A line passing through a point M on segment $AB$ intersects segment $AC$ at point N. a) How should M be chosen so that the perimeter of triangle $AMN$ is one-third of the perimeter of triangle $ABC$? b) How should M be chosen so that the area of triangle $AMN$ is one-quarter of the area of triangle $ABC$?

1. Problem Description

A line passing through a point M on segment $AB$ intersects segment $AC$ at point N.

a) How should M be chosen so that the perimeter of triangle $AMN$ is one-third of the perimeter of triangle $ABC$?

b) How should M be chosen so that the area of triangle $AMN$ is one-quarter of the area of triangle $ABC$?

2. Solution Steps

a) Let $P_{AMN}$ be the perimeter of triangle $AMN$, and $P_{ABC}$ be the perimeter of triangle $ABC$.

We are given that $P_{AMN} = \frac{1}{3} P_{ABC}$.

Let $AM = x \cdot AB$ and $AN = y \cdot AC$. Since the line $MN$ intersects $AB$ and $AC$, the triangles $AMN$ and $ABC$ are not necessarily similar. Thus, let $AM=k AB$, $AN=k AC$ and $MN=k BC$. So, $AM = xAB$ and $AN = xAC$, and it is given that $MN \parallel BC$.

If we choose $M$ such that the triangles $AMN$ and $ABC$ are similar, then $AM = xAB$, $AN = xAC$, and $MN = xBC$ for some scalar $x$.

Then $P_{AMN} = AM + AN + MN = xAB + xAC + xBC = x(AB + AC + BC) = x P_{ABC}$.

We want $P_{AMN} = \frac{1}{3} P_{ABC}$, so $x P_{ABC} = \frac{1}{3} P_{ABC}$, which means $x = \frac{1}{3}$.

Therefore, $AM = \frac{1}{3} AB$, i.e., $M$ should be chosen such that $AM$ is one-third of $AB$.

b) Let $Area(AMN)$ be the area of triangle $AMN$, and $Area(ABC)$ be the area of triangle $ABC$.

We are given that $Area(AMN) = \frac{1}{4} Area(ABC)$.

We know that $Area(AMN) = \frac{1}{2} AM \cdot AN \cdot \sin(\angle A)$, and $Area(ABC) = \frac{1}{2} AB \cdot AC \cdot \sin(\angle A)$.

Therefore, $\frac{1}{2} AM \cdot AN \cdot \sin(\angle A) = \frac{1}{4} (\frac{1}{2} AB \cdot AC \cdot \sin(\angle A))$.

This simplifies to $AM \cdot AN = \frac{1}{4} AB \cdot AC$.

Let $AM = x AB$ and $AN = y AC$. Then $x AB \cdot y AC = \frac{1}{4} AB \cdot AC$, so $xy = \frac{1}{4}$.

If we choose $M$ such that $MN \parallel BC$, then $\frac{AM}{AB} = \frac{AN}{AC}$, so $x = y$.

Then $x^2 = \frac{1}{4}$, so $x = \frac{1}{2}$.

In this case, $AM = \frac{1}{2} AB$, which means $M$ is the midpoint of $AB$.

3. Final Answer

a) $M$ should be chosen such that $AM = \frac{1}{3} AB$.

b) $M$ should be chosen such that $AM = \frac{1}{2} AB$, i.e., $M$ is the midpoint of $AB$.