Locus of the orthocentre...

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Locus of the orthocentre...

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Source: Bulgarian NMO 2014, p1

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2459 posts

#1 h Jul 22, 2014, 5:46 AM • 4 Y

Y by Davi-8191, centslordm, Adventure10, Mango247

Let $k$ be a given circle and $A$ is a fixed point outside $k$ . $BC$ is a diameter of $k$ . Find the locus of the orthocentre of $\triangle ABC$ when $BC$ varies.

Proposed by T. Vitanov, E. Kolev

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XmL

#2 h Jul 22, 2014, 7:53 AM • 3 Y

Y by centslordm, Adventure10, Mango247

Clearly the locus is the polar of $A$ wrt $k$

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73 posts

#3 h Jul 24, 2014, 2:59 AM • 2 Y

Y by centslordm, Adventure10

put circle center on $O, A(a,0),B(\cos{t},\sin{t}) \implies C(-\cos{t},-sin{t})$ , the orthocentre is the intersect point of following two lines::
$y=-\dfrac{x-a}{\tan{t}}$
$y-\sin{t}=-\dfrac{\cos{t}+a}{\sin{t}}(x-\cos{t})$

$\implies x=\dfrac{1}{a}$

so it is a vertical line of $AO$

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341 posts

#5 h Jul 24, 2014, 3:51 PM • 3 Y

Y by centslordm, Adventure10, Mango247

This is a sub-problem of the general problem which has the same locus:
"From a point $A$ outside a circle draw a line to intersect the circle at $B$ and $C$ ( $B$ between $A$ and $C$ ), and another line to intersect the circle at $D$ and $E$ ( $D$ between $A$ and $E$ ). $BE$ and $CD$ intersect at point $P$ . Find the locus of point $P$ ."
on page $13$ of the book http://www.amazon.com/Practice-Problems-Mathematical-Olympiad-Competitions-ebook/dp/B00EHU28C2 published in 2012, except that the general problem is more difficult.

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#6 h Jul 25, 2014, 7:28 AM • 2 Y

Y by centslordm, Adventure10

I don't think they are same problems. Your problem is not so hard but need more calculation:

this time we put $A$ on $O$ , the circle center on $(a,0)$ , two lines: $y=k_1x,y=k_2k,$ , circle $(x-a)^2+y^2=1$

$B,C : x=\dfrac{a\pm D_1}{1+k_1^2},D_1=\sqrt{1+k_1^2-(k_1a)^2},y=k_1x$
$D,E: x=\dfrac{a\pm D_2}{1+k_2^2},D_2=\sqrt{1+k_2^2-(k_2a)^2},y=k_2x$

$BE: \dfrac{y-\dfrac{k_2(a+D_2)}{1+k_2^2}}{x-\dfrac{(a+D_2)}{1+k_2^2}}=\dfrac{\dfrac{k_2(a+D_2)}{1+k_2^2}-\dfrac{k_1(a-D_1)}{1+k_1^2}}{\dfrac{(a+D_2)}{1+k_2^2}-\dfrac{(a-D_1)}{1+k_1^2}}=\dfrac{(k_1-k_2)(k_1k_2-1)a+(k_1D_1+k_2D_2+k_1k_2(k_1D_2+k_2D_1))}{(k_1^2-k_2^2)a+k_1^2D_2+k_2^2D_1}=t_1$

$CD:\dfrac{y-\dfrac{k_1(a+D_1)}{1+k_1^2}}{x-\dfrac{(a+D_1)}{1+k_1^2}}=\dfrac{\dfrac{k_1(a+D_1)}{1+k_1^2}-\dfrac{k_2(a-D_2)}{1+k_2^2}}{\dfrac{(a+D_1)}{1+k_1^2}-\dfrac{(a-D_2)}{1+k_2^2}}=\dfrac{(k_2-k_1)(k_1k_2-1)a+(k_1D_1+k_2D_2+k_1k_2(k_1D_2+k_2D_1))}{(k_2^2-k_1^2)a+k_1^2D_2+k_2^2D_1}=t_2$

$(t_1-t_2)x=(k_1-t_2)\dfrac{(a+D_1)}{1+k_1^2}-(k_2-t_1)\dfrac{(a+D_2)}{1+k_2^2}$

the process of calculation is heavy and ugly so I don't post it. but we can have the result is very simple:

$x=a-\dfrac{1}{a}$

it has the same position of first question, but it is a segment in the circle and first one is a line.

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Arab

#7 h Jul 25, 2014, 9:19 AM • 3 Y

Y by centslordm, Adventure10, Mango247

Vo Duc Dien is right.That is one of the special cases of the problem he gives,and the general situation is well-known and can be proved in many ways without calculation.

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341 posts

#8 h Jul 25, 2014, 3:20 PM • 2 Y

Y by centslordm, Adventure10

That's right. Problem can be proven without calculation. In the configuration I gave, if $K$ and $L$ are on $AC$ and $AE$ , respectively such that $A$ , $B$ , $K$ and $C$ form a harmonic division and $A$ , $D$ , $L$ , $E$ form another harmonic division, the points $P$ , $K$ and $L$ are on the line that is the locus of point $P$ .

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759 posts

#9 h Nov 19, 2014, 5:06 AM • 3 Y

Y by centslordm, Adventure10, Mango247

Let $D, E, F$ be the altitudes of $A, B, C$ , so $AD, BE, CF$ intersect at $H$ , the orthocenter. We want to find the locus of $H$ . I claim that the locus is the polar of $A$ wrt $k$ .

Let $O$ be the center of $k$ . Let the projection of $H$ onto $AO$ be $M$ . Then, $OMHD$ is cyclic so $AM \cdot AO = AH \cdot AD = AF \cdot AB = \text{constant}$ . Therefore, $AM$ is constant, so we know that the locus is a line. Furthermore, we can confirm that this line is indeed the polar by letting $C$ be on $k$ such that $AC$ is tangent to $k$ .

To show that every point on the line is in the locus, suppose we choose a point $X$ on the line. Draw circle with diameter $XO$ , and where $AX$ intersects that circle again is going to the projection of $A$ onto $BC$ . Therefore, $BC$ can be determined, and we have that $X$ is the orthocenter of $ABC$ .

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112 posts

#10 h Jan 6, 2016, 7:21 PM • 3 Y

Y by centslordm, Adventure10, Mango247

let the 2 extremities of the diameter be $(cosA, sinA) ; (-cosA, -sinA)$ let $A= (a,0)$
let orthocentre be $H (h,k)$
so, AH perpendicular on BC or $k/(h-a)= -cotA$
CH perpendicular to AB or $(k+sinA)/(h+cosA)=(a-cosA)/sinA$
solving, $h=1/a$

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7936 posts

#11 h Jan 6, 2016, 7:32 PM • 3 Y

Y by anantmudgal09, centslordm, Adventure10

Just a solution for fun more than anything else.

Let $P$ and $Q$ be points on the circle such that $AP$ and $AQ$ are tangent to the circle. By China 1996.1, $P$ , $H$ , and $Q$ are collinear. Furthermore, note that when $P$ is one of the endpoints of the diameter $BC$ , the orthocenter is $P$ , while if $Q$ is one of the endpoints, the orthocenter is $Q$ . From this it's clear that any point $X\in PQ$ can work as the orthocenter (just take the diameter perpendicular to $AX$ ), so the locus is $\overline{PQ}$ .

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201 posts

#12 h Mar 30, 2017, 5:05 AM • 2 Y

Y by centslordm, Adventure10

XmL wrote:

Clearly the locus is the polar of $A$ wrt $k$

Here's the projective solution Xml referred to:

Let $E = AC \cap k$ and $F = AB \cap k$ . Also let $P_1$ and $P_2$ be the tangency points of A with k.

Note that $BE \cap CF$ = H, the orthocenter of triangle ABC. Also note that $P_1P_2$ is the polar of A with respect to circle k. Hence it suffices to show that H lies on the polar to show that $P_1, P_2$ and H are collinear.

To do this, use Brocard's. The rest follows.

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249 posts

#13 h May 21, 2021, 5:37 AM • 4 Y

Y by centslordm, Mango247, Mango247, Mango247

Easy!
Let $X, Y$ be the intersections of $AB$ and $AC$ with $k$ . Then Brocard's on $BCYX$ gives the intersection of diagonals, the orthocentre of $\triangle ABC$ is on the polar of intersection of diagonals $BX$ and $CY$ , $A$ . Hence the locus is the polar of $A$ with respect to the circle $k$ .

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#14 h Feb 11, 2022, 4:40 PM

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Here is a trigonometric solution, which is pretty short

Let $D$ and $E$ be the feet from $B$ and $C$ onto $AC$ and $AB$ , and let $AF$ and $AG$ be tangents onto $(BCDE)$ .

Then notice that $BH.HD=4R^2\cos{\angle A}\cos{\angle B}\cos{\angle C}$ and that $AE.AC=4R^2\sin{\angle C}\sin{\angle B}\cos{\angle A} = AF^2$
Then the PoP of $H$ with respect to the circle $(A,AF)$ is $AF^2-AH^2$

When all is plugged in we have that $AF^2-AH^2=BH.HD$ , here $AH=2R\cos{\angle A}$ ,thus $H$ is on the radical axis of $(BCDE)$ and $(A,AF)$ . which is $GF$ , a fixed line.
Thus the locus of $H$ is the line $GF$ .

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#15 h Dec 29, 2023, 12:46 PM • 2 Y

Y by GeoKing, Assassino9931

Let $O$ be the center of $k$ . The centroid of the triangle is fixed as two thirds of the way from $A$ to $O$ . Thus, it suffices to find the locus of the circumcenter since a homothety of $-2$ about the centroid sends the circumcenter to the orthocenter. Let $X$ be the second intersection of line $AO$ with $(ABC)$ . By power of a point, $AX$ is fixed; the circumcenter of $\triangle ABC$ must lie on the perpendicular bisector of $AX$ , hence the circumcenter varies on a line.

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#16 h Dec 29, 2023, 6:14 PM

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Let the orthocentre be H.
Let AH intersects BC at G, BH intersects AC at M, CH intersects AB at N.

M and N are both on k because of right angles.
Now AM x AC = AN x AB = constant = l^2 (let l be the distance between A and its tangent point on k).

Considering that AHM and ACG are similar, we know that AH x AG = AM x AC (or AN x AB) = l^2.

Now connect AO and construct AHF so that F is on AO and AHF is similar to AOG. That is, AF and HF are perpendicular to each other. Once again, AF x AO = AH x AG = l^2. This way, we have decided on F.

Better still, since AF and HF are perpendicular, the locus of H is decided as well. It is a straight line that is perpendicular to AO and passes F.

Finally, let's examine the length of AF. AF = l^2/AO. In this case, the straight line in question is the one that connects both A's tangent points on k.

This post has been edited 1 time. Last edited by hidummies, Dec 29, 2023, 6:29 PM
Reason: To complete an unfinished post.

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Eka01

#17 h Nov 1, 2024, 4:37 PM

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Funny. Let $AB \cap k=D$ and $AC \cap k=E$ then the orthocenter is $BE \cap CD$ which lies on polar of $A$ with respect to $k$ due to Brocard's theorem.

This post has been edited 1 time. Last edited by Eka01, Nov 1, 2024, 4:38 PM

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