Let's break down the problem step by step: ## 1. **Understanding the problem** - Triangle \( ABC \) has integer side lengths. - Perimeter \( = 35 \). - Let \( AB = c \), \( BC = a \), \( CA = b \) (using standard notation). - The **incenter** \( I \) is the intersection of the angle bisectors. - The **centroid** \( G \) is the intersection of the medians. - The line \( IG \) is **perpendicular** to \( CI \), where \( C \) is a vertex, \( I \) is the incenter. We are to **find the length \( AB \)**. --- ## 2. **List all integer-sided triangles with perimeter 35** Let \( a \leq b \leq c \), \( a + b + c = 35 \), and triangle inequalities: - \( a + b > c \) - \( b + c > a \) - \( c + a > b \) But since \( a + b + c = 35 \), \( a + b = 35 - c \), so \( a + b > c \implies 35 - c > c \implies c < 17.5 \) So \( c \leq 17 \). Similarly, \( a \leq b \leq c \), so \( a \geq 35 - 2c \). Let's try possible \( c \) values from 17 down. ### For \( c = 17 \): \( a + b = 18 \), \( a \leq b \leq 17 \), \( a \geq 1 \). Check triangle inequalities: - \( a + b > 17 \implies 18 > 17 \) ✓ - \( b + 17 > a \implies b > a - 17 \) (always true since \( b \geq a \)) - \( 17 + a > b \implies a > b - 17 \) (always true) So \( a \geq 1 \), \( b = 18 - a \), \( b \leq 17 \implies a \geq 1 \). But \( a \leq b \implies a \leq 9 \). So possible pairs: \( (a, b, c) = (1, 17, 17), (2, 16, 17), \ldots, (9, 9, 17) \). But triangle sides must satisfy \( a + b > c \implies 18 > 17 \) ✓ But \( a + c > b \implies a + 17 > b \implies a > b - 17 \implies a \geq 1 \). So all these are possible triangles. But let's do the same for all possible \( c \). --- Continue for \( c = 16 \): \( a + b = 19 \), \( a \leq b \leq 16 \), \( a \geq 3 \) (since \( b \leq 16 \implies a \geq 3 \)). Similarly for \( c = 15, 14, ... \) But let's focus on the possible triangles and try to find which of them fit the additional geometric condition. --- ## 3. **Use the geometric condition** The **line connecting the incenter \( I \) and the centroid \( G \) is perpendicular to \( CI \).** Let’s recall: - The **centroid** \( G \) divides the median from \( C \) to \( AB \) in a \( 2:1 \) ratio. - The **incenter** \( I \) is at \((a x_A + b x_B + c x_C)/(a+b+c)\) for coordinates, and similarly for \( y \). But to proceed, let's try the following: ### Choose coordinates: Let’s place \( C \) at \( (0, 0) \), \( B \) at \( (a, 0) \), \( A \) at \( (d, h) \). - \( AB = c \) - \( CA = b \) - \( BC = a \) Let’s write \( AB = c \), \( BC = a \), \( CA = b \). Let’s use standard coordinates: Let \( C = (0, 0) \), \( B = (a, 0) \), \( A = (x, y) \). Then, \( CA = b = \sqrt{x^2 + y^2} \implies x^2 + y^2 = b^2 \) \( AB = c = \sqrt{(x-a)^2 + y^2} \implies (x-a)^2 + y^2 = c^2 \) Subtract: \[ (x^2 + y^2) - ((x-a)^2 + y^2) = b^2 - c^2 \implies x^2 - (x-a)^2 = b^2 - c^2 \implies x^2 - (x^2 - 2a x + a^2) = b^2 - c^2 \implies 2a x - a^2 = b^2 - c^2 \implies x = \frac{b^2 - c^2 + a^2}{2a} \] And then \( y^2 = b^2 - x^2 \). --- ### Coordinates of points: - \( C = (0, 0) \) - \( B = (a, 0) \) - \( A = (x, y) \) --- ### **Centroid** \( G \): Coordinates: \[ G = \left( \frac{0 + a + x}{3}, \frac{0 + 0 + y}{3} \right ) = \left( \frac{a + x}{3}, \frac{y}{3} \right ) \] --- ### **Incenter** \( I \): Coordinates: \[ I = \frac{a \cdot A + b \cdot B + c \cdot C}{a + b + c} \] Let’s compute: - \( A = (x, y) \) - \( B = (a, 0) \) - \( C = (0, 0) \) \[ I = \frac{a(x, y) + b(a, 0) + c(0, 0)}{a + b + c} = \frac{(a x + b a, a y)}{a + b + c} \] So, \[ I_x = \frac{a x + b a}{a + b + c} \] \[ I_y = \frac{a y}{a + b + c} \] --- ### **CI vector**: \[ \overrightarrow{CI} = (I_x, I_y) \] --- ### **IG vector**: \[ \overrightarrow{IG} = (G_x - I_x, G_y - I_y) \] --- ### **Condition: \( IG \perp CI \)** That is, their dot product is zero: \[ \overrightarrow{CI} \cdot \overrightarrow{IG} = 0 \] So \[ I_x (G_x - I_x) + I_y (G_y - I_y) = 0 \] --- Let’s substitute: - \( G_x = \frac{a + x}{3} \) - \( G_y = \frac{y}{3} \) - \( I_x = \frac{a x + b a}{a + b + c} \) - \( I_y = \frac{a y}{a + b + c} \) Let’s denote \( s = a + b + c = 35 \). --- \[ I_x = \frac{a x + b a}{s} \] \[ I_y = \frac{a y}{s} \] \[ G_x = \frac{a + x}{3} \] \[ G_y = \frac{y}{3} \] \[ G_x - I_x = \frac{a + x}{3} - \frac{a x + b a}{s} \] \[ G_y - I_y = \frac{y}{3} - \frac{a y}{s} \] Thus, \[ I_x (G_x - I_x) + I_y (G_y - I_y) = 0 \] Let’s expand: \[ I_x (G_x - I_x) = I_x G_x - I_x^2 \] \[ I_y (G_y - I_y) = I_y G_y - I_y^2 \] So, \[ I_x G_x + I_y G_y = I_x^2 + I_y^2 \] --- Let’s compute \( I_x^2 + I_y^2 \): \[ I_x^2 + I_y^2 = \left( \frac{a x + b a}{s} \right)^2 + \left( \frac{a y}{s} \right)^2 = \frac{ (a x + b a)^