $AB, BC, CA$ respectively:
$D = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right) = (1, 2, -3) \Rightarrow x_1+x_2=2, y_1+y_2=4, z_1+z_2=-6$
$E = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}\right) = (3, 0, 1) \Rightarrow x_2+x_3=6, y_2+y_3=0, z_2+z_3=2$
$F = \left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}, \frac{z_1+z_3}{2}\right) = (-1, 1, -4) \Rightarrow x_1+x_3=-2, y_1+y_3=2, z_1+z_3=-8$
Solving for $x_1, x_2, x_3$: $(x_1+x_2)+(x_2+x_3)+(x_1+x_3) = 2+6-2 = 6 \Rightarrow 2(x_1+x_2+x_3)=6 \Rightarrow x_1+x_2+x_3=3$. Thus $x_3=3-2=1, x_1=3-6=-3, x_2=3-(-2)=5$.
Similarly for $y$: $y_1+y_2+y_3 = \frac{4+0+2}{2} = 3$. Thus $y_3=3-4=-1, y_1=3-0=3, y_2=3-2=1$.
Similarly for $z$: $z_1+z_2+z_3 = \frac{-6+2-8}{2} = -6$. Thus $z_3=-6-(-6)=0, z_1=-6-2=-8, z_2=-6-(-8)=2$.
So,$A(-3, 3, -8)$,$D(1, 2, -3)$,and $F(-1, 1, -4)$.
The centroid of $\triangle ADF$ is $\left(\frac{-3+1-1}{3}, \frac{3+2+1}{3}, \frac{-8-3-4}{3}\right) = \left(\frac{-3}{3}, \frac{6}{3}, \frac{-15}{3}\right) = (-1, 2, -5)$.

(A) The centroid $(G)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the formula $G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$.
Given vertices are $A(3,4,5)$,$B(6,7,2)$,and $C(x, y, z)$,and the centroid is $(3,2,3)$.
Equating the coordinates:
$3 = \frac{3+6+x}{3} \implies 9 = 9+x \implies x = 0$.
$2 = \frac{4+7+y}{3} \implies 6 = 11+y \implies y = -5$.
$3 = \frac{5+2+z}{3} \implies 9 = 7+z \implies z = 2$.
Therefore,$x+y+z = 0 + (-5) + 2 = -3$.
Thus,the correct option is $A$.

(C) Let the vertices be $A(1,2,5), B(-1,6,1), C(3,4,-3)$ and $D(5,0,1)$.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(-1-1)^2 + (6-2)^2 + (1-5)^2} = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$BC = \sqrt{(3-(-1))^2 + (4-6)^2 + (-3-1)^2} = \sqrt{4^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
$CD = \sqrt{(5-3)^2 + (0-4)^2 + (1-(-3))^2} = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$DA = \sqrt{(1-5)^2 + (2-0)^2 + (5-1)^2} = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Since all sides are equal $(AB = BC = CD = DA = 6)$,the quadrilateral is a rhombus.
Now,we check the diagonals:
$AC = \sqrt{(3-1)^2 + (4-2)^2 + (-3-5)^2} = \sqrt{2^2 + 2^2 + (-8)^2} = \sqrt{4 + 4 + 64} = \sqrt{72} = 6\sqrt{2}$
$BD = \sqrt{(5-(-1))^2 + (0-6)^2 + (1-1)^2} = \sqrt{6^2 + (-6)^2 + 0^2} = \sqrt{36 + 36 + 0} = \sqrt{72} = 6\sqrt{2}$
Since the diagonals are also equal $(AC = BD = 6\sqrt{2})$,the quadrilateral is a square.

(B) Let the mid-points of the sides of $\triangle ABC$ be $D(1,0,3)$,$E(2,1,5)$,and $F(-2,3,6)$.
The centroid of the triangle formed by the mid-points of the sides of a triangle is the same as the centroid of the original triangle.
Therefore,the centroid of $\triangle ABC$ is the same as the centroid of $\triangle DEF$.
The formula for the centroid of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the given mid-points:
Centroid $= \left(\frac{1+2-2}{3}, \frac{0+1+3}{3}, \frac{3+5+6}{3}\right)$
$= \left(\frac{1}{3}, \frac{4}{3}, \frac{14}{3}\right)$.

(C) Let the coordinates of the third vertex $C$ be $(x, y, z)$.
Given vertices are $A(5, 4, 6)$ and $B(1, -1, 3)$.
The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given centroid $G = \left(\frac{10}{3}, 2, \frac{11}{3}\right)$.
Equating the coordinates:
$\frac{5+1+x}{3} = \frac{10}{3} \Rightarrow 6+x = 10 \Rightarrow x = 4$.
$\frac{4-1+y}{3} = 2 \Rightarrow 3+y = 6 \Rightarrow y = 3$.
$\frac{6+3+z}{3} = \frac{11}{3} \Rightarrow 9+z = 11 \Rightarrow z = 2$.
Therefore,the third vertex $C$ is $(4, 3, 2)$.

(B) Let the vertices of $\triangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Since $D(2, 1, 0)$ is the mid-point of $BC$,we have:
$\frac{x_2+x_3}{2} = 2 \implies x_2+x_3 = 4$
$\frac{y_2+y_3}{2} = 1 \implies y_2+y_3 = 2$
$\frac{z_2+z_3}{2} = 0 \implies z_2+z_3 = 0$
Since $E(2, 0, 0)$ is the mid-point of $AC$,we have:
$\frac{x_1+x_3}{2} = 2 \implies x_1+x_3 = 4$
$\frac{y_1+y_3}{2} = 0 \implies y_1+y_3 = 0$
$\frac{z_1+z_3}{2} = 0 \implies z_1+z_3 = 0$
Since $F(0, 1, 0)$ is the mid-point of $AB$,we have:
$\frac{x_1+x_2}{2} = 0 \implies x_1+x_2 = 0$
$\frac{y_1+y_2}{2} = 1 \implies y_1+y_2 = 2$
$\frac{z_1+z_2}{2} = 0 \implies z_1+z_2 = 0$
Adding these equations:
$2(x_1+x_2+x_3) = 4+4+0 = 8 \implies x_1+x_2+x_3 = 4$
$2(y_1+y_2+y_3) = 2+0+2 = 4 \implies y_1+y_2+y_3 = 2$
$2(z_1+z_2+z_3) = 0+0+0 = 0 \implies z_1+z_2+z_3 = 0$
The centroid of $\triangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the sums,the centroid is $\left(\frac{4}{3}, \frac{2}{3}, 0\right)$.

(D) Let the vertices of the triangle be $A = (1, 0, 0)$,$B = (0, 1, 0)$,and $C = (0, 0, 1)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$,we calculate the lengths of the sides:
$AB = \sqrt{(0-1)^2 + (1-0)^2 + (0-0)^2} = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$.
$BC = \sqrt{(0-0)^2 + (0-1)^2 + (1-0)^2} = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.
$CA = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
The perimeter of the triangle is the sum of the lengths of its sides:
$\text{Perimeter} = AB + BC + CA = \sqrt{2} + \sqrt{2} + \sqrt{2} = 3\sqrt{2}$.

(A) Let the vertices of the parallelogram be $O(0,0)$,$A(a \cos \theta, b \sin \theta)$,$B(a \cos \theta, -b \sin \theta)$,and $C(x, y)$.
Since $OABC$ is a parallelogram,the diagonals $OB$ and $AC$ bisect each other at the same midpoint.
Alternatively,the area of a parallelogram with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ can be calculated using the cross product of vectors.
The area of $\triangle OAB = \frac{1}{2} |x_A y_B - x_B y_A| = \frac{1}{2} |(a \cos \theta)(-b \sin \theta) - (a \cos \theta)(b \sin \theta)| = \frac{1}{2} |-2ab \sin \theta \cos \theta| = ab |\sin \theta \cos \theta| = \frac{ab}{2} |\sin 2\theta|$.
The area of the parallelogram $OABC = 2 \times \text{Area}(\triangle OAB) = 2 \times \frac{ab}{2} |\sin 2\theta| = ab |\sin 2\theta|$.
The maximum value of $|\sin 2\theta|$ is $1$,which occurs when $2\theta = \pi / 2$,i.e.,$\theta = \pi / 4$.
Thus,the maximum area is $ab$ when $\theta = \pi / 4$.