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Construct $\triangle ABC$ such that its orthocenter ($H$) and circumcenter ($O$) are on its incircle.

I've tried something by inverting everything WRT circumcircle but don't have proper idea... Or since $O$ and $H$ are isogonal conjugates trying to reflect them WRT sides of triangles and find something but nothing tried many different approaches... Does anyone has some idea how to do that?

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(Image from @Blue, using proportions calculated algebraically. (See comments, but ignore my non-constructibility nonsense.) It may-or-may-not be helpful to note that points $B$, $C$, $O$, $I$, $H$ lie on a circle congruent to the circumcircle. Specifically, $O$ is the midpoint of $\stackrel{\frown}{BC}$ on that circle, and $I$ is in turn the midpoint of $\stackrel{\frown}{OH}$. One can show that this property is a consequence of $\angle A=60^\circ$ alone. The construction corresponding to the additional geometric condition that causes $O$ and $H$ to lie on the incircle remains elusive.)

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  • $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: we want to see that you have put significant work into the problem. $\endgroup$ – Rory Daulton Feb 21 '16 at 1:37
  • $\begingroup$ If my calculations are correct, such a triangle has side-lengths in proportion $$0.680\dots\;:\;1.0\;:\;1.148\dots$$ However, the first and last values are roots of irreducible fourth-degree polynomials and are therefore not "constructible" in the ruler-and-compass sense. (Interestingly, one of the angles is precisely $60^\circ$.) $\endgroup$ – Blue Feb 21 '16 at 2:03
  • $\begingroup$ Thanks a lot $Blue$, if you have time could you explain how do you get those values? $\endgroup$ – jowanar Feb 21 '16 at 2:07
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    $\begingroup$ @RoryDaulton This does not look like a hw problem to me... $\endgroup$ – Lev Borisov Feb 21 '16 at 2:26
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    $\begingroup$ Some of the discussion here is about the constructibility of irreducible-quartic roots. Turns out there is a relatively easy test. If the resolvent cubic has a rational root, you're good! Clearly that happens here. $\endgroup$ – Oscar Lanzi Feb 21 '16 at 10:57
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We begin by proving the claims made by Blue in the edit to the question.

Let $I$ be the incentre of $ABC$, and let $R$ be its circumradius. Since $O$ and $H$ lie inside $ABC$, the triangle must be acute.

The incircle is divided into three arcs by its points of contact with the sides of $ABC$. At least one of these arcs, say the one nearest vertex $A$, contains neither $O$ nor $H$. Thus when the rays $AO$ and $AH$ meet the incircle at $O$ and $H$, respectively, each of these rays is intersecting the incircle for the second time. Moreover, as in any triangle, $AI$ bisects angle $OAH$. It follows that the points $O$ and $H$ are symmetric about $AI$. In particular, $AH = AO = R$.

In any triangle, $\overrightarrow{AH} = 2\overrightarrow{OA'}$, where $A'$ is the midpoint of $BC$. Hence $OA' = R/2$. It follows from this that $\angle BOC = 120^{\circ}$, hence that $\angle BAC = 60^{\circ}$.

If we introduce $O'$ as in the figure (the reflection of $O$ through $A'$), then $O$, $B$ and $C$ belong to the circle with radius $R$ centred at $O'$. Since $\overrightarrow{AH} = \overrightarrow{OO'}$, the quadrilateral $OAHO'$ is a rhombus with side $R$. Consequently, $H$ also belongs to circle $BOC$.

If $J$ is the point halfway along arc $OH$ on circle $BOC$, then $BJ$ bisects $\angle OBH$, hence $J$ lies on $BI$. Similarly, $J$ lies on $CI$. Hence $J = I$, and $I$ lies on circle $BOC$. Since $AI$ bisects $\angle BAC$, it meets the circumcircle of $ABC$ again at $O'$, which is midway between $B$ and $C$.

Conversely, we carry out a construction corresponding to the above requirements. Start with a circle centred at $O$ with radius $1$. Mark two points $B$ and $C$ on the circle so that $\angle BOC = 120^{\circ}$. Let $O'$ be the reflection of $O$ through $BC$. Then $O'$ is on the circle. Now let $I$ be any point on circle $BOC$, on the same side of $BC$ as $O$. (We will specify $I$ further below.) Let $O'I$ cut $BO'C$ again at $A$. Let $H$ be the reflection of $O$ through $O'I$. Then reversing the arguments above, we find that $H$ is the orthocentre and $I$ the incentre of triangle $ABC$, and that $IH = IO$.

The only question that remains is how to choose $I$ on circle $BOC$ so that $IO$ is equal to the inradius of $ABC$. If we let $x$ be the inradius, then $x$ is the distance from $I$ to line $BC$. We also have $IO^2 = (1/2 - x)^2 + 1 - (x+1/2)^2 = 1-2x$. The condition $OI = x$ is equivalent to $x^2 = 1 - 2x$, or $x = \sqrt{2} - 1$.

Thus the construction can be completed by letting $I$ be a point of intersection of circle $BOC$ with a circle centred at $O$ with radius $\sqrt{2}-1$.

I'm not sure how to motivate this last step geometrically.

Summary of my construction Given two points $O$ and $O'$, write $R = OO'$. Construct the circles $K$ and $K'$ of radius $R$ centred at $O$ and $O'$, respectively. Let $B$ and $C$ be the points of intersection. Let $I$ be a point of intersection of $K'$ with the circle of radius $(\sqrt{2}-1)R$ centred at $O$. Then let $A$ be the point of intersection of $O'I$ with $K$.

Alternative construction (using $IA = 2IO$, proved by dxiv below) Instead of constructing $I$, construct $A$ directly by intersecting $K$ with the circle of radius $(2\sqrt{2} - 1)R$ centred at $O'$.

Summary of dxiv's construction Construct a triangle $AIO$ with $IO= r$, $IA = 2r$, $OA = (\sqrt{2}+1)r$. Let $K$ be the circle centred at $O$ passing through $A$. Construct angles of $30^{\circ}$ on either side of $AI$. Let $B$ and $C$ be the intersections with $K$ of the outer sides of these angles.

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  • $\begingroup$ Following up on my comment under the original post that $R = (\sqrt{2} + 1) r$ by Euler's theorem. Once proven that $\angle A = 60°$ the construction can be simplified as following. Given points $I, O$ with $|IO| = r$ draw the circle with radius $2 r$ centered at $I$, and the circle with radius $(\sqrt{2} + 1) r$ centered at $O$. Then $A$ is one of the points of intersection of the two circles, and the rest is trivial. Note that (like the proof above) this assumes that $\Delta ABC$ exists with the given properties. $\endgroup$ – dxiv Feb 22 '16 at 16:59
  • $\begingroup$ @dxiv Actually, I do prove that the triangle exists, although I leave out some of the details because the proof is similar to the necessity part. I like your idea, but how do you prove that $AI = 2r$? $\endgroup$ – David Feb 22 '16 at 18:14
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    $\begingroup$ Let $I_c$ be the point of tangency of $AB$ to the incircle. Then $\Delta AI_cI$ is a right triangle with $\angle IAI_c = 30°$ and the opposite side $|I_cI| = r$. It follows that the hypothenuse $|IA| = 2 r$. As noted, this assumes $\angle A = 60°$ having been proved already. $\endgroup$ – dxiv Feb 22 '16 at 18:20
  • $\begingroup$ @dxiv Good.${{{{{{}}}}}}$ $\endgroup$ – David Feb 22 '16 at 18:21
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    $\begingroup$ @dxiv I've written what I think your construction is above. Please let me know if I've made a mistake. $\endgroup$ – David Feb 22 '16 at 19:24

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