# Distance between vertex and orthocenter

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?

• You are not really using the $\angle CAP=45^\circ$ are you? Commented Aug 3, 2020 at 14:50
• 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? Commented Aug 3, 2020 at 14:58
• I haven't tried this problem , but BH=2RcosB is a formula .. Commented Aug 3, 2020 at 15:52
• 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. Commented Aug 3, 2020 at 23:54

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 $$\boxed{BH=\frac{BX\cdot\sin{18^\circ}}{\sin{45^\circ}}=\frac{BC\cdot\sin{18^\circ}}{\sqrt{2}}}$$ 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 $$P$$).