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if $\bigodot P\bigcap \bigodot Q=A,B$,and the common tangent is $C,D$,and $E\in BA$,and $EC\bigcap \bigodot P=F,ED\bigcap \bigodot Q=G$,and if $\angle FAH=\angle HAG$

show that $$\angle FCH=\angle GDH$$enter image description here

it seem hard, I can't get this answer

For Weijie Chen answer,then I have add a fig,let we clear understand enter image description here

math110
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1 Answers1

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enter image description hereIt's not that difficult but it took me 1h.

Here's my solution (Sadly I don't even found the contest at AoPS):

Let $M=CD\cap FG$, $X=MA\cap \bigodot P$ and $Y=MA\cap \bigodot Q$ different from A.

Observe that if $\bigodot (CHD)$ is tangent to $FG$ we whould finish (angle chasing).

Notice that the tangent from $M$ to any circle that passes through $CD$ has the same lenght. Because it is the power from the point $M$. And nos I claim that the lenght is $MA$.

Proof: $MC^2=MX\cdot MA$ and $MD^2=MA\cdot MY$ it's easy to show that $XCDY$ is cyclic so we have $MX\cdot MY=MC\cdot MD$ hence $MC\cdot MD=MA^2$

So if we proof that $MA=MH$ we would finish. That is easy since $\bigodot(FAG)$ is tangent to $MA$ because $CDFG$ is cyclic (quite obvious) hence $MC\cdot MD=MA^2=MF\cdot MG$. This means that $\angle FAM=\angle AGF$ now by angle chasing we can show that $\angle MHA=\angle MAH$ hence $MH=MA$ and done.

If there's anything that is unclear please let me know.

Weijie Chen
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  • My geometry is a bit rusty. Can you please explain how to show that $XCDY$ and $CDFG$ are cyclic? – Batominovski Jul 25 '16 at 12:24
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    @Batominovski $CDFG$ is cyclic because $AB$ is the radical axis of $\bigodot P$ and $\bigodot G$ so we have that $EC\cdot EF=EA\cdot EB=ED\cdot EG$. $XCDY$ is cyclic because $M$ is the center of homethety that carries $\bigodot P$ to $\bigodot G$ now with angle chasing you have that $XCDY$ is cyclic. I hope this can clear your doubt. – Weijie Chen Jul 25 '16 at 12:41
  • @WeijieChen,I have add the answer with fig,can you explain why is $FG$ tangent $CHD$,then finsh it? – math110 Jul 25 '16 at 14:45
  • @functionsug Not sure if your figure is correct. It looks like $H$ in your second figure is the intersection of $AB$ and $FG$. (And I thank Weijie Chen---everything is clear now.) – Batominovski Jul 25 '16 at 14:56
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    @functionsug By angle chasing if $FG$ is tangent to $\bigodot (CHD)$ then $\angle CHF=\angle CDH$ and $\angle DHG=\angle DCH$. Since $CDFG$ is cyclic $\angle EFG=\angle EDC$ and $\angle EGF=\angle ECD$. Now $\angle HCF=180^o-\angle HCE=180^o-(\angle HCD+\angle DCE)=180^o-(\angle CHF+\angle CFH)$ So: $$\angle HCD+\angle DCE=\angle CHF+\angle CFH$$ and by the ther hand we have with the same reasoning: $$\angle HDC+\angle CDE=\angle DHG+\angle DGH$$ Now computing you get that $\angle ECH=\angle EDH$. If you have any other question just ask me. – Weijie Chen Jul 25 '16 at 14:57
  • Thanks,you can post add your anwers,or can you add your fig?Thanks – math110 Jul 25 '16 at 14:59
  • @functionsug Ok no problem – Weijie Chen Jul 25 '16 at 15:00
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    @functionsug As you wanted I added a Figure, but since the answeres from the questions are here on the comment seccion I won't add now (Probably later) since it is a lot of work and I don't have enough time – Weijie Chen Jul 25 '16 at 15:11