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.

We are given triangle $ ABC $ with side lengths:

- $ AB = c = 13 $
- $ BC = a = 14 $
- $ CA = b = 15 $

Let’s denote the standard notation: side $ a = BC $, opposite vertex $ A $; side $ b = CA $, opposite vertex $ B $; side $ c = AB $, opposite vertex $ C $.

We are told:

- The incircle touches $ AB $ at $ F $, and $ CA $ at $ E $.
- Line $ EF $ is extended to meet line $ BC $ at point $ P $.
- We need to find $ BP $.

---

### Step 1: Coordinates

Let’s place triangle $ ABC $ in the coordinate plane for convenience.

Let:
- $ B = (0, 0) $
- $ C = (14, 0) $ since $ BC = 14 $ lies on x-axis.
- Let $ A = (x, y) $, with $ y > 0 $.

Then:

- $ AB = 13 \Rightarrow x^2 + y^2 = 169 $
- $ AC = 15 \Rightarrow (x - 14)^2 + y^2 = 225 $

Subtract first equation from second:

$$
(x - 14)^2 + y^2 - (x^2 + y^2) = 225 - 169 = 56
$$
$$
x^2 - 28x + 196 + y^2 - x^2 - y^2 = 56 \Rightarrow -28x + 196 = 56 \Rightarrow -28x = -140 \Rightarrow x = 5
$$

Then $ x = 5 $, so from $ x^2 + y^2 = 169 $:

$$
25 + y^2 = 169 \Rightarrow y^2 = 144 \Rightarrow y = 12
$$

So $ A = (5, 12) $

---

### Step 2: Find coordinates of points $ E $ and $ F $

Recall: The incircle touches side $ AB $ at point $ F $, and side $ AC $ at point $ E $. The distances from vertices to touchpoints are related to the semiperimeter.

Let the semiperimeter be:

$$
s = \frac{a + b + c}{2} = \frac{14 + 15 + 13}{2} = \frac{42}{2} = 21
$$

Then:

- Distance from $ A $ to touchpoint on $ AB $ (i.e., $ AF $) = $ s - a = 21 - 14 = 7 $
- Distance from $ B $ to touchpoint on $ AB $ (i.e., $ BF $) = $ s - b = 21 - 15 = 6 $

Check: $ AF + BF = 7 + 6 = 13 = AB $, correct.

Similarly:

- On side $ AC $: $ AE = s - a = 7 $, $ EC = s - c = 21 - 13 = 8 $, and $ AE + EC = 7 + 8 = 15 = AC $, correct.

So:

- $ F $ lies on $ AB $, with $ AF = 7 $, $ FB = 6 $
- $ E $ lies on $ AC $, with $ AE = 7 $, $ EC = 8 $

We can find coordinates using section formula.

#### Coordinates of $ F $ on $ AB $:

$ A = (5,12) $, $ B = (0,0) $, and $ AF:FB = 7:6 $. So $ F $ divides $ AB $ in ratio $ AF:FB = 7:6 $ from $ A $ to $ B $.

So using internal division formula:

$$
F = \left( \frac{6 \cdot 5 + 7 \cdot 0}{7+6}, \frac{6 \cdot 12 + 7 \cdot 0}{13} \right) = \left( \frac{30}{13}, \frac{72}{13} \right)
$$

Wait: careful! If point divides segment $ AB $ in ratio $ m:n $ where $ m = AF $, $ n = FB $, then coordinates are:

$$
F = \frac{n A + m B}{m + n}
$$

Because closer to $ B $ if $ m > n $? Let's double-check.

General formula: If point $ P $ divides segment $ XY $ in ratio $ XP : PY = m:n $, then:

$$
P = \frac{n X + m Y}{m + n}
$$

So here, $ AF:FB = 7:6 $, so $ F $ divides $ AB $ such that from $ A $ to $ B $, ratio is $ 7:6 $. So:

$$
F = \frac{6 A + 7 B}{13} = \frac{6(5,12) + 7(0,0)}{13} = \left( \frac{30}{13}, \frac{72}{13} \right)
$$

Yes, that matches.

#### Coordinates of $ E $ on $ AC $:

$ A = (5,12) $, $ C = (14,0) $, and $ AE:EC = 7:8 $

So:

$$
E = \frac{8 A + 7 C}{15} = \frac{8(5,12) + 7(14,0)}{15} = \left( \frac{40 + 98}{15}, \frac{96 + 0}{15} \right) = \left( \frac{138}{15}, \frac{96}{15} \right) = \left( \frac{46}{5}, \frac{32}{5} \right)
$$

Simplify: $ \frac{138}{15} = \frac{46}{5} $, yes.

So:

- $ F = \left( \frac{30}{13}, \frac{72}{13} \right) $
- $ E = \left( \frac{46}{5}, \frac{32}{5} \right) $

---

### Step 3: Find equation of line $ EF $, then find intersection with line $ BC $ (x-axis, since $ B = (0,0), C = (14,0) $)

Line $ BC $ is the x-axis: $ y = 0 $. So point $ P $ is where line $ EF $ intersects $ y = 0 $.

So find equation of line through $ E $ and $ F $, then set $ y = 0 $ to solve for $ x $, which gives $ P = (x_P, 0) $, then $ BP = |x_P - 0| = x_P $ (since likely $ x_P > 0 $).

Let’s compute slope $ m $:

$$
m = \frac{y_E - y_F}{x_E - x_F} = \frac{\frac{32}{5} - \frac{72}{13}}{\frac{46}{5} - \frac{30}{13}}
$$

Compute numerator:

$$
\frac{32}{5} - \frac{72}{13} = \frac{32 \cdot 13 - 72 \cdot 5}{65} = \frac{416 - 360}{65} = \frac{56}{65}
$$

Denominator:

$$
\frac{46}{5} -