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I'm a newbie in this site. I tried to search if this question was already answered but I'm not sure on how to do it. The problem is: given three distincts points $O,H,I$ namely the circumcenter, the orthocenter and the incenter of a triangle $\triangle ABC$, construct $\triangle ABC$.

enter image description here What have I tried: $$\left| \overline{OI} \right| ^2 = R^2 - 2Rr$$ Taking the midpoint $N$ of $\ \overline{OH} $ (center of the nine point circle) we have: $\left| \overline{NI} \right| = \frac{R-2r}{2}$ So it seems to me we can obtain $R$ and $r$ from these points. I also tried to solve it using complex numbers it looked like this:

Let $a^2,b^2,c^2$ be the vertexes of $\triangle ABC$ in the complex plane in which the origin is the circumcenter let us also define $R=1$ (so $|a|=|b|=|c|=1)$ to simplify. Then we have that the orthocenter $H=a^2+b^2+c^2$ and the incenter $I=ab-ac +bc$ to find a solution to my problem is to find $a^2,b^2,c^2$ in terms of $H$ and $I$ but it was too complex to me. This complex approach indicates that the reflection of the incenter with respect to the circumcenter is important. Another cool distance formula is $\left| \overline{OH} \right|^2 = 9R^2 - (a^2+b^2+c^2)$ with $a,b,c$ being the sides of $\triangle ABC$

MarianD
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onlyme
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  • What have you already tried? – daOnlyBG Jan 29 '15 at 18:46
  • Do you understand what all of the points $O, H, I$ represent? – John Jan 29 '15 at 18:52
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    well at first I've tried to use some distance relations: OI² = R² - 2Rr , and if you take the midpoint M of OH than MI = (R-2r)/2 so it's possible to have the Radius of the circumcircle. Then I tried to use complex numbers to see if this has a solution: let a²,b² and c² be the vertices of the triangle in the complex plane, with the origin in the circumcenter, then the orthocenter H is a²+b²+c² and the incenter is I=ab-ac +bc, i couldn't figure it out if it's posible to find a,b,c in terms of H and I – onlyme Jan 29 '15 at 18:57
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    @onlyme Good to hear that you've tried some things. Please edit your question to put that information in the body. Doing this will increase the likelihood of getting answers, as well as their quality. Welcome to Math.SE. – John Jan 29 '15 at 18:59
  • @John, how do I use Latex here ? It is like [tex] [math] [/tex] [/math] ? – onlyme Jan 29 '15 at 19:03
  • @onlyme Check here but the short version is enclose the LaTeX in single dollar signs for inline, and double dollar signs for putting them on their own line. – John Jan 29 '15 at 19:05
  • Please take a look on the following question: https://math.stackexchange.com/questions/2321816/proving-lines-are-perpendicular – Blind Jun 15 '17 at 21:13

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I asked the same question to myself a while ago, and now I'm glad to share my conclusion.

To build a triangle given $O,H,I$ is substantially equivalent to solving a general cubic, hence it cannot be done with straightedge and compass only. To prove it, notice that we know $$ (R,r,a^2+b^2+c^2) $$ and we have to find $(a,b,c)$ or, equivalently, the exradii $(r_a,r_b,r_c)$. However, the relations: $$ \sum_{cyc} r_a = 4R+r,\quad \sum_{cyc}\frac{1}{r_a}=\frac{1}{r},$$ $$ \sigma_2 = \sum_{cyc} r_a r_b = \frac{a^2+b^2+c^2}{2}+r(4R+r) $$ give that the exradii are the roots of the polynomial: $$ z^3-(4R+r)z^2+\sigma_2 z - r\sigma_2.$$

Jack D'Aurizio
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  • Is it immediately obvious that you can generate an arbitrary cubic polynomial for some choices of $O,H,I$? To me, it seems that if you use isometries to fix $O = 0 + 0i, H = 1 + 0i$, and restrict $I$ to the upper half plane, then that gives a two-dimensional space of equivalence classes of problems. It's certainly still plausible that some of the image of rational points could give irreducible cubics with rational coefficients, but to me it's not obvious whether that will happen or not. – Daniel Schepler Sep 08 '23 at 17:33
  • Oh, I guess you could also say each cubic equation is equivalent, modulo translation and scaling by a constructible number, to a problem $x^3 \pm x + \lambda = 0$... – Daniel Schepler Sep 08 '23 at 17:42