A cool property regarding circumcenter

Let OO be the circumcenter of ABCABC. Reflect OO over BCBC to obtain OO'. Through OO' construct lines parallel to AC,ABAC,AB which respectively meet AB,ACAB,AC at F,EF,E. Define OFOB=Y,OECO=XO'F\cap OB=Y, O'E\cap CO=X. Prove XYEFXY||EF

I personally think this configuration is very rich and can be exploited to create difficult olympiad geo problems.

Note by Xuming Liang
4 years 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

Notice that BOCOBOCO' is a parallelogram. Since BPOCBP || O'C and ABOEAB || O'E, we see that FBX=EOC\angle FBX = \angle EO'C. It is easy to see that A=BFX=YEC\angle A = \angle BFX = \angle YEC. Therefore, BFXOEC\triangle BFX \sim \triangle O'EC. This implies that BFFX=OEEC\frac{BF}{FX} = \frac{O'E}{EC} .

Similarly, we have that YCEBOF    OFFB=CEEY\triangle YCE \sim \triangle BO'F \implies \frac{O'F}{FB} = \frac{CE}{EY}.

Multiplying the two ratios completes the proof.

Alan Yan - 4 years ago

Log in to reply

Yes! The parallel property will hold as long as BOCOBOCO' is a parallelogram. There are a couple of typos in your proof, otherwise you got it spot on.

Xuming Liang - 4 years ago

Log in to reply

PS: The problem can be generalized.

Xuming Liang - 4 years ago

Log in to reply

Can you post the proof or hint after a few days?

Alan Yan - 4 years ago

Log in to reply

Yes. :)

Xuming Liang - 4 years ago

Log in to reply

@Alan Yan Hints: We want to prove some two ratios are equal, perhaps look for some similar triangles from all the parallels.

Generalize this property if you get it.

Xuming Liang - 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...