1. **Problem statement:** Given parallelogram ABCD with $DA=4$, $DC=5$, and angle $\angle ADC=60^\circ$. (a) Prove that $\vec{AB} + \vec{AC} = 2\vec{AC}$. (b) Calculate the length of the vector $\vec{AB} + \vec{AC} + \vec{BC}$. (c) Point $M$ lies on diagonal $AC$, distinct from $A$ and $C$. Points $P$ and $Q$ lie on $AB$ and $BC$ respectively such that $MP \parallel BC$ and $MQ \parallel AB$. Let $N$ be the intersection of $AQ$ and $CP$. Given $\vec{DN} = m\vec{DA} + n\vec{DC}$, find the maximum value of $m+n$. --- 2. **Recall vector properties in a parallelogram:** - Opposite sides are equal and parallel: $\vec{AB} = \vec{DC}$, $\vec{AD} = \vec{BC}$. - Diagonals: $\vec{AC} = \vec{AB} + \vec{BC}$. --- 3. **Part (a) Proof:** We want to prove $\vec{AB} + \vec{AC} = 2\vec{AC}$. Note that $\vec{AC} = \vec{AB} + \vec{BC}$. So, $$\vec{AB} + \vec{AC} = \vec{AB} + (\vec{AB} + \vec{BC}) = 2\vec{AB} + \vec{BC}.$$ But since $\vec{BC} = \vec{AD}$ and $\vec{AB} = \vec{DC}$, this expression is not equal to $2\vec{AC}$ unless $\vec{AB} = \vec{AC}$, which is generally false. **Re-examining the problem statement:** It likely means to prove $\vec{AB} + \vec{AC} = 2\vec{AD}$ or a similar relation. Since the problem states $\vec{AB} + \vec{AC} = 2\vec{AC}$, this is only true if $\vec{AB} = \vec{AC}$, which is not true in a parallelogram. Assuming a typo and the intended statement is $\vec{AB} + \vec{AC} = 2\vec{AC}$, this is false. Instead, we can prove $\vec{AB} + \vec{AC} = 2\vec{AC}$ if $\vec{AB} = \vec{AC}$, which is not the case. Hence, the correct identity is likely $\vec{AB} + \vec{AC} = 2\vec{AC}$ is false. If the problem meant $\vec{AB} + \vec{AC} = 2\vec{AD}$, then: Since $\vec{AC} = \vec{AB} + \vec{BC}$, then $$\vec{AB} + \vec{AC} = \vec{AB} + \vec{AB} + \vec{BC} = 2\vec{AB} + \vec{BC}.$$ This is not equal to $2\vec{AD}$ either. **Therefore, we proceed to part (b) and (c) as given.** --- 4. **Part (b) Calculate $|\vec{AB} + \vec{AC} + \vec{BC}|$** Recall $\vec{AC} = \vec{AB} + \vec{BC}$. So, $$\vec{AB} + \vec{AC} + \vec{BC} = \vec{AB} + (\vec{AB} + \vec{BC}) + \vec{BC} = 2\vec{AB} + 2\vec{BC} = 2(\vec{AB} + \vec{BC}) = 2\vec{AC}.$$ Therefore, $$|\vec{AB} + \vec{AC} + \vec{BC}| = |2\vec{AC}| = 2|\vec{AC}|.$$ We need to find $|\vec{AC}|$. --- 5. **Calculate $|\vec{AC}|$ using the law of cosines:** In triangle $ADC$, sides: - $DA = 4$ - $DC = 5$ - $\angle ADC = 60^\circ$ Using law of cosines for $AC$: $$|\vec{AC}|^2 = DA^2 + DC^2 - 2 \cdot DA \cdot DC \cdot \cos 60^\circ = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cdot \frac{1}{2} = 16 + 25 - 20 = 21.$$ So, $$|\vec{AC}| = \sqrt{21}.$$ Hence, $$|\vec{AB} + \vec{AC} + \vec{BC}| = 2\sqrt{21}.$$ --- 6. **Part (c) Find max value of $m+n$ where $\vec{DN} = m\vec{DA} + n\vec{DC}$** - Let $M$ be on $AC$, so $\vec{AM} = t\vec{AC}$ with $t \in (0,1)$. - Points $P$ on $AB$, $Q$ on $BC$ such that $MP \parallel BC$ and $MQ \parallel AB$. Express vectors: - $\vec{AB} = \vec{b}$ - $\vec{AD} = \vec{a}$ - $\vec{DC} = \vec{b}$ (since $AB \parallel DC$ and equal length) - $\vec{AC} = \vec{a} + \vec{b}$ Coordinates: - $A = \vec{0}$ - $B = \vec{b}$ - $D = \vec{a}$ - $C = \vec{a} + \vec{b}$ Then, - $M = t(\vec{a} + \vec{b})$ Since $P$ lies on $AB$, $P = s\vec{b}$ for some $s \in [0,1]$. Since $Q$ lies on $BC$, $Q = \vec{b} + r\vec{a}$ for some $r \in [0,1]$. Conditions: - $MP \parallel BC \Rightarrow \vec{MP} = \lambda \vec{a}$ (since $BC = \vec{a}$) - $MQ \parallel AB \Rightarrow \vec{MQ} = \mu \vec{b}$ Calculate: $$\vec{MP} = P - M = s\vec{b} - t(\vec{a} + \vec{b}) = -t\vec{a} + (s - t)\vec{b}.$$ For $MP \parallel BC = \vec{a}$, the $\vec{b}$ component must be zero: $$s - t = 0 \Rightarrow s = t.$$ Then, $$\vec{MP} = -t\vec{a}.$$ Similarly, $$\vec{MQ} = Q - M = (\vec{b} + r\vec{a}) - t(\vec{a} + \vec{b}) = (1 - t)\vec{b} + (r - t)\vec{a}.$$ For $MQ \parallel AB = \vec{b}$, the $\vec{a}$ component must be zero: $$r - t = 0 \Rightarrow r = t.$$ So, - $P = t\vec{b}$ - $Q = \vec{b} + t\vec{a}$ --- 7. **Find $N = AQ \cap CP$** Parametric forms: - $AQ$: $\vec{A} + \alpha (Q - A) = \alpha (\vec{b} + t\vec{a})$ - $CP$: $\vec{C} + \beta (P - C) = (\vec{a} + \vec{b}) + \beta (t\vec{b} - (\vec{a} + \vec{b})) = (\vec{a} + \vec{b}) + \beta (-\vec{a} + (t-1)\vec{b})$ Set equal: $$\alpha (\vec{b} + t\vec{a}) = (\vec{a} + \vec{b}) + \beta (-\vec{a} + (t-1)\vec{b}).$$ Equate components: - $\vec{a}$: $\alpha t = 1 - \beta$ - $\vec{b}$: $\alpha = 1 + \beta (t - 1)$ From first: $$\beta = 1 - \alpha t.$$ Substitute into second: $$\alpha = 1 + (1 - \alpha t)(t - 1) = 1 + (t - 1) - \alpha t (t - 1) = t - \alpha t (t - 1).$$ Bring terms with $\alpha$ to one side: $$\alpha + \alpha t (t - 1) = t$$ $$\alpha (1 + t (t - 1)) = t$$ Calculate: $$1 + t (t - 1) = 1 + t^2 - t = t^2 - t + 1.$$ So, $$\alpha = \frac{t}{t^2 - t + 1}.$$ Then, $$\beta = 1 - \alpha t = 1 - \frac{t^2}{t^2 - t + 1} = \frac{t - t^2 + 1 - t^2}{t^2 - t + 1} = \frac{1 - t^2}{t^2 - t + 1}.$$ --- 8. **Express $\vec{DN}$ in terms of $\vec{DA} = \vec{a}$ and $\vec{DC} = \vec{b}$** Recall: $$\vec{D} = \vec{a}, \quad \vec{N} = \vec{A} + \alpha (\vec{b} + t\vec{a}) = \alpha \vec{b} + \alpha t \vec{a}.$$ So, $$\vec{DN} = \vec{N} - \vec{D} = \alpha t \vec{a} + \alpha \vec{b} - \vec{a} = (\alpha t - 1) \vec{a} + \alpha \vec{b}.$$ Therefore, $$m = \alpha t - 1, \quad n = \alpha.$$ Sum: $$m + n = (\alpha t - 1) + \alpha = \alpha (t + 1) - 1 = \frac{t}{t^2 - t + 1} (t + 1) - 1.$$ Simplify numerator: $$t(t + 1) = t^2 + t.$$ So, $$m + n = \frac{t^2 + t}{t^2 - t + 1} - 1 = \frac{t^2 + t - (t^2 - t + 1)}{t^2 - t + 1} = \frac{t^2 + t - t^2 + t - 1}{t^2 - t + 1} = \frac{2t - 1}{t^2 - t + 1}.$$ --- 9. **Find maximum of** $$f(t) = \frac{2t - 1}{t^2 - t + 1}, \quad t \in (0,1).$$ Denominator is always positive since $$t^2 - t + 1 = (t - \frac{1}{2})^2 + \frac{3}{4} > 0.$$ Find critical points by setting derivative to zero. Derivative: $$f'(t) = \frac{(2)(t^2 - t + 1) - (2t - 1)(2t - 1)}{(t^2 - t + 1)^2}.$$ Calculate numerator: $$2(t^2 - t + 1) - (2t - 1)^2 = 2t^2 - 2t + 2 - (4t^2 - 4t + 1) = 2t^2 - 2t + 2 - 4t^2 + 4t - 1 = -2t^2 + 2t + 1.$$ Set numerator zero: $$-2t^2 + 2t + 1 = 0 \Rightarrow 2t^2 - 2t - 1 = 0.$$ Solve quadratic: $$t = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}.$$ Only $t = \frac{1 + \sqrt{3}}{2} > 1$ is outside domain. $t = \frac{1 - \sqrt{3}}{2} < 0$ is outside domain. No critical points in $(0,1)$. Check endpoints: - At $t \to 0^+$: $$f(0) = \frac{2\cdot0 - 1}{0 - 0 + 1} = -1.$$ - At $t = 1$: $$f(1) = \frac{2\cdot1 - 1}{1 - 1 + 1} = \frac{1}{1} = 1.$$ Since $f(t)$ is continuous and no critical points inside $(0,1)$, maximum is at $t=1$. But $M$ cannot be $C$ ($t=1$), so consider limit approaching 1 from below: $$\lim_{t \to 1^-} f(t) = 1.$$ Hence, the maximum value of $m+n$ is $1$. --- **Final answers:** (a) The given identity $\vec{AB} + \vec{AC} = 2\vec{AC}$ is generally false for a parallelogram. (b) $|\vec{AB} + \vec{AC} + \vec{BC}| = 2\sqrt{21}$. (c) The maximum value of $m + n$ is $1$.