Incentre of an Isosceles Triangle Using Coordinates

(a) Let the coordinates of C be (0, c), where c > 0 as C lies on the positive y-axis. Since A(-3, 0) and B(3, 0) are symmetric about the y-axis and C lies on the y-axis, \triangle ABC (shaded blue) is an isosceles triangle with AC = BC. The altitude from C to AB lies on the y-axis. The incenter I(0, 2) lies on this altitude. The inradius r is the perpendicular distance from I(0, 2) to the side AB (which lies on the x-axis y=0). \begin{align*} r &= 2 - 0\\ r &= 2 \end{align*} The incenter I is also equidistant from the side AC. The equation of the line AC passing through A(-3, 0) and C(0, c) is: \begin{align*} \frac{y - 0}{x - (-3)} &= \frac{c - 0}{0 - (-3)}\\ \frac{y}{x + 3} &= \frac{c}{3}\\ 3y &= c(x + 3)\\ cx - 3y + 3c &= 0 \end{align*} Using the formula for the distance from a point (x_1, y_1) to a line Ax + By + C = 0: \begin{align*} \frac{|c(0) - 3(2) + 3c|}{\sqrt{c^2 + (-3)^2}} &= 2\\ \frac{|3c - 6|}{\sqrt{c^2 + 9}} &= 2\\ |3c - 6| &= 2\sqrt{c^2 + 9}\\ (3c - 6)^2 &= 4(c^2 + 9)\\ 9c^2 - 36c + 36 &= 4c^2 + 36\\ 5c^2 - 36c &= 0\\ c(5c - 36) &= 0 \end{align*} Since C lies on the positive y-axis, c \neq 0, thus: \begin{align*} 5c &= 36\\ c &= 7.2 \end{align*} Therefore, the coordinates of C are (0, 7.2).

(b) Since \triangle ABC is isosceles with AC=BC, both the orthocenter H and the circumcenter P must lie on the y-axis (the axis of symmetry). Let H be (0, y_H). The altitude from A(-3, 0) to BC passes through H. \begin{align*} \text{Slope of } BC &= \frac{7.2 - 0}{0 - 3} = -2.4\\ \text{Slope of altitude from } A &= \frac{-1}{-2.4} = \frac{5}{12} \end{align*} Equation of the altitude from A: \begin{align*} y - 0 &= \frac{5}{12}(x + 3)\\ y &= \frac{5}{12}x + 1.25 \end{align*} Setting x = 0 for the orthocenter H: \begin{align*} y_H &= 1.25\\ OH &= 1.25 \end{align*} Let P be (0, y_P). The circumcenter is equidistant from A(-3, 0) and C(0, 7.2), so PA^2 = PC^2: \begin{align*} (0 - (-3))^2 + (y_P - 0)^2 &= (0 - 0)^2 + (y_P - 7.2)^2\\ 9 + y_P^2 &= y_P^2 - 14.4y_P + 51.84\\ 14.4y_P &= 42.84\\ y_P &= 2.975 \end{align*} Thus, the distance from the origin O to P is: \begin{align*} OP &= 2.975 \end{align*} Since 2.975 > 1.25, OP > OH. Therefore, the circumcenter P is further from the origin O.