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. The incircle is tangent to AB at F and to CA at E. The line EF is extended to meet BC at P. I need to find the length of BP.

### Step 1: Draw the Triangle and Understand the Incircle
First, I need to draw the triangle ABC with the given sides:
- AB = 13
- BC = 14
- CA = 15

I recall that the incircle is the largest circle that fits inside the triangle and is tangent to all three sides. Points E and F are the points of tangency on CA and AB, respectively.

### Step 2: Find the Semiperimeter and Inradius
To find the incircle's properties, I need the semiperimeter (s) of the triangle.

The perimeter P is 13 + 14 + 15 = 42.

So, s = P/2 = 21.

Now, I recall the formula for the area (A) of a triangle using Heron's formula:
A = √[s(s - a)(s - b)(s - c)] = √[21(21-13)(21-14)(21-15)] = √[21 * 8 * 7 * 6] = √[21 * 8 * 7 * 6].

Let's break it down:
- 21 * 6 = 126
- 8 * 7 = 56
- Now, 126 * 56 = (100 + 26) * 56 = 5600 + 1456 = 7056.

So, A = √7056 = 84.

Now, the area can also be expressed as A = r * s, where r is the inradius.

So, r = A / s = 84 / 21 = 4.

### Step 3: Find the Points of Tangency
The incircle touches the sides at points E, F, and D (on BC).

I recall that the lengths of the tangents from the vertices to the points of tangency are equal.

Let's denote:
- AE = AF = x (since both are tangents from A to the incircle)
- BD = BF = y (tangents from B)
- CD = CE = z (tangents from C)

Now, the sides can be expressed as:
- AB = AF + BF = x + y = 13
- BC = BD + CD = y + z = 14
- CA = CE + AE = z + x = 15

Now, we have a system of equations:
1. x + y = 13
2. y + z = 14
3. z + x = 15

Let's solve this system.

First, add all three equations:
(x + y) + (y + z) + (z + x) = 13 + 14 + 15 → 2x + 2y + 2z = 42 → x + y + z = 21.

Now, subtract equation 1 from the sum:
(x + y + z) - (x + y) = 21 - 13 → z = 8.

Subtract equation 2 from the sum:
(x + y + z) - (y + z) = 21 - 14 → x = 7.

Subtract equation 3 from the sum:
(x + y + z) - (z + x) = 21 - 15 → y = 6.

So, the lengths are:
- AE = AF = 7
- BD = BF = 6
- CD = CE = 8

### Step 4: Find Coordinates for Easier Calculation
Since the problem involves a line EF extended to meet BC at P, maybe coordinate geometry can help.

First, I need to place the triangle in a coordinate system.

Let's say:
- Point B is at the origin (0,0).
- Point C is along the x-axis at (14,0).

Now, I need to find the coordinates of A.

Using the distance formula:
- AB = √(x_A² + y_A²) = 13 → x_A² + y_A² = 169.
- AC = √((x_A - 14)² + y_A²) = 15 → (x_A - 14)² + y_A² = 225.

Subtract the first equation from the second:
(x_A - 14)² - x_A² = 225 - 169 → x_A² - 28x_A + 196 - x_A² = 56 → -28x_A + 196 = 56 → -28x_A = -140 → x_A = 5.

Now, plug x_A back into the first equation:
5² + y_A² = 169 → 25 + y_A² = 169 → y_A² = 144 → y_A = ±12.

Since the triangle is above the x-axis, y_A = 12.

So, A is at (5,12).

### Step 5: Find Points E and F
From Step 3:
- AF = 7 → F is 7 units from A towards B.
- AE = 7 → E is 7 units from A towards C.

First, find the parametric positions.

**For F:**
F divides AB in the ratio AF:FB = 7:6.

Using the section formula:
F_x = (6*5 + 7*0)/(7+6) = 30/13 ≈ 2.3077
F_y = (6*12 + 7*0)/(7+6) = 72/13 ≈ 5.5385

**For E:**
E divides AC in the ratio AE:EC = 7:8.

Using the section formula:
E_x = (8*5 + 7*14)/(7+8) = (40 + 98)/15 = 138/15 = 46/5 = 9.2
E_y = (8*12 + 7*0)/(7+8) = 96/15 = 32/5 = 6.4

### Step 6: Find the Equation of Line EF
Now, we have points E (9.2, 6.4) and F (30/13, 72/13).

First, find the slope (m) of EF:
m = (6.4 - 72/13) / (9.2 - 30/13)

Let's convert everything to fractions:
6.4 = 32/5
72/13 remains the same.

Numerator: 32/5 - 72/13 = (32*13 - 72*5)/65 = (416 - 360)/65 = 56/65.

Denominator: 9.2 = 46/5
46/5 - 30/13 = (46*13 - 30*5)/65 = (598 - 150)/65 = 448/65.

So, m = (56/65) / (448/65) = 56/448 = 1/8.

Now, the equation of the line EF using point-slope form:
y - y1 = m(x - x1).

Using point E