# Construct a triangle with its orthocenter and circumcenter on its incircle.

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?

(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.)

• 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. – Rory Daulton Feb 21 '16 at 1:37
• 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 $60^\circ$.) – Blue Feb 21 '16 at 2:03
• Thanks a lot $Blue$, if you have time could you explain how do you get those values? – jowanar Feb 21 '16 at 2:07
• @RoryDaulton This does not look like a hw problem to me... – Lev Borisov Feb 21 '16 at 2:26
• 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. – Oscar Lanzi Feb 21 '16 at 10:57

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.

• 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. – dxiv Feb 22 '16 at 16:59
• @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 $AI = 2r$? – David Feb 22 '16 at 18:14
• 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. – dxiv Feb 22 '16 at 18:20
• @dxiv Good.${{{{{{}}}}}}$ – David Feb 22 '16 at 18:21
• @dxiv I've written what I think your construction is above. Please let me know if I've made a mistake. – David Feb 22 '16 at 19:24