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Let a Triangle $\triangle ABC$ be inscribed in a circle, along the arc $\overset{\frown}{BC}$ lies a point $P$ such as, $BP=4\sqrt{2}$.

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Compute the distance between the two orthocenters of the triangles $\triangle ABC$ and $\triangle APC $.

  • As you can see form the picture above the segment $BP$ seems to be parallel with "$H_1H_2$". But I don't know how to prove it. I believe something about the nine point circle might be useful.
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  • $\begingroup$ It's impossible to prove $\;BP||AC\;$ (and it doesn't look like that, either), and that length $\;4\sqrt2\;$ can hardly make any difference here, imo, if we're not told something else, say: the circumcircle's radius... $\endgroup$ – DonAntonio Jun 13 '14 at 21:47
  • $\begingroup$ You must mean that $BP$ is parallel to $H_1H_2$ (in which case $BPH_2H_1$ is a parallelogram and $|H_1H_2|=BP$) $\endgroup$ – sds Jun 13 '14 at 21:52
  • $\begingroup$ Sorry edited, I meant that $BP$ is parallel to $H_1H_2$ $\endgroup$ – Keith Jun 13 '14 at 22:21
  • $\begingroup$ One thing that was obvious from start is that $BH_1=2R\cos B=PH_2$. $\endgroup$ – Sawarnik Aug 14 '14 at 16:01
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Let

  • $D=AC\cap BH_1$ be the foot of the altitude from $B$
  • $E$ - the other intersection of of the altitude $BH_1$ with the circle

and similarly

  • $Q=AC\cap PH_2$ be the foot of the altitude from $P$
  • $R$ - the other intersection of of the altitude $PH_2$ with the circle

Then $BPRE$ is a symmetric (since it is inscribed) trapezoid (since $BE\perp AC \perp PR \implies BE\parallel PR$) and thus angles $ERP$ and $DPR$ are the same.

Also, $H_1H_2RE$ is a symmetric trapezoid (its axis of symmetry is $AC$) therefore $H_1H_2 \parallel BP$, $BPH_1H_2$ is a parallelogram and $H_1H_2=BP$.

PS. Please tell me if any steps are unclear.

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  • $\begingroup$ How did you conclude that $BPRE$ is a symmetric trapezoid ? $\endgroup$ – Keith Jun 13 '14 at 22:28
  • $\begingroup$ Sorry, but How did you conclude that $H_1H_2RE$ is a symmetric trapezoid ? Answ: The point $Q$ and $D$ are Euler points. Hence $H_1D \cong DE$ and $H_2Q \cong QR$ and the conclusion follows $\endgroup$ – Keith Jun 13 '14 at 23:42
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It's not difficult to show that, in $\triangle XYZ$ with orthocenter $W$ and circumdiameter $d$, we have this nice counterpart to the Law of Sines: $$\frac{|\overline{WX}|}{|\cos X|} = \frac{|\overline{WY}|}{|\cos Y|} = \frac{|\overline{WZ}|}{|\cos Z|} = d$$


In your $\triangle ABC$ and $\triangle APC$, the angles at $B$ and $P$ subtend the same chord, $\overline{AC}$, and are therefore congruent (by the Inscribed Angle Theorem). Writing $d$ for the common circumdiameter, we have $$|\overline{BH_1}| = d\;|\cos B| = d\;|\cos P| = |\overline{PH_2}| \qquad\qquad(\star)$$

Since $\overline{BH_1}$ and $\overline{PH_2}$ are clearly parallel, and now also congruent, it follows that $\square BH_1H_2P$ is a parallelogram. Thus, $\overline{BP}\cong\overline{H_1H_2}$.


Note that $(\star)$ says, for fixed $A$ and $C$ on a circle, the distance from a moving $B$ on that circle to the orthocenter of the corresponding $\triangle ABC$ is constant; the locus of those orthocenters, then, is a translate of the circle. This is especially-easy to see when $\overline{AC}$ is a diameter of the circle.

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