Extend sides and find the locus of a point

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Extend sides and find the locus of a point

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Source: China TST 1997, problem 1

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orl

#1 h May 22, 2005, 3:33 PM • 2 Y

Y by Adventure10, Mango247

Given a real number $\lambda > 1$ , let $P$ be a point on the arc $BAC$ of the circumcircle of $\bigtriangleup ABC$ . Extend $BP$ and $CP$ to $U$ and $V$ respectively such that $BU = \lambda BA$ , $CV = \lambda CA$ . Then extend $UV$ to $Q$ such that $UQ = \lambda UV$ . Find the locus of point $Q$ .

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Amir.S

#2 h May 27, 2005, 7:05 AM • 2 Y

Y by Adventure10, Mango247

$\frac{BU}{AB}=\frac{CV}{AC}=\lambda$ and $\angle UBA=\angle ACV$ there for $\Delta ABU \cong \Delta ACV$
it means that $\Delta AUV \cong \Delta ABC$ then $\frac{AV}{AC}=\frac{AU}{AB}=\frac{UV}{BC}$
we can get from this: $UV,UQ$ are constant now

$AQ=\frac{\sin \angle B.QU}{\sin \angle QAY}$ and it's constant so the locus of Q is a circle by center A.

Amir

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yetti

#3 h May 28, 2005, 4:01 AM • 2 Y

Y by Adventure10, Mango247

Hello Amir,

I do not think that you are right.

amir2 wrote:

$\frac{BU}{AB}=\frac{CV}{AC}=\lambda$ and $\angle UBA=\angle ACV$ , therefore $\Delta ABU \cong \Delta ACV$
it means that $\Delta AUV \cong \Delta ABC$ ...

This is good stuff, but I do not get the rest. In fact, the triangles $\triangle ABU \sim \triangle ACV$ and the triangles $\triangle AUV \sim \triangle ABC$ are just similar and not congruent. I would add that the last pair is similar, because from similarity of the triangles $\triangle ABU \sim \triangle ACV$ , we have

$\frac{AB}{AC} = \frac{AU}{AV}$

$\angle UAV = \angle UAB - (\angle CAV - \angle CAB) = \angle CAB$

The triangles $\triangle AUV, \triangle ABC$ have a common vertex $A$ and they can be made centrally similar with positive coefficient of similarity by rotating the triangle $\triangle AUV$ counter-clockwise by the angle $\angle UAB$ or centrally similar with negative coefficient of similarity by rotating the triangle $\triangle AUV$ clockwise by the angle $180^o - \angle UAB$ . Hence, they are spirally similar with ceneter of similarity $A$ . Let $D$ be a point on the extended line $BC$ , such that $BD = \lambda BC$ and the points $B, C, D$ follow on the line $BC$ in this order. Since $\frac{BD}{BC} = \frac{UQ}{UV} = \lambda$ and since the triangles $\triangle ABC \sim \triangle AUV$ are similar, the quadrilateral $ABCD \sim AUVQ$ (each with one internal angle equal to 180^o) are also spirally similar with the same center of similarity $A$ and the same coefficient of similarity. As the point $P$ moves on the arc $BAC$ of the circumcircle $(O)$ of the triangle $\triangle ABC$ , the locus of points $U$ is an arc of a circle $(B)$ centered at the vertex $B$ and radius $r_B = \lambda BA$ . Similarly, the locus of points $V$ is an arc of a circle $(C)$ centered at the vertex $C$ and radius $r_C = \lambda CA$ . When the point $P$ coincides with the vertex $B$ , the locus arc $(B)$ of the points $U$ is limited by the intersection $S$ of the circumcircle tangent at the vertex $B$ with the circle $(B)$ and the locus arc $(C)$ of the points $V$ by the intersection $B'$ of the ray $CB$ with the circle $(C)$ . When the point $P$ coincides with the vertex $C$ , the locus arc $(C)$ of the points $V$ is limited by the intersection $T$ of the circumcircle tangent at the vertex $C$ with the circle $(C)$ and the locus arc $(B)$ of the points $U$ by the intersection $C'$ of the ray $BC$ with the circle $(B)$ . Thus the central angles of the locus arcs $C'S \equiv (B)$ , $B'T \equiv (C)$ of the points $U, V$ are

$\angle C'BS = 90^o + \angle OBC = 90^o + 90^o - \angle A = \angle B + \angle C$

$\angle B'CT = 90^o + \angle OCB = 90^o + 90^o - \angle A = \angle B + \angle C$

Theorem. Let the figure $\Phi$ move in such a way that it is always directly similar to a figure $\Phi_0$ and one common point $A$ of the similar figures $\Phi \sim \Phi_0$ does not move at all. Let the locus of a point $B$ of the figure $\Phi$ be a curve $\Gamma$ . Then the locus of any other point $C$ (except $A$ ) of the figure $\Phi$ is a curve $\Gamma'$ similar to the curve $\Gamma$ .

For a simple proof, see Yaglom, Geometric Transformations II (chapter 1, \1, pp. 68-69, theorem without a number between theorems 2 and 3). It follows that the locus of points $Q$ is an arc similar to the arcs $C'S \equiv (B)$ , $B'T \equiv (C)$ of the points $U, V$ . Since these 2 arcs are centered at the vertices $B, C$ of the quadrilateral $ABCD$ , the locus arc $(D)$ of the points $Q$ is centered at the point $D$ corresponding to the point $Q$ in the spiral similarity of the quadrilaterals $ABCD \sim AUVQ$ and with the same central angle $\angle B + \angle C$ . Radius $r_D$ of the circle $(D)$ is easily found when the point $P$ coincides with the vertex $A$ and the triangles $\triangle AUV \sim \triangle ABC$ and the quadrilaterals $ABCD \sim AUVQ$ become centrally similar with homothety center $A$ and homothety coefficient $\frac{AU}{AB} = \frac{\lambda - 1}{\lambda}$ . Denote $a = BC, b = CA, c = AB$ . Using the cosine theorem for the triangle $\triangle ABD$ ,
$DA = \sqrt{AB^2 + BD^2 - 2AB \cdot BD \cos{\widehat B}} = \sqrt{c^2 + \lambda^2 a^2 - 2\lambda ca \cos{\widehat B}}$
$r_D = DQ = DA + AQ = DA \left(1 + \frac{\lambda - 1}{\lambda}\right) = \lambda DA$
Substituting $\cos{\widehat B} = \frac{c^2 + a^2 - b^2}{2ca}$ , we get
$r_D = \sqrt{c^2 + \lambda^2 a^2 - \lambda(c^2 + a^2 - b^2)} = \sqrt{\lambda(\lambda - 1) a^2 + \lambda b^2 - (\lambda - 1)c^2}$
The locus arc $(D)$ of the points $Q$ is limited by the appropriate intersection $S'$ of the line $C'S$ with the circle $(D)$ , when the point $P$ becomes identical with the vertex $B$ and by the appropriate intersection $T'$ of the line $B'T$ with the circle $(D)$ , when the point $P$ becomes identical with the vertex $C$ .

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This post has been edited 3 times. Last edited by Luis González, May 15, 2022, 5:39 PM
Reason: Fixing LaTeX error

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Amir.S

#4 h May 28, 2005, 8:07 AM • 2 Y

Y by Adventure10, Mango247

ooops,sorry really I had a bad day, I wasn't OK.

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#5 h May 28, 2005, 8:39 PM • 2 Y

Y by Adventure10, Mango247

Oh, c'm on Yetti, it is so uncharacteristic of you!
Do not drop the ball half-court, take it to the hoop.

Describe the bunches of circles centered at B and C,
reveal the envelopes of the UV line for LAM=1, LAM>1, LAM<1,
disclose the role of vertex A, pinpoint the exact location of center D,
identify and explain the locus of intersection of two UV lines for different LAMs,
and I hope you can elaborate (but not too much, just a bit!) on the connection
(if there is any?) of this problem to the Simson line of Poncelet rotation problem.

That would be a worth while reply, provided you keep it under 4 pages

Maj. Pestich.

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yetti

#6 h May 28, 2005, 9:45 PM • 2 Y

Y by Adventure10, Mango247

pestich wrote:

Oh, c'm on Yetti, it is so uncharacteristic of you!
Do not drop the ball half-court, take it to the hoop.

Describe the bunches of circles centered at B and C,
reveal the envelopes of the UV line for LAM=1, LAM>1, LAM<1,
disclose the role of vertex A, pinpoint the exact location of center D,
identify and explain the locus of intersection of two UV lines for different LAMs,
and I hope you can elaborate (but not too much, just a bit!) on the connection
(if there is any?) of this problem to the Simson line of Poncelet rotation problem.

That would be a worth while reply, provided you keep it under 4 pages

Maj. Pestich.

Any geometric ankers (sic) ?

What Mathlinks really should do is to set up a new section Psychology or better yet, a new section Psychology of Yetti. That way, your life would become organized, because you could keep all your messages in one place.

Yetti

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