enmobladatn

2022-07-23

Greatest distance one point can have from a vertice of a square given following conditions

A point P lies in the same plane as a given square of side 1.Let the vertices of the square,taken counterclockwise,be A,B,C, and D.Also,let the distances from P to A,B, and C, respectively, be u,v and w.

What is the greatest distance that P can be from D if ${u}^{2}+{v}^{2}={w}^{2}$?

Some thoughts I had:

1) Given a pair of vertices I could construct an ellipse with P as a point on the ellipse.

2) From the equality ${u}^{2}+{v}^{2}={w}^{2}$. I think that I have to consider the case where the angle between u and v is ${90}^{\circ}$. In this case I would have $w=1$ and $PD<2$.

That being said,I still fail to come at a concrete solution of the problem,it might be that none of my thoughts are right...

A point P lies in the same plane as a given square of side 1.Let the vertices of the square,taken counterclockwise,be A,B,C, and D.Also,let the distances from P to A,B, and C, respectively, be u,v and w.

What is the greatest distance that P can be from D if ${u}^{2}+{v}^{2}={w}^{2}$?

Some thoughts I had:

1) Given a pair of vertices I could construct an ellipse with P as a point on the ellipse.

2) From the equality ${u}^{2}+{v}^{2}={w}^{2}$. I think that I have to consider the case where the angle between u and v is ${90}^{\circ}$. In this case I would have $w=1$ and $PD<2$.

That being said,I still fail to come at a concrete solution of the problem,it might be that none of my thoughts are right...

Rihanna Robles

Beginner2022-07-24Added 18 answers

Step 1

We may suppose that $A(1,0),B(1,1),C(0,1),D(0,0),P(x,y).$

Then, we have ${u}^{2}=(x-1{)}^{2}+{y}^{2}$

${v}^{2}=(x-1{)}^{2}+(y-1{)}^{2}$

${w}^{2}={x}^{2}+(y-1{)}^{2}$

Step 2

So, we have ${u}^{2}+{v}^{2}={w}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}2(x-1{)}^{2}+{y}^{2}+(y-1{)}^{2}={x}^{2}+(y-1{)}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-4x+2+{y}^{2}=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(x-2{)}^{2}+{y}^{2}=2$

Hence, we want to find the greatest distance from the origin to a point on a circle $(x-2{)}^{2}+{y}^{2}=2$.

Thus, the answer is ${2+\sqrt{2}}$ when $P(2+\sqrt{2},0)$

We may suppose that $A(1,0),B(1,1),C(0,1),D(0,0),P(x,y).$

Then, we have ${u}^{2}=(x-1{)}^{2}+{y}^{2}$

${v}^{2}=(x-1{)}^{2}+(y-1{)}^{2}$

${w}^{2}={x}^{2}+(y-1{)}^{2}$

Step 2

So, we have ${u}^{2}+{v}^{2}={w}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}2(x-1{)}^{2}+{y}^{2}+(y-1{)}^{2}={x}^{2}+(y-1{)}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-4x+2+{y}^{2}=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(x-2{)}^{2}+{y}^{2}=2$

Hence, we want to find the greatest distance from the origin to a point on a circle $(x-2{)}^{2}+{y}^{2}=2$.

Thus, the answer is ${2+\sqrt{2}}$ when $P(2+\sqrt{2},0)$

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