Question: Given a parallelogram ABCD. The points E and K lie on the line segments BC and CD (respectively), so that BE/EC = 3, CK/KD = 2. Let [AEK]/[ABCD] = m/n, where m and n are relatively prime natural numbers. Find m + n.

1. **Problem Statement:** We have a parallelogram ABCD with points E on BC and K on CD such that $\frac{BE}{EC} = 3$ and $\frac{CK}{KD} = 2$. We want to find the ratio of the area of triangle AEK to the area of parallelogram ABCD, expressed as $\frac{m}{n}$ where $m$ and $n$ are coprime, and then find $m+n$. 2. **Set up coordinates:** Let’s place the parallelogram in the coordinate plane for convenience. - Let $A = (0,0)$. - Let $B = (b,0)$. - Let $D = (0,d)$. Then $C = B + D = (b,d)$. 3. **Find coordinates of E and K:** - Point $E$ lies on $BC$ such that $\frac{BE}{EC} = 3$. Since $B=(b,0)$ and $C=(b,d)$, segment $BC$ is vertical. The ratio means $BE:EC = 3:1$, so $E$ divides $BC$ in ratio 3:1 starting from $B$. Using section formula for internal division: $$E = \left(b, \frac{3 \cdot d + 1 \cdot 0}{3+1}\right) = (b, \frac{3d}{4})$$ - Point $K$ lies on $CD$ such that $\frac{CK}{KD} = 2$. Since $C=(b,d)$ and $D=(0,d)$, segment $CD$ is horizontal. The ratio means $CK:KD = 2:1$, so $K$ divides $CD$ in ratio 2:1 starting from $C$. Using section formula: $$K = \left(\frac{2 \cdot 0 + 1 \cdot b}{2+1}, d\right) = \left(\frac{b}{3}, d\right)$$ 4. **Coordinates summary:** $$A = (0,0), E = \left(b, \frac{3d}{4}\right), K = \left(\frac{b}{3}, d\right)$$ 5. **Area of parallelogram ABCD:** The area is base times height: $$[ABCD] = b \times d$$ 6. **Area of triangle AEK:** Use the determinant formula for area of triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$x_1=0, y_1=0$$ $$x_2=b, y_2=\frac{3d}{4}$$ $$x_3=\frac{b}{3}, y_3=d$$ Calculate: $$\text{Area}(AEK) = \frac{1}{2} |0 \cdot (\frac{3d}{4} - d) + b \cdot (d - 0) + \frac{b}{3} \cdot (0 - \frac{3d}{4})|$$ $$= \frac{1}{2} |0 + bd - \frac{b}{3} \cdot \frac{3d}{4}| = \frac{1}{2} |bd - \frac{b d}{4}| = \frac{1}{2} \cdot \frac{3 b d}{4} = \frac{3 b d}{8}$$ 7. **Ratio of areas:** $$\frac{[AEK]}{[ABCD]} = \frac{\frac{3 b d}{8}}{b d} = \frac{3}{8}$$ 8. **Simplify ratio:** $m=3$, $n=8$, which are relatively prime. 9. **Find $m+n$:** $$m + n = 3 + 8 = 11$$ **Final answer:** $11$