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
3 years, 9 months ago

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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 - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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PS: The problem can be generalized.

Xuming Liang - 3 years, 9 months ago

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

Alan Yan - 3 years, 9 months ago

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Yes. :)

Xuming Liang - 3 years, 9 months ago

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@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 - 3 years, 9 months ago

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