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In the figure, G and H are the circum-center and the orthocenter of ⊿ABC respectively. AH produced meets BC at O. GR ┴ BC at R. BS is the diameter of the circumscribed circle. Show that B, O, H, and G are not concyclic.

The derived facts are (1) AHCS is a parallelogram: and (2) AH = 2GR.

My questions are:

(1) Fact #1 helps, but does fact #2 also help?

(2) Does “if they are concyclic, then BH = HS ……” help?

(3) I can prove it, in a clumsy way, through angle-chasing. Would like to see if there is a more elegant proof.

VividD
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Mick
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  • If $B$, $O$, $H$, $G$ were concyclic, then $\angle HGB$ would have to be a right angle (in order to be supplementary to $\angle HOB$). So, can you prove that $HG$ is not perpendicular to $BS$? – Blue Jun 15 '14 at 05:44
  • Note that if the triangle is equilateral, then $G$ and $H$ coincide; in that case, $B$, $O$, $H$, $G$ are concyclic. – Blue Jun 15 '14 at 06:02
  • @Blue (1) About your 2nd comment, I think the proof is not that direct. (2) For your 1st comment, I have thought about it, but not successful. – Mick Jun 15 '14 at 07:03
  • @Blue What I can do is to show that angle GHA is, somehow, larger than angle GBO. – Mick Jun 15 '14 at 07:05
  • Write $T$ for the foot of the perpendicular from $C$ to $AB$. We certainly have concyclic $B$, $O$, $H$, $T$, say on circle $\gamma$. ($BH$ is diameter of $\gamma$.) Fixing $B$ & $C$ to make a central angle of at least $90^\circ$, move $A$ along the circumcircle. When $\angle ABC = 90^\circ$, circle $\gamma$ collapses to a pt at $B$; pt $G$ clearly lies outside this circle. As $A$ moves toward $S$, circle $\gamma$ grows; pt $G$ lies inside the circle $\gamma$ when $A=S$. By continuity, there's a location for $A$ such that $G$ lies on $\gamma$. We can't prove non-concyclicity in general. – Blue Jun 15 '14 at 07:06
  • Agree that "We can't prove non-concyclicity in general." I guess the only way is to prove that by contradiction. – Mick Jun 15 '14 at 07:48
  • Since the assertion you want to prove is false (the points are not (always) non-concyclic), I'm curious to know where this problem arose. If it's a textbook or contest problem, there seem to have been some missing conditions. If it's a conjecture on your part, a note to that effect is appreciated; answerers deserve fair warning before spending effort trying to prove something that might not be true. – Blue Jun 15 '14 at 08:32

2 Answers2

2

Elaborating on my comment ...

Consider $B$ and $C$ fixed, with $\angle BGC \geq 90^\circ$, so that $\angle A \geq 45^\circ$. Let $T$ be the foot of the perpendicular from $C$ to $AB$. Note that $B$, $O$, $H$, $T$ are concyclic (since opposite angles at $O$ and $T$ are supplementary); call the circle $\gamma$.

For $A$ such that $\angle ABC \approx 90^\circ$, circle $\gamma$ is nearly a point-circle at $B$. Circumcenter $G$ lies outside $\gamma$.

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For $A$ close to $S$, circle $\gamma$ pass through a point of $\overline{GS}$ (in fact, $\gamma$ contains $G$ itself, when $\angle A = 45^\circ$ and $A=S$). Circumcenter $G$ lies inside $\gamma$.

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By continuity, we can position $A$ such that $G$ lies on $\gamma$.

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Therefore, the assertion that $B$, $O$, $H$, $G$ are non-concyclic is false, in general.

We can be precise about this.

Let $G^\prime$ be the foot of the perpendicular from $H$ to $\overline{BS}$. Certainly, $B$, $O$, $H$, $G^\prime$ are concyclic; moreover, $B$, $O$, $H$, $G$ will be concyclic if and only if $G=G^\prime$.

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Writing $d := |\overline{BS}|$ for the circumdiameter, we have $$\begin{align} |BG| &= \frac{1}{2}d \\[6pt] |BG^\prime| &= |BH| \;\cos\angle HBG^\prime \\ &= |BH|\;\cos(\angle CBS - \angle CBG) \\ &= |BH|\;\cos((90^\circ - A)-(90^\circ-C)) \\ &= |BH|\;\cos(C-A) \end{align}$$

(With appropriate adjustment when $C-A > 90^\circ$.)

Now, as I pointed out recently in this answer, vertex-to-orthocenter lengths obey a Law-of-Sines-like relation, but with cosines; here, in particular, we have $|BH| = d\;|\cos B|$.

Therefore, $$\begin{align} G = G^\prime \qquad&\Leftrightarrow\qquad |\overline{BG}| = |\overline{BG^\prime}| \\ &\Leftrightarrow\qquad \frac{1}{2}d = d\;|\cos B|\;\cos(C-A) \\[6pt] &\Leftrightarrow\qquad 1 = 2\;|\;\cos(C+A)\;\cos(C-A)\;| \\[6pt] &\Leftrightarrow\qquad 1 = |\;\cos 2A + \cos 2C\;| \qquad (\star) \end{align}$$

We can see that condition $(\star)$ holds in some special cases I've mentioned: $(1)$ equilateral $\triangle ABC$ (ie, $A=B=C=60^\circ$), and $(2)$ $A=S$ (ie, $C=90^\circ$) with $A=45^\circ$.

Thus, there are plenty of triangles for which the four points in question are concyclic; and there are plenty of triangles for which the points are non-concyclic.

Community
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Blue
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  • I am posted mine. Please see if it is ok. Your final finding could be correct too. It probably depends on where A is. – Mick Jun 15 '14 at 11:01
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To simplify the explanation, I have the angles labeled.

My proof is by the following contradiction.

If GBOH is a cyclic quadrilateral, then

k + h = p … [ext. angle of ‘cyclic quad’.]

= q … [angles in the same seg.]

= r … [alt. angles AS// HC]

∴ k + h + s = r + s

But h = r + s … [ext. angle of triangle]

∴ k + s = 0 which is impossible (unless TC on the other side of G)

[The ’unless’ part can be forced to not going to happen if A is located even closer to S.]

Mick
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  • This argument is clever for what it does, but you effectively come to this conclusion: The statement is true, unless it's not. That's not terribly insightful. :) There's nothing in your question that suggests $TC$ could not be "on the other side of $G$", so by avoiding this possibility, you leave a huge hole in your argument. – Blue Jun 15 '14 at 11:18