How do I find the modulus of a sum of vectors which is part of a triangle embedded in a quarter a circle?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

In the figure from below, calculate the modulus of $\vec{x}+\vec{y}$. $P$ is tangential point. Show the answer in terms of $R$.



The alternatives are as follows:

$\begin{array}{ll}
1.&1R\\
2.&0.41R\\
3.&0.59R\\
4.&1.41R\\
5.&2.12R\\
\end{array}$

The only thing which I was able to spot here was to establish that

$x=\frac{(R+a)\sqrt{2}}{2}+a$

$y=\frac{(R+a)\sqrt{2}}{2}+a$

But this doesn't seem very convincing to me. How exactly can I use the vector decomposition in this set of vectors?
 

romsek

Math Team
Sep 2015
2,875
1,608
USA
what is $a$ ??

$P=R\left(\dfrac{1}{\sqrt{2}},~\dfrac{1}{\sqrt{2}}\right)$

$\vec{x} = P-(0,R) $
$\vec{y} = P-(R,0)$

you should be able to find $\|\vec{x}+\vec{y}\|$ given this.

It is one of the listed choices.
 
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skipjack

Forum Staff
Dec 2006
21,379
2,410
The given angle implies that the diagram is symmetrical, so the coordinates of P can be found easily.

The required modulus turns out to be $(2 - √2)R$.
 
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Jun 2017
337
6
Lima, Peru
$|x+y| = R(2-\sqrt{2})$

View attachment 10816
How did you get to $135^{\circ}$ on $P$? I'm still confused on what sort of relation does it exist between $R$ and $\vec{y}$? I'm lost at geometry. Can you add the justification for the $R$? Please, I'm still stuck on that part.

The only thing which it comes to my mind is that a ray between the center of the quarter of the circle connects it to $P$ and this ray divides the right angle in two equal halves $45^{\circ}$ and $45^{\circ}$, but why? What's the geometrical justification for this?. Given this condition, I'm assuming that this also repeats for $\vec{x}$, which makes it the same size (or norm) for $\vec{y}$.

In any case I'm getting:

$\left\|\vec{x}\right\|=\left\|\vec{y}\right\|=R\sqrt{2-\sqrt{2}}$

Then repeating the cosines law I'm getting:

$\left\|x+y\right\|^2=\left(R\sqrt{2-\sqrt{2}}\right)^2+\left(R\sqrt{2-\sqrt{2}}\right)^2-2\left(R\sqrt{2-\sqrt{2}}\right)^2\cos 45^{\circ}$

$\left\|x+y\right\|^2=2(R^2)(2-\sqrt{2})\left(1-\frac{\sqrt{2}}{2}\right)$

$\left\|x+y\right\|^2=R(2-\sqrt{2})$

This is the only method which I've found to get to the answer, but I'd like to know a geometrical justification for this. Please.
 

skipjack

Forum Staff
Dec 2006
21,379
2,410
The tangent through $P$ makes an angle of 45° with one radius, so it makes the same angle with the other radius (so that the sum of these angles and the angle of 90° between the radii is 180°). This implies the diagram is symmetrical about the radius through $P$. One can use this to obtain the angle at $P$.

However, that angle is 135° regardless of where $P$ lies on the quarter circle shown. To show this, draw the radius through $P$ and consider the angles in the two isosceles triangles formed. In an isosceles triangle, the angles opposite the equal length sides have equal magnitude. The sum of the remaining angles is given as 90°.

If you want a purely geometrical method of obtaining the final answer, consider the shorter part of the radius through $P$ in the diagram below.
Quadrant.PNG