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This problem is so hard that I cannot figure it out. I hope you guys can give me a small push on how to tackle this problem, as I have been thinking about this for, like a week. Here's the problem:

Problem

Given a circle with center O, radius R (denoted as (O; R)), and a point A exterior to (O). From A, draw 2 separate tangent line segments AB, and AC to (O) (B, C are tangent points). Let D be the intersection of the line segment AO, and (O). Choose a point E arbitrarily on the small arc BC. Let F, G be the intersection points of DE, with AB, and AC respectively. Let I be the intersection of CF, and BG; H be the intersection of AO, and BC.

Prove that H, I, and E are collinear points.

Picture

Picture 1

Any help would be greatly appreciated.


Why can't I embed pictures in my post? :((

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  • $\begingroup$ Can we apply co-ordinate geometry? $\endgroup$
    – Tapu
    Feb 5, 2013 at 11:01
  • $\begingroup$ Yes, any way would be nice, but, personally, I think applying co-ordinate geometry can be a little bit hard, as there are so many free parimeters in this problem, like $R$, and the co-ordinate of $E$. $\endgroup$
    – user49685
    Feb 5, 2013 at 11:05
  • $\begingroup$ Choose $B=(0,1)$, $C=(0,,-1)$, and $E=(u,v)$; then everything is determined. $\endgroup$ Feb 7, 2013 at 10:25
  • $\begingroup$ I posted an answer below. It's not a push but a complete solution. I can't figure out which part of it would be enough of a push. Probably the first two paragraphs after the diagram. $\endgroup$
    – Dan Shved
    Feb 7, 2013 at 12:43
  • $\begingroup$ I found an pure geometry solution using Menelaus's theorem, harmonic quadrilateral, harmonic range, but it's a little bit complicated. If you still need it, you can tell me! $\endgroup$ Apr 23, 2016 at 17:09

2 Answers 2

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Let $X$ be the point where lines $BC$ and $FG$ meet. Let $J$ be the point where BC intersects $AE$. Let's forget about point $I$ for a moment. Here is the diagram: Diagram with additional points

Observe that $BD$ is the bisector of angle $HBA$, therefore $HB/BA=HD/DA$. Similarly, $HC/CA=HD/DA$. We see that points $B$, $C$ and $D$ all have the same ratio of distances from points $H$ and $A$. It follows that $(O)$ is an Apollonian circle of segment $HA$. Since $E$ is on this circle too, $HE/EA=HD/DA$. It follows that $ED$ is the bisector of angle $HEA$.

Now let us have a look at lines $EX$, $EB$, $EJ$, $EH$ and $EC$ and some of their cross-ratios. Let $f$ be the reflection across line $EX$. What we have established above is that $f(EJ)=EH$. It is also an easy exercise to see that $f(EB)=EC$ and $f(EX)=EX$. So, $f$ maps the four lines $EJ$, $EB$, $EC$ and $EX$ to lines $EH$, $EC$, $EB$ and $EX$ respectively. It follows that the corresponding cross-ratios are the same: $$ (EJ,EB;EC,EX) = (EH,EC;EB,EX). $$ Therefore, we have an equality for cross-ratios of points on line $BC$: $$ (J,B;C,X) = (H,C;B,X). $$

If we look at the perspective projection from line $BC$ to line $FG$ with center $A$, it sends points $J$, $B$, $C$ and $X$ to $E$, $F$, $G$ and $X$ respectively. Since perspective projections preserve cross-ratios, we have: $$ (E,F;G,X) = (H, C; B, X). $$

Now let us look at point $I$ (not shown on my diagram) where $BG$ and $FC$ intersect. Let $g$ be the perspective projection from line $BC$ to line $FG$ with center $I$. It is clear that $g(X)=X$, $g(B)=G$ and $g(C)=F$. $g$ preserves cross-ratios, therefore $$ (H, C; B, X) = (g(H), F; G, X), $$ and so $$ (E,F;G,X) = (g(H), F; G, X). $$ It follows that $g(H)=E$, which means that $H$, $E$ and $I$ are collinear, QED.

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  • $\begingroup$ Hi, thank you very much for your help. I understand the first paragraph perfectly. However, I'm a little bit lost from the "mirror symmetry" part. I did do some wikipedia search, and it returned something on String Theory @.@ Can you explain briefly what mirror symmetry is (like what does $f$ do?...), or can you give me some reference on it. Thanks very much in advance. :) $\endgroup$
    – user49685
    Feb 7, 2013 at 14:50
  • $\begingroup$ @user49685 I mean a reflection. "Mirror symmetry" was a bad translation of the same thing into English. $\endgroup$
    – Dan Shved
    Feb 7, 2013 at 15:19
  • $\begingroup$ I know reflection across a point, and a line; but this doesn't seem like any of the above two. :( Can you show me what $f$ does, how can $f$ map $EJ$ to $EH$? $\endgroup$
    – user49685
    Feb 7, 2013 at 15:25
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    $\begingroup$ @user49685 Of course, $f$ doesn't map point $J$ to point $H$, or segment $EJ$ to segment $EH$. But $f$ does map line $EJ$ to line $EH$, because $\angle JEX = \angle HEG$. $\endgroup$
    – Dan Shved
    Feb 7, 2013 at 15:32
  • $\begingroup$ Yay, I get it. Thanks very much. :* Studying your work, I've learnt the second definition on circles (Apollonian circle), so, that's great. Thanks. I'll leave this bounty open for several more days, hope you don't mind. :) $\endgroup$
    – user49685
    Feb 7, 2013 at 15:43
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We do not need appolonian circles:

Let $P_{XY}$ be the point at infinity on $XY$. Let the parallel through $B$ to $EC$ meet $AC$ at $P$ and define $Q$ similarly. We have,

$B(P_{EC}, G; E, C) = E(P, D; B, C) = E(Q, D; C, B) = C(P_{EB}, F; E, B)$

So if $BG \cap EC = X, CF \cap EB=Y$ then $XY \parallel BC \implies E, I, H$ are collinear.

Note: If one considers $EBC$, and takes the isogonals of the problem, it becomes similar to a problem used in the team selection tests for Bulgaria.

Furthermore, if the symmedian point of $BEC$ is $K$ then $K, I, B, C, A$ all lie on a conic. I believe this problem can thus be generalised and probably has already been.

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