math-geo-01

# Incircle Contact Chord Meets the Base

In triangle $ABC$ we have $AB = 13$, $BC = 14$, and $CA = 15$.

The incircle of triangle $ABC$ is tangent to side $AB$ at $F$ and tangent to side $CA$ at $E$. Line $EF$ is extended until it meets line $BC$ at a point $P$.

Find the length $BP$.

Give your final answer as an integer.

Alright, I have this problem about a triangle ABC with sides AB = 13, BC = 14, and CA = 15. There's an incircle that touches AB at F and CA at E. Then, EF is extended to meet BC at P, and I need to find BP.

### Step 1: Understand the Triangle and Its Properties
First, I need to draw the triangle ABC based on the given side lengths. It's a scalene triangle since all sides are different.

### Step 2: Find the Semiperimeter and Area
I recall that for any triangle, the semiperimeter (s) is half the perimeter. So:
- Perimeter = AB + BC + CA = 13 + 14 + 15 = 42
- Semiperimeter, s = 42 / 2 = 21

Now, to find the area, I can use Heron's formula:
Area = √[s(s - a)(s - b)(s - c)] = √[21(21 - 13)(21 - 14)(21 - 15)] = √[21 * 8 * 7 * 6]

Let's break it down:
- 21 * 8 = 168
- 7 * 6 = 42
- 168 * 42 = (170 - 2)(40 + 2) = 170*40 + 170*2 - 2*40 - 2*2 = 6800 + 340 - 80 - 4 = 7056

So, Area = √7056 = 84

### Step 3: Find the Inradius
The area of the triangle is also equal to the product of the semiperimeter and the inradius (r). So:
84 = 21 * r ⇒ r = 84 / 21 = 4

### Step 4: Find the Points of Tangency E and F
The incircle is tangent to the sides. The lengths of the tangents from the vertices to the points of tangency can be found using the formula:
- For vertex A: AF = AE = s - BC = 21 - 14 = 7
- For vertex B: BF = BD = s - CA = 21 - 15 = 6
- For vertex C: CE = CD = s - AB = 21 - 13 = 8

But in the problem, the incircle is tangent to AB at F and to CA at E. So:
- AF = 7 ⇒ F is 7 units from A on AB.
- AE = 7 ⇒ E is 7 units from A on CA.

### Step 5: Find Coordinates to Visualize the Problem
To make it easier, I think I can place the triangle in a coordinate system.

Let's say:
- Point A is at the origin (0,0).
- Side AB is along the x-axis. So, point B is at (13,0) since AB = 13.
- Point C is somewhere in the plane. I need to find its coordinates.

First, find the angle at A using the cosine rule:
cos(A) = (AB² + AC² - BC²) / (2 * AB * AC) = (13² + 15² - 14²) / (2 * 13 * 15) = (169 + 225 - 196) / 390 = (394 - 196)/390 = 198/390 = 33/65

Now, the coordinates of C can be found using the angle A and the lengths AB and AC.
- C_x = AC * cos(A) = 15 * (33/65) = 495/65 = 7.6
- C_y = AC * sin(A) = 15 * √(1 - (33/65)²) = 15 * √[(65² - 33²)/65²] = 15 * √[(4225 - 1089)/4225] = 15 * √(3136/4225) = 15 * 56/65 = 840/65 = 12.942857...

### Step 6: Find the Equation of EF
Points E and F are the points of tangency on CA and AB, respectively.

- F is on AB, 7 units from A ⇒ F is at (7,0).
- E is on CA, 7 units from A ⇒ E is 7/15 of the way from A to C.

But C is at (7.6, 12.942857...), so E's coordinates are:
- E_x = A_x + (7/15) * (C_x - A_x) = 0 + (7/15)*7.6 ≈ 3.542857
- E_y = A_y + (7/15) * (C_y - A_y) ≈ 0 + (7/15)*12.942857 ≈ 6.285714

Now, the line EF goes from F (7,0) to E (3.542857, 6.285714).

First, find the slope (m) of EF:
m = (6.285714 - 0) / (3.542857 - 7) ≈ 6.285714 / (-3.457143) ≈ -1.818182

Now, the equation of the line EF in point-slope form using point F (7,0):
y - 0 = m (x - 7) ⇒ y = -1.818182 (x - 7)

### Step 7: Find the Equation of BC
Points B and C are at (13,0) and (7.6, 12.942857...).

Slope of BC:
m_BC = (12.942857 - 0) / (7.6 - 13) ≈ 12.942857 / (-5.4) ≈ -2.396822

Equation of BC using point B (13,0):
y - 0 = m_BC (x - 13) ⇒ y ≈ -2.396822 (x - 13)

### Step 8: Find the Intersection Point P
P is the intersection of EF and BC.

From EF: y ≈ -1.818182 (x - 7)
From BC: y ≈ -2.396822 (x - 13)

Set them equal:
-1.818182 (x - 7) ≈ -2.396822 (x - 13)

Let's solve for x:
-1.818182x + 12.727274 ≈ -2.396822x + 31.158686
-1.818182x + 2.396822x ≈ 31.158686 - 12.727274
0.57864x ≈ 18.431412
x ≈ 18.431412 / 0.57864 ≈ 31.85

Now, find y using EF's equation:
y ≈ -1.818182 (31.85 - 7) ≈ -1.818182 * 24.85 ≈ -45.19

But P is supposed to be on BC, and BC goes from (13,0) to (7.6,12.94). The y-coordinate of P is negative, but C has a positive y-coordinate. This suggests P is outside the segment BC, but the problem says "meets line BC at P," so it's acceptable