Points related to triangle Questions in English

(D) The vertices of the triangle are $A(0, 0)$,$B(3, 0)$,and $C(0, 4)$.
Let the side lengths be $a$,$b$,and $c$ opposite to vertices $A$,$B$,and $C$ respectively.
$a = BC = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$b = AC = \sqrt{(0-0)^2 + (4-0)^2} = 4$.
$c = AB = \sqrt{(3-0)^2 + (0-0)^2} = 3$.
The coordinates of the incentre $I$ are given by $\left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right)$.
Here,$(x_1, y_1) = (0, 0)$,$(x_2, y_2) = (3, 0)$,and $(x_3, y_3) = (0, 4)$.
$I = \left(\frac{5(0) + 4(3) + 3(0)}{5+4+3}, \frac{5(0) + 4(0) + 3(4)}{5+4+3}\right)$.
$I = \left(\frac{12}{12}, \frac{12}{12}\right) = (1, 1)$.

(D) Let the vertices of the triangle be $A(1,3), B(-3,5)$,and $C(5,-1)$.
Slope of $AC = \frac{-1-3}{5-1} = \frac{-4}{4} = -1$.
The altitude from $B$ to $AC$ is perpendicular to $AC$. Let its slope be $m_1$. Since $m_1 \times (-1) = -1$,$m_1 = 1$.
The equation of the altitude from $B(-3,5)$ is $(y-5) = 1(x+3) \Rightarrow x-y = -8$ (Equation $i$).
Slope of $BC = \frac{-1-5}{5-(-3)} = \frac{-6}{8} = -\frac{3}{4}$.
The altitude from $A$ to $BC$ is perpendicular to $BC$. Let its slope be $m_2$. Since $m_2 \times (-\frac{3}{4}) = -1$,$m_2 = \frac{4}{3}$.
The equation of the altitude from $A(1,3)$ is $(y-3) = \frac{4}{3}(x-1)$ $\Rightarrow 3y-9 = 4x-4$ $\Rightarrow 4x-3y = -5$ (Equation $ii$).
To find the orthocentre,solve the system of equations:
$x-y = -8 \Rightarrow 3x-3y = -24$ (Equation $iii$).
Subtracting (ii) from (iii): $(3x-3y) - (4x-3y) = -24 - (-5)$ $\Rightarrow -x = -19$ $\Rightarrow x = 19$.
Substituting $x=19$ into $(i)$: $19-y = -8 \Rightarrow y = 27$.
Thus,the orthocentre is $(19,27)$.

(D) Let $G$ be the centroid of $\triangle ABC$. The medians $AD$ and $BE$ intersect at $G$.
Given $AD=4$,$\angle GAB = \frac{\pi}{6}$,and $\angle GBA = \frac{\pi}{3}$.
In $\triangle AGB$,the sum of angles is $\pi$,so $\angle AGB = \pi - (\frac{\pi}{6} + \frac{\pi}{3}) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.
Since $G$ is the centroid,$AG = \frac{2}{3} AD = \frac{2}{3} \times 4 = \frac{8}{3}$.
In right-angled $\triangle AGB$,$\sin(\angle GBA) = \frac{AG}{AB} \implies \sin(\frac{\pi}{3}) = \frac{8/3}{AB} \implies \frac{\sqrt{3}}{2} = \frac{8}{3AB} \implies AB = \frac{16}{3\sqrt{3}}$.
Also,$\tan(\angle GBA) = \frac{AG}{BG} \implies \tan(\frac{\pi}{3}) = \frac{8/3}{BG} \implies \sqrt{3} = \frac{8}{3BG} \implies BG = \frac{8}{3\sqrt{3}}$.
Area of $\triangle AGB = \frac{1}{2} \times AG \times BG = \frac{1}{2} \times \frac{8}{3} \times \frac{8}{3\sqrt{3}} = \frac{32}{9\sqrt{3}}$.
Since the centroid divides the triangle into three triangles of equal area,Area$(\triangle ABC)$ = $3 \times$ Area$(\triangle AGB)$ = $3 \times \frac{32}{9\sqrt{3}} = \frac{32}{3\sqrt{3}}$ sq. units.

(C) The slope of $BC$ is $\frac{-3 - 3}{-1 - (-2)} = \frac{-6}{1} = -6$.
The slope of the altitude $L_1$ from $A$ to $BC$ is the negative reciprocal of the slope of $BC$,which is $\frac{1}{6}$.
The equation of $L_1$ passing through $A(1, -2)$ is $y - (-2) = \frac{1}{6}(x - 1) \Rightarrow y + 2 = \frac{x}{6} - \frac{1}{6} \Rightarrow y = \frac{x}{6} - \frac{13}{6}$ ... $(i)$
The midpoint of $AB$ is $\left(\frac{1 - 2}{2}, \frac{-2 + 3}{2}\right) = \left(-\frac{1}{2}, \frac{1}{2}\right)$.
The slope of $AB$ is $\frac{3 - (-2)}{-2 - 1} = \frac{5}{-3} = -\frac{5}{3}$.
The slope of the perpendicular bisector $L_2$ is $\frac{3}{5}$.
The equation of $L_2$ is $y - \frac{1}{2} = \frac{3}{5}(x + \frac{1}{2}) \Rightarrow y = \frac{3x}{5} + \frac{3}{10} + \frac{5}{10} \Rightarrow y = \frac{3x}{5} + \frac{4}{5}$ ... (ii)
Equating $(i)$ and (ii) to find the intersection point $(l, m)$:
$\frac{l}{6} - \frac{13}{6} = \frac{3l}{5} + \frac{4}{5} \Rightarrow \frac{l}{6} - \frac{3l}{5} = \frac{4}{5} + \frac{13}{6} \Rightarrow \frac{5l - 18l}{30} = \frac{24 + 65}{30} \Rightarrow -13l = 89 \Rightarrow 13l = -89$.
Since $13l = -89$,we have $l = -\frac{89}{13}$.
Substituting $l$ into $(i)$: $m = \frac{-89/13}{6} - \frac{13}{6} = \frac{-89}{78} - \frac{169}{78} = \frac{-258}{78} = -\frac{43}{13}$.
Then $26m - 3 = 26(-\frac{43}{13}) - 3 = 2(-43) - 3 = -86 - 3 = -89$.
Since $13l = -89$,we have $26m - 3 = 13l$.

(D) Let $G$ be the centroid of $\triangle ABC$. The centroid is the intersection of the medians. Given the equations of the medians $x+y=5$ and $x=4$. Substituting $x=4$ into $x+y=5$,we get $4+y=5$,so $y=1$. Thus,the centroid $G$ is $(4, 1)$.
Let $A = (1, 2)$ and $D = (\alpha, \beta)$ be the midpoint of $BC$. The centroid $G$ divides the median $AD$ in the ratio $2:1$.
Using the section formula,$G = \left(\frac{2\alpha + 1}{3}, \frac{2\beta + 2}{3}\right) = (4, 1)$.
Equating the coordinates:
$\frac{2\alpha + 1}{3} = 4 \implies 2\alpha + 1 = 12 \implies 2\alpha = 11 \implies \alpha = \frac{11}{2}$.
$\frac{2\beta + 2}{3} = 1 \implies 2\beta + 2 = 3 \implies 2\beta = 1 \implies \beta = \frac{1}{2}$.
Therefore,the midpoint $D$ of $BC$ is $\left(\frac{11}{2}, \frac{1}{2}\right)$.

(D) In an equilateral triangle,the orthocenter coincides with the centroid. Let $O(0,0)$ be the orthocenter. The altitude $AD$ passes through $O$. The slope of $BC$ is $m_{BC} = -\frac{1}{2\sqrt{2}}$. Since $AD \perp BC$,the slope of $AD$ is $m_{AD} = 2\sqrt{2}$. The equation of $AD$ is $y = 2\sqrt{2}x$,so $\beta = 2\sqrt{2}\alpha$.
The distance from $O(0,0)$ to $BC$ $(x+2\sqrt{2}y-4=0)$ is $OD = \frac{|0+0-4|}{\sqrt{1^2+(2\sqrt{2})^2}} = \frac{4}{\sqrt{9}} = \frac{4}{3}$.
In an equilateral triangle,the centroid divides the altitude in the ratio $2:1$. Thus,$AO = 2OD = 2(\frac{4}{3}) = \frac{8}{3}$. The total length of the altitude $AD = AO + OD = \frac{8}{3} + \frac{4}{3} = 4$.
The vertex $A$ lies on the line $y = 2\sqrt{2}x$ at a distance $4$ from the line $BC$. The coordinates of $A$ are $(\alpha, 2\sqrt{2}\alpha)$. The distance from $A$ to $x+2\sqrt{2}y-4=0$ is $\frac{|\alpha+2\sqrt{2}(2\sqrt{2}\alpha)-4|}{\sqrt{1+8}} = \frac{|9\alpha-4|}{3} = 4$.
This gives $9\alpha-4 = 12$ or $9\alpha-4 = -12$. So $\alpha = \frac{16}{9}$ or $\alpha = -\frac{8}{9}$.
Since $O(0,0)$ and $A$ must be on the same side of $BC$ (as $O$ is the centroid),we test the sign of the expression $x+2\sqrt{2}y-4$ at $(0,0)$,which is $-4$. For $A(\alpha, 2\sqrt{2}\alpha)$,the expression is $9\alpha-4$. Thus $9\alpha-4 < 0$,which implies $\alpha < \frac{4}{9}$. Therefore,$\alpha = -\frac{8}{9}$ and $\beta = 2\sqrt{2}(-\frac{8}{9}) = -\frac{16\sqrt{2}}{9}$.
We need to find the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta| = |-\frac{8}{9} + \sqrt{2}(-\frac{16\sqrt{2}}{9})| = |-\frac{8}{9} - \frac{32}{9}| = |-\frac{40}{9}| = \frac{40}{9} \approx 4.44$.
The greatest integer is $4$.

(C) Given $A = (1, 2)$ and midpoint of $AB$ is $M = (5, -1)$.
Since $M = (A+B)/2$,we have $(1+B_x)/2 = 5 \implies B_x = 9$ and $(2+B_y)/2 = -1 \implies B_y = -4$. Thus,$B = (9, -4)$.
Given centroid $G = (3, 4)$,we have $(A+B+C)/3 = G$,so $A+B+C = 3G = (9, 12)$.
$C = (9, 12) - (1, 2) - (9, -4) = (-1, 14)$.
Let circumcenter be $O = (\alpha, \beta)$. $O$ is equidistant from vertices $A, B, C$,so $OA^2 = OB^2 = OC^2$.
$OA^2 = (\alpha-1)^2 + (\beta-2)^2$
$OB^2 = (\alpha-9)^2 + (\beta+4)^2$
$OC^2 = (\alpha+1)^2 + (\beta-14)^2$
Equating $OA^2 = OB^2$: $(\alpha-1)^2 - (\alpha-9)^2 = (\beta+4)^2 - (\beta-2)^2$
$16\alpha - 80 = 12\beta + 12 \implies 4\alpha - 3\beta = 23$.
Equating $OB^2 = OC^2$: $(\alpha-9)^2 - (\alpha+1)^2 = (\beta-14)^2 - (\beta+4)^2$
$-20\alpha + 80 = -36\beta + 180 \implies 5\alpha - 9\beta = -25$.
Solving the system: $12\alpha - 9\beta = 69$ and $5\alpha - 9\beta = -25$.
Subtracting gives $7\alpha = 94 \implies \alpha = 94/7$.
Substituting $\alpha$: $4(94/7) - 3\beta = 23 \implies 376/7 - 23 = 3\beta \implies 215/7 = 3\beta \implies \beta = 215/21$.
Then $\alpha + \beta = 94/7 + 215/21 = (282+215)/21 = 497/21$.
$21(\alpha + \beta) = 497$.