A rectangle $HOME$ has sides $HO=11$ and $OM=5$. A triangle $ABC$ has $H$ as intersection of altitudes, $O$ as the circumcentre, $M$ as the midpoint of $BC$ and $E$ as the foot of altitude from $A$. Find the length of $BC$.

I need some hints to start off with the problem. Also, a diagram would help. Thanks.

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    $\begingroup$ For a fixed segment BC, the geometric locus of point A such that HOME is a rectangle, i.e., the Euler line is parallel to the base, is the ellipse having BC as minor axis, and the segment determined by the two positions of A such that the triangle ABC is equilateral, as major axis. $\endgroup$ – Lucian Nov 7 '15 at 23:50

Here is a diagram.

enter image description here

The altitudes are obvious. $M$ is the midpoint of $BC$, $G$ is the midpoint of $AB$, and $I$ is the midpoint of $AC$.

Here is a hint to solve your problem. We construct that diagram in a different way, starting with the rectangle (whose details we know).

enter image description here

Construct rectangle $HOME$ with $HO=EM=11$, $HE=MO=5$. Construct lines $\overleftrightarrow{HE}$ and $\overleftrightarrow{ME}$.

Construct arbitrary point $B$ on ray $\overrightarrow{ME}$. We need $M$ to be the midpoint of $\overline{BC}$, so construct circle $c$ with center $M$ and radius $BM$ and point $C$ the intersection of circle $c$ with ray $\overrightarrow{EM}$.

Point $O$ needs to be the circumcenter of $\triangle ABC$ and thus equidistant from points $A$ and $C$. Therefore construct circle $d$ with center $O$ and radius $CO$ and point $A$ the intersection of circle $d$ with line $\overleftrightarrow{HE}$. Construct $\triangle ABC$. We are guaranteed that point $O$ is its circumcenter.

But we also want point $H$ to be its orthocenter (intersection of the altitudes). Construct point $F$ as the intersection of lines $\overleftrightarrow{CH}$ and $\overleftrightarrow{AB}$. If lines $\overleftrightarrow{CH}$ and $\overleftrightarrow{AB}$ are perpendicular at $F$, $H$ is indeed the intersection of the altitudes.

We want to find the position of point $B$ that makes this so. Therefore define $r=BE$. You can now find the values of $CM$, $CE$, $OC$ and $OA$, $AH$, and $AE$. (You asked for only a hint, so I'll leave that to you.) Those desired lines are perpendicular iff triangles $\triangle BEA$ and $\triangle HEC$ are similar, so set the proportion


and solve for $r$. (That involves solving a factorable fourth-degree polynomial equation.) From that you easily find the value of $BC$.

Ask if you need the details or the final answer.

  • $\begingroup$ Is answer of above question $3$? $\endgroup$ – mnulb Dec 27 '15 at 8:35
  • $\begingroup$ @Ayushakj: No, $HO$ is $11$ and $BC$ is larger than that. $\endgroup$ – Rory Daulton Dec 27 '15 at 8:36
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    $\begingroup$ Sorry my previous comment was incomplete I want to ask BE=3? $\endgroup$ – mnulb Dec 27 '15 at 8:39
  • $\begingroup$ @Ayushakj: Yes, $r=BE=3$, so the final answer is $BC=28$. $\endgroup$ – Rory Daulton Dec 27 '15 at 20:12

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