3
$\begingroup$

To a circle of radius $1$, two tangents are drawn from any point $P$ on a line $3$ units away from its center. They touch the circle in $A$ and $B$.

Find the locus of the orthocenter of $\triangle PAB$.

I have tried various ways to approach this problem, assuming the coordinates of (P), drawing the tangents, finding the coordinates of (A) and (B), so on.

All this consumes a lot of time, and even then, I failed to solve it. Is there a short and efficient way of solving this problem?

Source - This problem appeared in our test, and I was not able to solve it.

$\endgroup$
2
$\begingroup$

Wrong question

At first I missed the “on a line” part of your question, and assumed that $P$ would be on a circle of radius $3$ around the center of the stated circle. The computation below originally refers to that.

This problem has perfect rotational symmetry: you have that circle of radius $1$ and another of radius $3$ and $P$ can be at any position on the latter. That position alone determines $A$, $B$ and the orthocenter. So if you rotate $P$ around the center, the orthocenter will rotate as well. Which means that the orthocenter has to be yet another concentric circle.

Now all that remains is finding its radius. For that we'll use coordinates. Assuming the origin as the center of all circles, choose $P=(3,0)$. Then its polar line, which passes through the touching points of the tangents, will be $x=\frac13$ which intersects the circle in two points:

$$\left(\frac13\right)^2+y^2=1\implies y=\pm\frac{2\sqrt2}3$$

So let's use $A=\left(\frac13,\frac{2\sqrt2}3\right)$ and $BP:x-2\sqrt2y=3$. The altitude between these two is $2\sqrt2x+y=\frac43\sqrt2$ which intersects $y=0$ (the altitude for $P$) in the orthocenter $\left(\frac23,0\right)$. There you have the radius of that locus.

Figure 1

An experiment

Then Mick's comment made me aware of my mistake.

$P$ is on a line and that line is at distance $3$ to the center of the circle. Of course, the situation used in the computation above will still remain a valid instance of this, if I choose the line to be the one at $x=3$. So we still know one point on the locus, but that locus won't be a concentric circle any more.

Doing a quick experiment with Cinderella, I found the locus to be a circle through the origin.

Figure 2

This is numerical evidence, but no proof. If you can show that the curve has to be a circle through the origin, then you can easily describe that circle using the point I computed above. Until then, this answer is incomplete.

Combining approaches

So you can do the same computation as above, this time assuming a variable parameter as one coordinate. Starting with $P=(3,a)$ you obtain $AB:3x+ay=1$. Intersecting that with the unit circle, you have to solve a quadratic equation. To shorten things, I'll use another letter for one such solution, namely $b=\sqrt{a^2+8}$. Using that you can describe the points of intersection as

$$A=\frac1{a-3b}\begin{bmatrix}3a-b\\-8\end{bmatrix}\qquad B=\frac1{a+3b}\begin{bmatrix}b+3a\\-8\end{bmatrix}$$

Then the line $BP$ becomes $(3ab + a^2 + 8)x - 8by = ab+3a^2+24$ and the altitude becomes $(3a^2-ab+24)x + (a^3+8a+3b)y = 2a^2 + 16$. Intersecting that with $OP:ax-3y=0$ you get the orthocenter

$$C=\frac1{a^2+9}\begin{bmatrix}6\\2a\end{bmatrix}$$

which is a point on the circle

$$3(x^2+y^2)=2x$$

I wouldn't want to do the above computations without the help of a computer algebra system, though.

Circle inversions

Figure 3

Brute force computation is the approach to use if you have a computer and don't want to think. But we can do better than this. Observe how a tangent is orthogonal to the radius at the touching point. So in the labels of my figure above, $AOBC$ will be a rhombus. If we want to show that the locus you are searching for is a circle of diameter $\frac23$ through the origin, you may as well demonstrate that the center of that rhombus $R$ lies on a circle of diameter $\frac13$, since the diagonals of a rhombus bisect one another.

Now consider inversions in the unit circle. These will leave $A$ and $B$ fixed. The line $OP$ will remain fixed as well. The line $AB$ can be considered a circle through $A,B$ and $\infty$, the point at infinity. Circle inversion will map that point at infinity to the circle center, so the image of the line $AB$ will be a circle through $O,A,B$. Since $AOBP$ is a cyclic quadrilateral, $P$ lies on that circle as well. The center of the rhombus $R$ is the intersection of the line $AB$ with the line $OP$. Therefore its inverse in the circle is the intersection of $\bigcirc ABP$ with $OP$, which is the point $P$. Therefore the locus of $R$ is the inverse of the locus of $P$, i.e. the inverse of the line. Since a line is a circle through $\infty$, its inverse is a circle through the midpoint. And since the point closest to the circle has distance $3$, its image at distance $\frac13$ has maximal distance from $O$, therefore the locus of $R$ has diameter $\frac13$ as we had to demonstrate.

$\endgroup$
5
  • $\begingroup$ Shouldn't a "line" be regarded as a straight line? $\endgroup$
    – Mick
    Oct 23 '14 at 9:15
  • $\begingroup$ @Mick: Yes, “line” generally refers to straight lines, as opposed to “curve”. I guess I simply missed the phrase “on a line” in the question. I just adapted my answer so it isn't completely wrong any more, but it's still incomplete now. $\endgroup$
    – MvG
    Oct 23 '14 at 9:31
  • $\begingroup$ Could we use the fact that distance between the orthocenter and the centroid is twice that between the centroid and the circumcenter? $\endgroup$ Oct 23 '14 at 9:54
  • $\begingroup$ @Pkwssis: I doubt it: the locus of the centroid looks pretty complicated. An observation that might be of use is the fact that the line $AB$ bisects the segment $OC$, so $AOBC$ is a rhombus. I guess I'll simply repeat my computation above with a variable as one coordinate. $\endgroup$
    – MvG
    Oct 23 '14 at 11:11
  • $\begingroup$ @Pkwssis: Yes, that rhombus approach works, see my updated answer. The computational approaches are no fun at all. I wonder whether I should remove them. Do you consider that last section to be understandable all by itself? If so, I'll likely delete the rest. $\endgroup$
    – MvG
    Oct 23 '14 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.