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# math help

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Consider parallelogram ABCD with points S and T chosen such that CS:SD = BT:TC = 2, as in the picture.

Let and overrightarrow{AB} = v and overrightarrow{AD} = w. Then there exist constants r, s, t, u such that   overrightarrow{AT} = r v + s w, overrightarrow{BS} = t v + u w.

Also, what is $$\frac{AT^2+BS^2}{AC^2+BD^2}$$equal to?

Aug 4, 2019

#1
+24389
+2

math help

Consider parallelogram ABCD with points S and T chosen such that $$CS:SD = BT:TC = 2:1$$, as in the picture.
Let $$\vec{AB} = v$$ and $$\vec{AD} = w$$.

1.
Then there exist constants r, s, t, u such that

$$\vec{AT} = r v + s w$$,
$$\vec{BS} = t v + u w$$.

$$\text{Let \vec{AB} = \vec{DC} } \\ \text{Let \vec{AD} = \vec{BC} }$$

$$\begin{array}{|rcll|} \hline \vec{AT} &=& v +\dfrac{2}{3}w \quad | \quad \vec{AT} = r v + s w \\ \mathbf{r} &=& \mathbf{1} \\ \mathbf{s} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{BS} &=& w -\dfrac{2}{3}v \quad | \quad \vec{BS} = t v + u w \\ \mathbf{t} &=& \mathbf{-\dfrac{2}{3}} \\ \mathbf{u} &=& \mathbf{1} \\ \hline \end{array}$$

Aug 4, 2019
#2
+24389
+1

math help

Consider parallelogram ABCD with points S and T chosen such that CS:SD = BT:TC = 2:1, as in the picture.
Let$$\vec{AB} = v$$ and $$\vec{AD} = w$$.
Then there exist constants r, s, t, u such that
$$\vec{AT} = r v + s w$$,
$$\vec{BS} = t v + u w$$.

2.

Also, what is$$\mathbf{\dfrac{AT^2+BS^2}{AC^2+BD^2}}$$ equal to?

$$\begin{array}{|rcll|} \hline \left(\vec{AT}\right)^2 &=& \left(v +\dfrac{2}{3}w \right)^2 \\ \mathbf{\left(\vec{AT}\right)^2} &=& \mathbf{v^2+2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}w^2 } \\\\ \left(\vec{BS}\right)^2 &=& \left(w -\dfrac{2}{3}v \right)^2 \\ \mathbf{\left(\vec{BS}\right)^2} &=& \mathbf{w^2 - 2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}v^2 } \\\\ \left(\vec{AC}\right)^2 &=& \left(v+w \right)^2 \quad | \quad \vec{AC}=v+w \\ \mathbf{\left(\vec{AC}\right)^2} &=& \mathbf{v^2 + 2vw + w^2 } \\\\ \left(\vec{BD}\right)^2 &=& \left(w-v \right)^2 \quad | \quad \vec{BD}=w-v \\ \mathbf{\left(\vec{BD}\right)^2} &=& \mathbf{w^2 - 2vw + v^2 } \\ \hline \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2+2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}w^2+w^2 - 2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}v^2} {v^2 + 2vw + w^2+w^2 - 2vw + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2+\dfrac{4}{3}vw +\dfrac{4}{9}w^2+w^2 - \dfrac{4}{3}vw +\dfrac{4}{9}v^2} {v^2 + w^2+w^2 + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2 +\dfrac{4}{9}w^2+w^2 +\dfrac{4}{9}v^2} {v^2 + w^2+w^2 + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2 +\dfrac{4}{9}v^2+w^2 +\dfrac{4}{9}w^2} {2(v^2 + w^2)} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{\dfrac{13}{9}(v^2+w^2)} {2(v^2 + w^2)} \\\\ \mathbf{\dfrac{AT^2+BS^2}{AC^2+BD^2}} &=& \mathbf{\dfrac{13}{18}} \\ \hline \end{array}$$

Aug 4, 2019