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Given the base of a triangle and ratio of other $2$ sides. Find the locus of the third vertex

To prove that a geometric shape is the correct locus for a given set of conditions. We have to divide the proof into two stages:

  1. Proof that all the points that satisfy the given conditions are on the given shape.
  2. Proof that all the points on the given shape satisfy the given conditions.

I am able to prove that the locus of a point which satisfy the satisfy the given conditions is a circle.

Let $BC$ be the base. And $A$ be the third vertex. We are given $AB:AC$ ratio. Angle Bisectors of $\angle A$ meet $BC$ at $P$ and $Q$. Since $\angle PAQ$ is a right angle. The locus of $A$ is a circle with $PQ$ as a diameter. This is first proof.

I couldn't obtain the solution for second proof. I want to prove that all the points on a circle with $PQ$ as a diameter is such that the ratio of the other two sides is constant that we initialised earlier. We initialised $AB:AC$ to be constant. Let a new point on the circle be $A'$. I want to prove that $A'B:A'C$ is same as $AB:AC$.

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  • $\begingroup$ Are you avoiding coordinate arguments? If not, it's fairly clear that if the ratio is denoted $k > 0$, then the circle in question is the level curve $f(X) = k$ of the real-valued function $f(X) = |X - B|/|X - C|$. (Even if you are avoiding coordinates, you can argue that for each point $X \neq B$, $C$, there is a well-defined ratio $f(X) = |X - B|/|X - C|$, and perhaps can show $f(X) = k$ if and only if $X$ lies on your circle.) $\endgroup$ – Andrew D. Hwang Jun 30 '17 at 11:18
  • $\begingroup$ @Andrew: I didn't get what you commented $\endgroup$ – isaac ismav Jun 30 '17 at 12:26
  • $\begingroup$ If you're willing to use Cartesian coordinates, it's not hard to show algebraically that$$\frac{(x - x_{B})^{2} + (y - y_{B})^{2}}{(x - x_{C})^{2} + (y - y_{C})^{2}} = k^{2}$$(i.e., the ratio $AB:AC = k$) if and only if $A = (x, y)$ lies on a circle of Apollonius, so that "all the points on the given shape satisfy the given conditions". On the other hand, if you do not want to use coordinates, you might still be able to use a coordinate proof as inspiration. It might help to notice that if $A$ is an arbitrary point distinct from $B$ and $C$, the ratio $AB:AC$ is some positive number. $\endgroup$ – Andrew D. Hwang Jun 30 '17 at 13:31
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Here's another way to get the same result. Let $M$ be the midpoint of $PQ$ (and center of the circle), $\theta=\angle AMB$ and $r=PM=QM=AM$. By the cosine rule applied to triangles $ABM$ and $ACM$ we have: $$ AB^2=BM^2+r^2-2r\, BM\,\cos\theta,\\ AC^2=CM^2+r^2-2r\, CM\,\cos\theta. $$ Substituting here $BM=r+BP$ and $CM=r-CP$, we get after some algebra: $$ AB^2={1\over2}(1+\cos\theta)BP^2+{1\over2}(1-\cos\theta)(2r+BP)^2,\\ AC^2={1\over2}(1+\cos\theta)CP^2+{1\over2}(1-\cos\theta)(2r-CP)^2. $$ But $2r+BP=QB$ and $2r-CP=QC$, thus we may rewrite the above equalities as follows: $$ AB^2={1\over2}BP^2\left[1+\cos\theta+ (1-\cos\theta)\,\left({QB\over BP}\right)^2\right],\\ AC^2={1\over2}CP^2\left[1+\cos\theta+ (1-\cos\theta)\,\left({QC\over CP}\right)^2\right]. $$ As we know by hypothesis that $PB:PC=QB:QC$, the two expressions in square brackets are equal between them, and we obtain: $$AB:AC=BP:CP,$$ which is what we wanted to prove.

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At this moment, I can only offer the following particular solution to your problem.

In $\triangle ABC$, we have $\theta = \theta’$ such that, by angle bisector theorem, AB : AC = BP : PC.

The black circle with PQ as diameter is constructed as described.

enter image description here

Let AC be extended to cut the black circle at A’ (which will be our particular point on the locus). Then, $\theta = \theta_1$.

Next A’P is extended to cut QA extended at R.

Then, $\angle 1 = \angle 2 = \angle 3 = \angle 4$ implies APBR is cyclic. Then, x = y and $\theta’ = \theta’_1$.

$\theta'_1 = ... = \theta_1$ further means BRQA’ is cyclic with x = z.

z = y means P is the in-center of $\triangle ABA’$. Result follows.

Of course, if we use $A_1$, the diametric replica of A, as the third vertex, we will generate the second instance, namely A’’.

For the future development, I think the line BR will be of great help because $BR \bot BQ$.


Added.

enter image description here

Let X be a point on the said locus (i.e. the black circle with PQ as a diameter). We further let X’ be the reflection of X about PQ. As a consequence of BQ being the perpendicular bisector of XX’, we have (1) the purple marked angles are equal; and (2) $\rho_1 = \rho_2$.

Form the rays XP and XC. Suppose that XC extended cuts BX’ at T. [You need to show that T is also a con-cyclic point of the black circle. Geogebra confirms that is true.]

Then, $\rho_3 = \rho_1 = \rho_2 = \rho_4$. This means p is the in-center of $\triangle XBT$. Result follows.

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  • $\begingroup$ thanks sir. i understood everything. But i didnt understand how angle 1=angle 2 $\endgroup$ – isaac ismav Jun 30 '17 at 15:09
  • $\begingroup$ $\angle 1 = 90^0 - \theta' = 90^0 - \theta = \angle 2$. $\endgroup$ – Mick Jun 30 '17 at 15:11
  • $\begingroup$ thanks a lot @Mick. i see a great beauty in your approach to the problem $\endgroup$ – isaac ismav Jun 30 '17 at 15:14
  • $\begingroup$ what does BR perpindicular to BQ imply? $\endgroup$ – isaac ismav Jun 30 '17 at 15:15
  • $\begingroup$ A' is a point on the black circle and in particular it is at the extension of AC too. The next step we need to do is to use any point on the black circle that also give the proposed result. I think, the fact that $BR \bot BQ$ maybe of some help if we really need to prove that. $\endgroup$ – Mick Jun 30 '17 at 15:21
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Let points $P$ and $Q$ on line $BC$ satisfy $PB/PC=QB/QC$ and construct the circle of diameter $PQ$ (red circle in diagram below). Draw now a circle of center $B$ and radius $1$ (dashed in the diagram) and consider the inversion transformation with respect to that circle. Points $P$ and $Q$ are transformed into $P'$ and $Q'$, the red circle of diameter $PQ$ into the blue circle of diameter $P'Q'$, and point $C$ is transformed into $C'$, midpoint of $P'Q'$ (this follows from the above relation among points $BCPQ$).

A point $A$ on the red circle is transformed into point $A'$ on the blue circle: from $BA\cdot BA'=BC\cdot BC'$ it follows $BA':BC'=BC:BA$, and triangles $BAC$ and $BC'A'$, having an angle in common, are similar. We then get $AC:A'C'=AB:BC'$, whence: $$ {AB\over AC}={BC'\over A'C'}={BC'\over P'C'}= {BC'\over BP'-BC'}={1/BC\over 1/BP-1/BC}= {BP\over BC-BP}={BP\over CP}, $$ which is what we wanted to prove.

enter image description here

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  • $\begingroup$ It just says BP:PC = AB:AC. When AP is biector. I have already used but it only helped me in first proof, not the second one $\endgroup$ – isaac ismav Jun 30 '17 at 10:35
  • $\begingroup$ Then I don't understand your method: $P$ and $Q$ are FIXED in you second proof, as are $B$ and $C$; it follows that $A'B/A'C=BP/BC=k$, the same value as in the case of $A$. $\endgroup$ – Aretino Jun 30 '17 at 11:20
  • $\begingroup$ And notice that the theorem also works for an exterior angle. $\endgroup$ – Aretino Jun 30 '17 at 11:23
  • $\begingroup$ In the 2nd proof, as u said, P and Q are fixed. But we cannot say A'B:A'C = BP:PC, because we haven't proved that A'P is the angle bisector $\endgroup$ – isaac ismav Jun 30 '17 at 11:25
  • $\begingroup$ We just assumed AP to be the bisector of angle A, but we haven't proved A'P to be the angle bisector of angle A. If this is proved then the question is done $\endgroup$ – isaac ismav Jun 30 '17 at 11:28

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