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.

To solve the problem, we start by noting the semiperimeter \( s \) of triangle \( ABC \) with sides \( AB = 13 \), \( BC = 14 \), and \( CA = 15 \). The semiperimeter is calculated as:

\[
s = \frac{13 + 14 + 15}{2} = 21
\]

Using the semiperimeter, we determine the lengths from the vertices to the points of tangency:

- \( AF = AE = s - BC = 21 - 14 = 7 \)
- \( BF = BD = s - AC = 21 - 15 = 6 \)
- \( CD = CE = s - AB = 21 - 13 = 8 \)

Next, we place the triangle in a coordinate system with \( B \) at \( (0, 0) \), \( C \) at \( (14, 0) \), and \( A \) at \( (5, 12) \). The coordinates of points \( E \) and \( F \) are found using the section formula:

- \( E \) divides \( AC \) in the ratio \( 7:8 \), giving \( E \left( \frac{46}{5}, \frac{32}{5} \right) \).
- \( F \) divides \( AB \) in the ratio \( 7:6 \), giving \( F \left( \frac{35}{13}, \frac{84}{13} \right) \).

The slope of line \( EF \) is calculated as:

\[
\text{slope} = \frac{\frac{32}{5} - \frac{84}{13}}{\frac{46}{5} - \frac{35}{13}} = \frac{-4/65}{423/65} = -\frac{4}{423}
\]

Using the point-slope form with point \( F \), the equation of line \( EF \) is:

\[
y - \frac{84}{13} = -\frac{4}{423} \left( x - \frac{35}{13} \right)
\]

Setting \( y = 0 \) to find the intersection with \( BC \):

\[
- \frac{84}{13} = -\frac{4}{423} \left( x - \frac{35}{13} \right)
\]

Solving for \( x \):

\[
x = 686
\]

Thus, point \( P \) is at \( (686, 0) \), and the length \( BP \) is:

\[
BP = 686
\]

The final answer is:

\[
\boxed{686}
\]