libra7z's blog

A problem of mathematics
By libra7z, history, 5 years ago, In English

(Actually a problem of geometry)

you are given a circle $$$O$$$ whose radius is $$$2$$$.

And there are two fixed points $$$A$$$ and $$$B$$$ out of the circle. it is known that $$$OA = 4, OB = 4\sqrt 2$$$, and also $$$\angle AOB = 45^\circ$$$.

There's another moving point $$$P$$$ on circle $$$O$$$ (which means that $$$OP = 2$$$).

You are required to write the minimum value of $$$2\sqrt 2PA + PB$$$.

(I guess I may not put the figure here?)

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5 years ago, # |
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Auto comment: topic has been updated by libra7z (previous revision, new revision, compare).

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5 years ago, # |
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Another note is that the problem is the last problem in our school's monthly test for Junior 3 (so I guess it won't be very hard, even if I haven't work it out

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5 years ago, # |
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How is an angle between a circle and two points (your $$$\angle AOB$$$) defined?

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    5 years ago, # ^ |
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    A and B are two exterior point and O is the center actually that forms a right triangle(right angle at A) and other two angles as 45

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      5 years ago, # ^ |
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      And there we have the importance of proper terminology. Distance from a circle exists and is different from distance from its center.

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    5 years ago, # ^ |
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    here, 'O' refers to centre of the circle.

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5 years ago, # |
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Is the answer (32)^(1/2) + (20)^(1/2)?

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5 years ago, # |
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it should be 10.

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    5 years ago, # ^ |
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    An approach would be helpful, I am not able to figure out an answer. Thank you!

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5 years ago, # |
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I did it in the least elegant way possible (by making the equation of circle, finding the formula for that value and differentiating it) and the answer is 10. I hope someone has a much better and easier approach.

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5 years ago, # |
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I know you were probably expecting an elegant solution. But the best I could come up with was:

Let ∠AOP = x, therefore ∠BOP = x + 45

So using cosine rule,

AP = sqrt(OA^2 + OP^2 — 2.OA.OP.cos(x))

BP = sqrt(OB^2 + OP^2 — 2.OB.OP.cos(x+45))

Plugging in all necessary values, gives us a single variable function:

2.sqrt(2)PA+PB = 4.sqrt(10 — 8.cosx) + 2.sqrt(9 — 4.cosx + 4.sinx)

You can differentiate it with x and find the minima, or just plug in a grapher and check the minimum point.

Comes out to be 10 at x = -6.58480112 degrees

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    5 years ago, # ^ |
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    Let ∠AOP = x, therefore ∠BOP = x + 45.
    These is not a case for all time.
    Sometimes, it will be x, 45 — x.

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      5 years ago, # ^ |
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      Without loss of generality u can assume anticlockwise angle to be positive, -ve values will take care of 45-x. The answer he got is in turn negative, implying that it is in the opposite direction of what he assumed it to be in.

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