Exercise 15. Let $A B C D$ be a parallelogram with area 1 and $M$ a point on the segment $[B D]$ such that $M D=3 M B$. We denote $N$ as the intersection point of the lines $(A M)$ and $(C B)$. Calculate the area of the triangle $M N D$.


## Solution to Exercise 15

![](https://cdn.mathpix.com/cropped/2024_06_04_02d16a1a39731d936aa9g-19.jpg?height=617&width=920&top_left_y=525&top_left_x=568)

In the following, we will denote $|X Y Z|$ as the area of triangle $X Y Z$.

We use the fact that the area of a triangle is equal to half the product of the lengths of its base and its height. Therefore, to compare the areas of different triangles, we can try to find a common base and compare the heights.

First, note that the area of parallelogram $A B C D$ is equal to twice the area of triangle $A B D$ because triangles $A B D$ and $C D B$ are isometric. Therefore, we have $|A B D| = \frac{1}{2}$.

We now want to compare the area of triangle $M N D$ and the area of triangle $A B D$.

Unfortunately, these two triangles do not have a common base. However, triangles $A N D$ and $A B D$ share the side $A D$. We draw the heights from vertices $N$ and $B$ in triangles $A N D$ and $A B D$, respectively. Let $K$ be the foot of the height from vertex $B$ in triangle $A B D$ and $J$ be the foot of the height from vertex $N$ in triangle $A D N$. We then have

$$
\frac{|A N D|}{|A B D|} = \frac{\frac{1}{2} \cdot A D \cdot N J}{\frac{1}{2} \cdot A D \cdot B K} = \frac{N J}{B K}
$$

Since the lines $(M I)$ and $(B K)$ are perpendicular to segment $[A D]$, they are parallel. The quadrilateral $J N B K$ is therefore a parallelogram with two right angles, making it a rectangle. Thus, $N J = B K$ and therefore

$$
|A N D| = |A B D| = \frac{1}{2}
$$

We apply the same method to compare the areas of triangles $A N D$ and $A M D$. We introduce $J$ as the foot of the height from vertex $N$ and observe that

$$
\frac{|A M D|}{|A N D|} = \frac{|A M D|}{|A B D|} = \frac{\frac{1}{2} \cdot A D \cdot M I}{\frac{1}{2} \cdot A D \cdot B K} = \frac{M I}{B K}
$$

The lines $(B K)$ and $(M I)$ are parallel, so by Thales' theorem for triangles $A B K$ and $A M I$, we find

$$
\frac{M I}{B K} = \frac{M D}{D B} = \frac{M D}{M B + M D} = \frac{3 M B}{M B + 3 M B} = \frac{3}{4}
$$

Thus, $|A M D| = \frac{3}{4} |A B D| = \frac{3}{8}$

We can then conclude since

$$
|M N D| = |A N D| - |A M D| = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}
$$

This allows us to conclude that the area of triangle $M N D$ is $\frac{1}{8}$.

Comment from the graders: This problem was difficult. The main idea (to remember) is that if two triangles have the same base, the ratio of their areas corresponds to the ratio of the lengths of the heights perpendicular to this base, or that if two triangles have heights of the same length, the ratio of their areas corresponds to the ratio of the lengths of the bases. Several students were able to use this observation to obtain interesting area equalities and solve the problem at least partially, and we congratulate them! Despite the difficulty of the problem, a significant number of students managed to find a complete solution, sometimes in a very original way. Here are, however, several frequent errors:

$\triangleright$ Many students placed points $M$ and $N$ incorrectly on the figure, such that $M B = 3 M D$ instead of $M D = 3 M B$ as required by the problem statement. We recommend carefully reading the problem statement to ensure the correct figure.

$\triangleright$ A significant number of students provided incorrect formulas for the area of a triangle or a parallelogram. Note that the area of a parallelogram is not equal to the product of its sides.

$\triangleright$ A significant number of students assumed (often implicitly) that a parallelogram with an area of 1 was a square with side 1 or a rectangle. While treating a specific case is always a good idea, the method used to determine the area of triangle $M N D$ for a square often did not allow solving the problem for any parallelogram.

$\triangleright$ A significant number of students assumed (sometimes implicitly) that the area of triangle $M N D$ did not depend on the chosen parallelogram. This assertion is difficult to prove and required a proof.

$\triangleright$ A handful of students claimed that, since $A D = 3 B N$, $M D = 3 M B$, and $A M = 3 M N$, we also have $A_{A D M} = 3 A_{B N M}$. Instead, we have $A_{A D M} = 9 A_{B N M}$.

$\triangleright$ A handful of students provided an answer based on measurements taken with a ruler on the figure. This was not what was expected in the problem.

$\triangleright$ A few students were very sparing with details in their writing. An unproven assertion does not earn points, so it is important to prove each assertion.

We remind the importance of drawing a large figure that is sufficiently general to avoid misleading the student (if the parallelogram looks too much like a rectangle, one is tempted to assume that certain angles are right angles). Some figures were a bit too small to serve as a basis for reflection. On the other hand, we appreciated several carefully drawn and intelligently annotated figures, demonstrating the student's understanding of the problem.