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This is a problem from an italian Olympiad contest: I don't need a solution of this problem, but I'd appreciate some hints, because all my attempts of solution have failed. Consider a triangle $ABC,$ with $AC>BC.$ Trace the circle $\Gamma_1$ passing through the points $A,M,N$ where $M,N$ are, respectively, the feet of the altitudes from $B$ and $C.$ Let $\Gamma_2$ be the circumcircle of $ABC:$ these two circles intersect in $A,P.$ We know the length of $BC$ and the two angles $$\angle{BCA}=27^\circ\qquad \angle{CAP}=45^\circ.$$ What is the distance between $B$ and the orthocenter $H$ of $ABC$?

I report a figure and my attempts of solution.

First of all, using the fact that angles insisting over the same arc are equal, I can compute the angles as in figure. I also considered the symmetric point of $H$ with respect to $AC,$ which belongs to the cicumcircle. Then I made a lot of attempts but I didn't get anything useless. With my data I can compute everything about the triangle $BCM$: I considered the formula $$BH^2=4R^2-AC^2,$$ where $R$ is the radius of the circumcircle, but I don't know how to compute $R,AC$. I even tried to compute some other angles, but the only interesting relation I have found is $$\angle{HAP}=18^\circ$$ (for instance considering the fact that $\{A,B,C,H\}$ is an orthocentric system). Finding $x$ I would be able to solve the triangle $ABC,$ and then the problem would be finished, but I don't find any other relation involving $x,y.$

I even know that the orthocenter of $ABC$ belongs to the circle passing through a vertex and the feet of the altitudes front the other two (hence $H\in\Gamma_1$) and that the symmetric points of H with respect the sides of $ABC$ belong to $\Gamma_2$ (like the point $L$ in figure) but I don't know hw to use these facts. At this point I'm stuck. Could you please give me some hints or some ideas to solve it?

enter image description here

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  • $\begingroup$ You are not really using the $\angle CAP=45^\circ$ are you? $\endgroup$ Commented Aug 3, 2020 at 14:50
  • $\begingroup$ I've already thought about how using that angle (to compute $\angle{HAP},$ and the only other idea was to reflect $P$ with respect to $AC$ in order to have an angle of $90$ degrees. But I don't see how to relate this fact with the other information... Can I use that angle in another way? $\endgroup$
    – PS48725
    Commented Aug 3, 2020 at 14:58
  • $\begingroup$ I haven't tried this problem , but BH=2RcosB is a formula .. $\endgroup$ Commented Aug 3, 2020 at 15:52
  • $\begingroup$ You could rotate the diagram by 90 degrees to match $P$ and $C$, but I don't see any immediate nice things you could say. On the other hand, if you bash it out in Cartesian coordinates, say $C=(1,0)$ and $P=(0,1)$, $B=(\cos 2\angle A,\sin 2\angle A)$ and $A=(\cos 2\angle B,-\sin 2\angle B)$, using $H=(\tan\angle A:\tan\angle B:\tan\angle C)$ in barycentric coordinates and $\angle C=27^\circ$, you will see $AP\perp PH$ imposes a trig equation on the angle $A$. It seems like there should be a nicer way to derive that though, given it is an olympiad problem. $\endgroup$ Commented Aug 3, 2020 at 23:54

1 Answer 1

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Below you will find a few hints and a solution.

Hint 1:

Consider the midpoint $X$ of the line $BC$.

Hint 2:

Try proving that $P$, $H$ and $X$ lie on a straight line.

Hint 2.1:

The reflection of $H$ over $X$ is the point diametrically opposite $A$

Hint 3:

Sine law in $\Delta BHX$

My solution:

Let $X$ be the midpoint of $BC$ and $Y$ the reflection of $H$ over $X$. We claim $P$, $H$, $Y$ lie on a straight line. Indeed, $$\angle APY=90^\circ=\angle AMH=\angle APH.$$ It follows that $P$, $H$ and $X$ are collinear. Now it's not hard to see that $$\angle BHX=\angle CAP=45^\circ\text{ and }\angle XBH=\angle 180^\circ-\angle MBX=117^\circ.$$ It follows that $\angle HXB=18^\circ.$ Using sine law in $\Delta BHX$, we get $$\boxed{BH=\frac{BX\cdot\sin{18^\circ}}{\sin{45^\circ}}=\frac{BC\cdot\sin{18^\circ}}{\sqrt{2}}}$$ Diagram Remark: The fact that $X$, $P$ and $H$ lie on a straight line can also be proven using the fact that the Miquel point of a cyclic quadrilateral is the image of the intersection of its diagonals under inversion with respect to the circumcircle of the quadrilateral (in this case the quadrilateral is $MNBC$ with Miquel point $P$).

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