IJ = AH wanted, parallel , starting with northocenter of an acute triangle

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IJ = AH wanted, parallel , starting with northocenter of an acute triangle

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Source: Switzerland - Swiss MO 2016 p8

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#1 h Jul 15, 2020, 6:02 PM

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Let $ABC$ be an acute-angled triangle with height intersection $H$ . Let $G$ be the intersection of parallel of $AB$ through $H$ with the parallel of $AH$ through $B$ . Let $I$ be the point on the line $GH$ , so that $AC$ bisects segment $HI$ . Let $J$ be the second intersection of $AC$ and the circumcircle of the triangle $CGI$ . Show that $IJ = AH$

This post has been edited 1 time. Last edited by parmenides51, Jul 16, 2020, 8:31 PM
Reason: hugt typo, see #7, corrected perimeter with circumcircle

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#3 h Jul 16, 2020, 5:29 AM • 1 Y

Y by parmenides51

I'm confused about the problem, isn't the intersection $J$ of $AC$ with the triangle $CGI$ just the same as the intersection of $HI$ with $AC$ , which is the midpoint of $HI$ (since $AC$ bisects $HI$ )? Could you draw a diagram?

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#5 h Jul 16, 2020, 5:44 AM • 1 Y

Y by parmenides51

I drew circle centered at $A$ with radius $IJ$ and it doesn't pass through $H$ !

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#6 h Jul 16, 2020, 6:44 AM • 1 Y

Y by parmenides51

It is 2015 ISL G1.

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#7 h Jul 16, 2020, 8:22 PM

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Thank you trumpeter. Now everything became clear.

parmenides51 wrote:

Let $J$ be the second intersection of $AC$ and the perimeter of the triangle $CGI$ .

Meanwhile at IMO SL 2015 G1.

Problem_Penetrator wrote:

Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$ .

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#8 h Jul 16, 2020, 8:30 PM • 3 Y

Y by WolfusA, Mango247, Mango247

oops, google translate betrayed me,

, I corrected the perimeter instead of circumcircle in 2 more problems from the same source (2016 p4, 2018 p4 )

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