Maths question help (with worked solutions) (1 Viewer)

xdxqx · Feb 21, 2021

Hi, I came across this question and I don’t understand why BX is parallel to VY and VB is parallel to XY, can someone please answer this? Thanks!

Pikapizza · Feb 21, 2021

From the question diagram and the solutions diagram, it seems that they have swapped position of A and B which might explain the confusion

CM_Tutor · Feb 21, 2021

Let $\overrightarrow{AU}=\overrightarrow{a} \implies \overrightarrow{AV}=2\overrightarrow{a} \quad \text{and} \quad \overrightarrow{AB}=3\overrightarrow{a}$

And, let $\overrightarrow{AZ}=\overrightarrow{b} \implies \overrightarrow{AY}=2\overrightarrow{b} \quad \text{and} \quad \overrightarrow{AC}=3\overrightarrow{b}$

So, $\overrightarrow{UZ}=\overrightarrow{UA} + \overrightarrow{AZ} = \overrightarrow{AZ} - \overrightarrow{AU} = \overrightarrow{b} - \overrightarrow{a}$

And, $\overrightarrow{VY}=\overrightarrow{VA} + \overrightarrow{AY} =\overrightarrow{AY} - \overrightarrow{AV} = 2\overrightarrow{b} - 2\overrightarrow{a} = 2\left(\overrightarrow{b} - \overrightarrow{a}\right) = 2\overrightarrow{UZ}$

And, $\overrightarrow{BC}=\overrightarrow{BA} + \overrightarrow{AC} =\overrightarrow{AC} - \overrightarrow{AB} = 3\overrightarrow{b} - 3\overrightarrow{a} = 3\left(\overrightarrow{b} - \overrightarrow{a}\right) = 3\overrightarrow{UZ}$

Given these three vectors are scalar multiples, they must be parallel to each other... $\left|\overrightarrow{UZ}\right| = \left|\overrightarrow{VY}\right| = \left|\overrightarrow{BC}\right|$

And similarly $\left|\overrightarrow{VW}\right| = \left|\overrightarrow{UX}\right| = \left|\overrightarrow{AC}\right|$ and $\left|\overrightarrow{YX}\right| = \left|\overrightarrow{ZW}\right| = \left|\overrightarrow{AB}\right|$ .

Each set of parallel lines produces a triad of similar triangles, $\Delta CXY ||| \Delta CWX ||| \Delta CBA$ , as they are equiangular, and with the sides in ratio 1:2:3, the areas are in ratio $1^2:2^2:3^2=1:4:9$ , and so if the area of $\Delta ABC = A$ then

$\frac{\text{Area } \Delta CXY}{\text{Area } \Delta ABC}=\frac{1}{9} \quad \implies \quad \text{Area } \Delta CXY=\frac{A}{9}$

The two areas marked $A_2$ are equal as they divide the parallelogram $BVYZ$ into two congruent triangles, $\Delta BVX \equiv \Delta YXV$ (AAS).

The two areas marked as $A_1$ are

$\frac{1}{2}\times\left|\overrightarrow{ZA}\right|\times\left|\overrightarrow{ZV}\right|\times\sin{\angle AZV}$

and

$\frac{1}{2}\times\left|\overrightarrow{ZV}\right|\times\left|\overrightarrow{ZY}\right|\times\sin{\angle VZY}$

These are equal as $\left|\overrightarrow{ZV}\right|$ is common, and

$\left|\overrightarrow{ZA}\right| = \left|\overrightarrow{ZY}\right| = \left|\overrightarrow{b}\right|$

and

$\angle AZV + \angle VZY = 180^o \quad \implies \quad \sin{\angle AZV} = \sin{\angle VZY}$ .

So, the total area of $\Delta ABC$ is

$\begin{align*} A &= \frac{A}{9} + A_1 \times 2 + A_2 \times 2 \\ \frac{8A}{9} &= 2\left(A_1 + A_2\right) \\ A_1 + A_2 &= \frac{4A}{9} \end{align*}$

So, the result we seek is...

$\frac{\text{Area} ZVXY}{\text{Area} ABC} = \frac{A_1 + A_2}{A} = \frac{\frac{4A}{9}}{A} \times \frac{9}{9} = \frac{4A}{9A} = \frac{4}{9}$

CM_Tutor · Feb 21, 2021

Pikapizza said:
From the question diagram and the solutions diagram, it seems that they have swapped position of A and B which might explain the confusion

FYI, my solution uses the positions of $A$ and $B$ in the solution diagram, but could be easily adapted to have the original form. Good catch @Pikapizza on noting the mistake in the given solution, though.

Maths question help (with worked solutions) (1 Viewer)

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