1. **Problem statement:** We have triangle OAB with points P and N on OA and OB respectively, M is midpoint of AB, and lines OPM and APN are straight. Given $OP : PM = 4 : 3$, find the ratio $ON : NB$. 2. **Given vectors:** $\overrightarrow{OA} = a$, $\overrightarrow{OB} = b$. 3. **Step 1: Find $\overrightarrow{OM}$:** Since M is midpoint of AB, $$\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{a + b}{2}.$$ 4. **Step 2: Express $\overrightarrow{OP}$ on line OA:** Given $OP : PM = 4 : 3$, point P divides segment OM in ratio 4:3. Since $O$, $P$, $M$ are collinear, $$\overrightarrow{OP} = \frac{4}{4+3} \overrightarrow{OM} = \frac{4}{7} \cdot \frac{a + b}{2} = \frac{2}{7}(a + b).$$ 5. **Step 3: Express $\overrightarrow{OP}$ also on OA:** But P lies on OA, so $$\overrightarrow{OP} = t a$$ for some $t$ between 0 and 1. Equate this to previous expression: $$t a = \frac{2}{7}(a + b).$$ 6. **Step 4: Solve for $t$:** Comparing components, $$t a = \frac{2}{7} a + \frac{2}{7} b.$$ Since $a$ and $b$ are independent vectors, this equality holds only if the $b$ component is zero, which is not possible unless $b=0$. So the assumption that $P$ lies on OA is incorrect if $\overrightarrow{OP} = \frac{2}{7}(a + b)$. Re-examine step 4: The ratio $OP : PM = 4 : 3$ applies on line OPM, so $P$ divides segment OM in ratio 4:3, but $P$ lies on OA, so $P$ divides OA in some ratio $s$. 7. **Step 5: Find $s$ such that $P = s a$ lies on OA and also lies on line OPM:** Since $M$ is midpoint of AB, $\overrightarrow{M} = \frac{a + b}{2}$. Line OPM is straight, so points O, P, M are collinear. Parametrize line OA: $\overrightarrow{OP} = s a$. Parametrize line OM: $\overrightarrow{OM} = \frac{a + b}{2}$. Since $P$ divides $OM$ in ratio $4:3$, $\overrightarrow{OP} = \frac{4}{7} \overrightarrow{OM} + \frac{3}{7} \overrightarrow{O} = \frac{4}{7} \cdot \frac{a + b}{2} = \frac{2}{7}(a + b)$. But $P$ lies on OA, so $\overrightarrow{OP} = s a$. Equate: $$s a = \frac{2}{7} a + \frac{2}{7} b.$$ This implies $\frac{2}{7} b = (s - \frac{2}{7}) a$, which is impossible unless $b$ is parallel to $a$, contradicting triangle. 8. **Step 6: Correct interpretation:** Since $P$ lies on OA, and $OP : PM = 4 : 3$, and $M$ lies on AB, the line OPM is a straight line passing through points O, P, M. Therefore, $P$ lies on OA, $M$ lies on AB, and $O$, $P$, $M$ are collinear. Express $P = s a$, $M = \frac{a + b}{2}$. Vectors $\overrightarrow{OP} = s a$, $\overrightarrow{OM} = \frac{a + b}{2}$. Since $O$, $P$, $M$ are collinear, vectors $\overrightarrow{OP}$ and $\overrightarrow{OM}$ are linearly dependent: $$\overrightarrow{OM} = \lambda \overrightarrow{OP}$$ for some $\lambda > 1$ (since $P$ lies between $O$ and $M$). So, $$\frac{a + b}{2} = \lambda s a.$$ Equate components: - Along $a$: $\frac{1}{2} = \lambda s$. - Along $b$: $\frac{1}{2} = 0$ (contradiction). So $a$ and $b$ are independent, so this is impossible unless $b=0$. 9. **Step 7: Use vector ratios to find $N$:** Since $N$ lies on OB, let $\overrightarrow{ON} = x b$. Since $M$ is midpoint of AB, $\overrightarrow{OM} = \frac{a + b}{2}$. Line APN is straight, so points A, P, N are collinear. Express $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = s a - a = (s - 1) a$. Express $\overrightarrow{AN} = \overrightarrow{ON} - \overrightarrow{OA} = x b - a$. Since A, P, N are collinear, vectors $\overrightarrow{AP}$ and $\overrightarrow{AN}$ are linearly dependent: $$\overrightarrow{AN} = \mu \overrightarrow{AP}$$ for some scalar $\mu$. So, $$x b - a = \mu (s - 1) a.$$ Equate components: - Along $a$: $-1 = \mu (s - 1)$ - Along $b$: $x = 0$ So $x = 0$ means $N = O$, which contradicts $N$ on OB between O and B. 10. **Step 8: Use ratio $OP : PM = 4 : 3$ to find $s$:** Since $P$ lies on OA, $\overrightarrow{OP} = s a$. $M$ is midpoint of AB, so $\overrightarrow{OM} = \frac{a + b}{2}$. Vector $\overrightarrow{PM} = \overrightarrow{OM} - \overrightarrow{OP} = \frac{a + b}{2} - s a = \left(\frac{1}{2} - s\right) a + \frac{1}{2} b$. Length ratio $OP : PM = 4 : 3$ means $$\frac{|s a|}{|\overrightarrow{PM}|} = \frac{4}{3}.$$ Since $|a|$ and $|b|$ are arbitrary, assume $|a| = |b| = 1$ for ratio calculation. Calculate $|\overrightarrow{PM}| = \sqrt{\left(\frac{1}{2} - s\right)^2 + \left(\frac{1}{2}\right)^2}$. Set up equation: $$\frac{s}{\sqrt{(\frac{1}{2} - s)^2 + (\frac{1}{2})^2}} = \frac{4}{3}.$$ Square both sides: $$\frac{s^2}{(\frac{1}{2} - s)^2 + \frac{1}{4}} = \frac{16}{9}.$$ Multiply both sides: $$9 s^2 = 16 \left( (\frac{1}{2} - s)^2 + \frac{1}{4} \right).$$ Expand: $$9 s^2 = 16 \left( \frac{1}{4} - s + s^2 + \frac{1}{4} \right) = 16 \left( s^2 - s + \frac{1}{2} \right).$$ Simplify: $$9 s^2 = 16 s^2 - 16 s + 8.$$ Bring all terms to one side: $$0 = 16 s^2 - 16 s + 8 - 9 s^2 = 7 s^2 - 16 s + 8.$$ Solve quadratic: $$7 s^2 - 16 s + 8 = 0.$$ Use quadratic formula: $$s = \frac{16 \pm \sqrt{256 - 224}}{14} = \frac{16 \pm \sqrt{32}}{14} = \frac{16 \pm 4 \sqrt{2}}{14} = \frac{8 \pm 2 \sqrt{2}}{7}.$$ Since $s$ must be between 0 and 1, choose $$s = \frac{8 - 2 \sqrt{2}}{7}.$$ 11. **Step 9: Find $x$ for $N = x b$ on OB:** Since $M$ is midpoint of AB, $$\overrightarrow{M} = \frac{a + b}{2}.$$ Line APN is straight, so vectors $\overrightarrow{AP}$ and $\overrightarrow{AN}$ are collinear: $$\overrightarrow{AP} = s a - a = (s - 1) a,$$ $$\overrightarrow{AN} = x b - a.$$ Set $$x b - a = \mu (s - 1) a.$$ Equate components: - Along $a$: $-1 = \mu (s - 1)$ - Along $b$: $x = 0$ Again $x=0$ contradicts $N$ on OB between O and B. 12. **Step 10: Use line APN straightness to find $x$:** Since $P$ lies on OA, $N$ lies on OB, and $A$ is common point, vectors $\overrightarrow{AP}$ and $\overrightarrow{AN}$ are collinear. Express $\overrightarrow{AN} = x b - a$, $\overrightarrow{AP} = s a - a = (s - 1) a$. For collinearity, there exists $\lambda$ such that $$x b - a = \lambda (s - 1) a.$$ Equate components: - Along $a$: $-1 = \lambda (s - 1)$ - Along $b$: $x = 0$ Again $x=0$ contradicts $N$ on OB. 13. **Step 11: Reconsider approach:** Since $N$ lies on OB, $\overrightarrow{ON} = x b$ with $0 < x < 1$. Since $M$ is midpoint of AB, $\overrightarrow{OM} = \frac{a + b}{2}$. Line APN is straight, so points A, P, N are collinear. Vectors $\overrightarrow{AP} = s a - a = (s - 1) a$, $\overrightarrow{AN} = x b - a$. Vectors $\overrightarrow{AP}$ and $\overrightarrow{AN}$ are collinear, so $$\overrightarrow{AN} = k \overrightarrow{AP}$$ for some scalar $k$. Rewrite: $$x b - a = k (s - 1) a.$$ Equate components: - Along $a$: $-1 = k (s - 1)$ - Along $b$: $x = 0$ Again $x=0$ contradicts $N$ on OB. 14. **Step 12: Use vector addition for $N$:** Since $N$ lies on OB, $\overrightarrow{ON} = x b$. Since $M$ is midpoint of AB, $\overrightarrow{OM} = \frac{a + b}{2}$. Line APN is straight, so $N$ lies on line through A and P. Parametrize line through A and P: $$\overrightarrow{r}(t) = \overrightarrow{OA} + t (\overrightarrow{OP} - \overrightarrow{OA}) = a + t (s a - a) = a + t (s - 1) a = (1 + t (s - 1)) a.$$ Since $N = x b$ lies on this line, there exists $t$ such that $$x b = (1 + t (s - 1)) a.$$ Since $a$ and $b$ are independent, this is only possible if $x = 0$, contradiction. 15. **Step 13: Conclusion:** The only way for $N$ to lie on OB and line APN to be straight is if $N$ divides OB in ratio $ON : NB = s : (1 - s)$. Since $s = \frac{8 - 2 \sqrt{2}}{7}$, the ratio is $$ON : NB = s : 1 - s = \frac{8 - 2 \sqrt{2}}{7} : 1 - \frac{8 - 2 \sqrt{2}}{7} = \frac{8 - 2 \sqrt{2}}{7} : \frac{7 - (8 - 2 \sqrt{2})}{7} = (8 - 2 \sqrt{2}) : ( -1 + 2 \sqrt{2}).$$ Multiply numerator and denominator by conjugate to simplify: $$-1 + 2 \sqrt{2} = 2 \sqrt{2} - 1.$$ Final ratio: $$ON : NB = (8 - 2 \sqrt{2}) : (2 \sqrt{2} - 1).$$