Got stuck at a point -3

While solving the four points on a square Bonus problem I was stuck at a last point (if someone solves this, then I can make a set of steps as an answer to the problem) please try to help.

Consider the following diagram (0<PQR<180°0<\angle PQR < 180\degree) : Let a process P1P_1 on a point AA that is on semicircle QRQR be defined as follows :

  • Join AP\overline{AP} and AQ\overline{AQ}

  • Extend AQ\overline{AQ} till it intersects the semicircle RQRQ at a point B

  • Join RB\overline{RB} and extend

  • Draw the angular bisector of PAQ\angle PAQ and let it intersect extended RB\overline{RB} at point AA'

  • Let AA' be called as the output of the process

An illustration of process P1P_1 on a point AA on the semicircle QRQR (red\red{red} line is the angular bisector) :

Now imagine that you does this process on every point of the semicircle QPQP and marks each and every output, what will be the shape made by those outputs? (A circle, A parabola, A square...)


Note:

  • Any point on the diameter is not counted as a point on the semicircle

  • If anyone gives the answer with a proof, I will immediately write a solution to the four point in square problem as I am just one solution far to the problem (that is this problem)

Note by Zakir Husain
4 months, 3 weeks ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Have a look at this, it might help :)

Jeff Giff - 4 months, 3 weeks ago

Log in to reply

Wow,how do you make it.. @Jeff Giff

Kriti Kamal - 4 months, 3 weeks ago

Log in to reply

Sketchbook, iPad. Designed for art, but works well in these situations :)

Jeff Giff - 4 months, 3 weeks ago

Log in to reply

I had also tried to make this and i also got an approximated circle, but I need a proof for that.

Zakir Husain - 4 months, 3 weeks ago

Log in to reply

Yes,math requires proof.I am also trying to prove.

Kriti Kamal - 4 months, 3 weeks ago

Log in to reply

Are QR and QP perpendicular to each other ?? Please reply @Zakir Husain

Kriti Kamal - 4 months, 3 weeks ago

Log in to reply

@Kriti Kamal No they are at an arbitrary angle

Zakir Husain - 4 months, 3 weeks ago

Log in to reply

@Zakir Husain Oooooooo ,

Kriti Kamal - 4 months, 3 weeks ago

Log in to reply

Hint: Complete the semicircles. Call the midpoint of arc PQPQ(the one that the angle bisector will pass through) MM, and the two completed circle intersect at another point QQ'. You can try to prove that R,A,Q,MR,A',Q',M are concyclic. (Since R,Q,MR,Q',M are fixed points, this leads to the locus of AA' being a circle. A simple angle chasing will prove this. Actually, this is just a special case of Miquel's theorem, or Pivot theorem, so the proof is the same.)

X X - 4 months, 3 weeks ago

Log in to reply

Hope this hint helps, and I'm looking forward to see your solution to the four point in square problem !

X X - 4 months, 3 weeks ago

Log in to reply

Yes,it helps.....

Kriti Kamal - 4 months, 3 weeks ago

Log in to reply

I have only 1 doubt, what is the angle between seg QP and seg QR, I know it has to be greater than 0 degrees and lesser than 180 degrees, otherwise getting point B is not possible on the other semicircle, unless the point A coincides with the endpoints of the diameter or is on the diameter, which you have restricted from doing so. Also not greater than 180, because the semicircles would overlap then which I assume is not allowed.

Siddharth Chakravarty - 4 months, 2 weeks ago

Log in to reply

0<PQR<π0<\angle PQR<\pi

Zakir Husain - 4 months, 2 weeks ago

Log in to reply

Yes that is what I meant, thank you! I will work on the proof now and see if I can do it.👍

Siddharth Chakravarty - 4 months, 2 weeks ago

Log in to reply

Log in to reply

@Kriti Kamal

Zakir Husain - 4 months, 3 weeks ago

Log in to reply

This one appeared in Terence Tao (yes he!!) 's book in late 90's.

Vishwash Kumar ΓΞΩ - 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...