In triangle ABO, M is the midpoint of \overline{AB}. Points C, D, and N lie on line segments $\overline{AO},$ $\overline{BO},$ and $\overline{MO}$ extended, respectively, so that $DO = BO,$ $NO = MO,$ and $CO = AO.$ If $K$ is the area of triangle $ABO,$ then what is the area of triangle CDN?

Answer 1

Let the area of triangle \( ABO \) be \( K \). We will find the area of triangle \( CDN \) in terms of \( K \).

Let \( A = (0, 0) \), \( B = (b, 0) \), and \( O = (0, h) \) for some \( b > 0 \) and \( h > 0 \). The coordinates of the points are as follows:

• \( M \), the midpoint of \( AB \), is given by

\[ M = \left( \frac{b}{2}, 0 \right). \]

Let \( C \) be on \( \overline{AO} \), so it can be expressed as:

\[ C = (0, ky) \quad \text{for some } 0 < k < 1, \]

where \( ky \) lies on the line segment \( \overline{AO} \).

Let \( D \) be on \( \overline{BO} \). Since \( DO = BO \), point \( D \) is twice the distance from \( B \) to \( O \). The coordinates of \( D \) can be calculated using the coordinates of \( B \) and \( O \):

\[ D = B + \overrightarrow{BO} = (b, 0) + \left( -b, h \right) = (0, h). \]

For point \( N \), which lies on \( \overline{MO} \) extended, since \( NO = MO \), we can determine \( N \) similarly. The depiction of \( M \) and \( O \) gives:

\[ MO \text{ is represented as } \overline{MO} = O - M = \left( 0 - \frac{b}{2}, h - 0 \right) = \left( -\frac{b}{2}, h \right) \]

The coordinates of \( N \) can thus be calculated. To find \( N \), we scale the vector \( \overrightarrow{MO} \):

\[ N = M + \overrightarrow{MO} = \left( \frac{b}{2}, 0 \right) + k \left( -\frac{b}{2}, h \right) = \left( \frac{b}{2} - \frac{kb}{2}, kh \right) = \left( \frac{b(1-k)}{2}, kh \right) \]

Now we have the coordinates:

• \( C = (0, ky) \)
• \( D = (0, h) \)
• \( N = \left( \frac{b(1-k)}{2}, kh \right) \)

Next, we calculate the area of triangle \( CDN \). Using the vertices \( C(0, ky) \), \( D(0, h) \), \( N \left( \frac{b(1-k)}{2}, kh \right) \), we apply the formula for the area of a triangle determined by three points:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates into the formula:

\[ \text{Area} = \frac{1}{2} \left| 0(h - kh) + 0(ky - h) + \frac{b(1-k)}{2}(ky - h) \right| \]

This simplifies to:

\[ = \frac{1}{2} \left| \frac{b(1-k)}{2}(ky - h) \right| = \frac{b(1-k)}{4} |ky - h|. \]

Next, we see that since \( K \) is the area of triangle \( ABO \):

\[ K = \frac{1}{2}bh. \]

Casting the area of triangle \( CDN \) in terms of \( K \), observe that the area relates directly through length ratios due to similarity of triangles.

Recognizing \( y = \frac{Ky}{h} \):

The area \( CDN \) becomes:

\[ \text{Area} = \frac{1}{2} \left| 0(h - kh) + 0(ky - h) + \frac{b(1-k)}{2}(ky - h) \right| = \frac{1}{2}K(1-k) \Rightarrow \text{Area of } CDN = K. \]

Thus the area of triangle \( CDN\):

\[ \text{Area of } CDN = 3K. \]

Finally, the area of triangle \( CDN \) is

\[ \boxed{3K}. \]