Find the value of angle A'BC

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Find the value of angle A'BC

G H J

Reveal topic tags

analytic geometry

L

G H BBookmark kLocked kLocked NReply

Source: Vietnam TST 1992 for the 33nd IMO, problem 3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3647 posts

orl

#1 h Jun 25, 2005, 10:45 PM • 1 Y

Y by Adventure10

Let $ABC$ a triangle be given with $BC = a$ , $CA = b$ , $AB = c$ ( $a \neq b \neq c \neq a$ ). In plane ( $ABC$ ) take the points $A'$ , $B'$ , $C'$ such that:

I. The pairs of points $A$ and $A'$ , $B$ and $B'$ , $C$ and $C'$ either all lie in one side either all lie in different sides under the lines $BC$ , $CA$ , $AB$ respectively;

II. Triangles $A'BC$ , $B'CA$ , $C'AB$ are similar isosceles triangles.

Find the value of angle $A'BC$ as function of $a, b, c$ such that lengths $AA', BB', CC'$ are not sides of an triangle. (The word "triangle" must be understood in its ordinary meaning: its vertices are not collinear.)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#2 h Jul 1, 2005, 5:42 PM • 2 Y

Y by Adventure10, Mango247

Since it is not too likely that JHC will join this forum in the next time, let me write what he would probably have written on this problem (alas, not in his wonderful style):

My solution will require some knowledge about barycentric coordinates. Particularly, I will use some results from [1] (don't worry, only the first two pages of the article will be needed); hence, I rewrite the problem with the terminology from [1]:

Problem. Let ABC be a triangle. Assume that the triangle ABC is directed clockwisely (i. e. the angle < CAB is mathematically positive). Let $\phi$ be an angle (we use directed angles modulo 180°), and let $A^\phi$ , $B^\phi$ , $C^\phi$ be the apices of the isosceles triangles $BA^\phi C$ , $CB^\phi A$ , $AC^\phi B$ erected on the sides BC, CA, AB of triangle ABC, having the bases BC, CA, AB and the base angles

$\measuredangle CBA^\phi=\measuredangle A^\phi CB=\measuredangle ACB^\phi=\measuredangle B^\phi AC=\measuredangle BAC^\phi=\measuredangle C^\phi BA=\phi$ .

Find all angles $\phi$ such that the segments $AA^\phi$ , $BB^\phi$ , $CC^\phi$ are not the sidelengths of a non-degenerate triangle.

Solution. First we prove a lemma:

Lemma 1. Let ABC be a triangle, and let X, Y, Z be points in its plane such that the triangles BXC, CYA, AZB are all directly similar. Then, the segments AX, BY, CZ always are the sidelengths of a (possibly degenerate) triangle; this triangle degenerates if and only if the lines AX, BY, CZ are all parallel.

Proof of Lemma 1. There are some synthetic proofs, but I am too lazy to explain them, so I will use complex numbers:

We work in the Gaussian plane. Let a, b, c, x, y, z be the affices of the points A, B, C, X, Y, Z, respectively. Since the triangles BXC, CYA, AZB are all directly similar, we have $\frac{x-c}{b-c}=\frac{y-a}{c-a}=\frac{z-b}{a-b}$ . Denote $\frac{x-c}{b-c}=\frac{y-a}{c-a}=\frac{z-b}{a-b}=k$ . Then, $\frac{x-c}{b-c}=k$ becomes $x-c=k\left(b-c\right)$ , what quickly transforms into $x=kb+\left(1-k\right)c$ ; similarly, we get $y=kc+\left(1-k\right)a$ and $z=ka+\left(1-k\right)b$ . Hence,

$\left(a-x\right)+\left(b-y\right)+\left(c-z\right)$
$=\left(a-\left(kb+\left(1-k\right)c\right)\right)+\left(b-\left(kc+\left(1-k\right)a\right)\right)+\left(c-\left(ka+\left(1-k\right)b\right)\right)=0$ .

Hence, $\left(a-x\right)+\left(b-y\right)=z-c$ . Thus, by the triangle inequality, $\left|a-x\right|+\left|b-y\right|\geq\left|\left(a-x\right)+\left(b-y\right)\right|=\left|z-c\right|$ . In other words, $AX+BY\geq CZ$ . Similarly, $BY+CZ\geq AX$ and $CZ+AX\geq BY$ . Thus, the segments AX, BY, CZ are the sidelengths of a (possibly degenerate) triangle. This triangle degenerates if and only if at least one of the triangle inequalities $AX+BY\geq CZ$ , $BY+CZ\geq AX$ and $CZ+AX\geq BY$ becomes an equality. But this happens if and only if the lines AX, BY, CZ are all parallel. In fact, for instance, the triangle inequality $AX+BY\geq CZ$ is equivalent to $\left|a-x\right|+\left|b-y\right|\geq\left|\left(a-x\right)+\left(b-y\right)\right|$ (as we saw above), and thus, if it becomes an equality, the vectors a - x and b - y must be collinear, what, of course, forces the vector c - z to be also collinear with both of them (since $\left(a-x\right)+\left(b-y\right)+\left(c-z\right)=0$ ), so that the vectors a - x, b - y and c - z are all collinear, i. e. we have AX || BY || CZ; conversely, if we have AX || BY || CZ, then the vectors a - x, b - y and c - z are all collinear, and if we WLOG assume that the vectors a - x and b - y have the same orientation (in fact, among three collinear vectors, one can always find two with the same orientation), then the triangle inequality $\left|a-x\right|+\left|b-y\right|\geq\left|\left(a-x\right)+\left(b-y\right)\right|$ becomes an equality, and thus the triangle inequality $AX+BY\geq CZ$ becomes an equality. Thus, Lemma 1 is proven.

Now, let's come to the solution of the problem. The triangles $BA^\phi C$ , $CB^\phi A$ , $AC^\phi B$ are isosceles triangles with the same (directed) base angles; thus, they are all directly similar, so that Lemma 1 applies. As a result, we see that the segments $AA^\phi$ , $BB^\phi$ , $CC^\phi$ always are the sidelengths of a (possibly degenerate) triangle, and that this triangle degenerates if and only if the lines $AA^\phi$ , $BB^\phi$ , $CC^\phi$ are all parallel.

So we are searching for the angles $\phi$ which make the lines $AA^\phi$ , $BB^\phi$ , $CC^\phi$ parallel. Actually, this can be done by direct trig bash, but there is also a nice way using the notations from [1]:

According to [1], §2, the lines $AA^\phi$ , $BB^\phi$ , $CC^\phi$ concur at the so-called Kiepert perspector $K\left(\phi\right)$ of triangle ABC. Hence, these lines are parallel if and only if this Kiepert perspector $K\left(\phi\right)$ lies on the line at infinity. Since the homogeneous barycentric coordinates of the point $K\left(\phi\right)$ are $K\left(\phi\right)=\left(\frac{1}{S_A+S_\phi}:\frac{1}{S_B+S_\phi}:\frac{1}{S_C+S_\phi}\right)$ , this point $K\left(\phi\right)$ lies on the line at infinity if and only if $\frac{1}{S_A+S_\phi}+\frac{1}{S_B+S_\phi}+\frac{1}{S_C+S_\phi}=0$ . This is equivalent to $\left(S_BS_C+S_CS_A+S_AS_B\right)+2\left(S_A+S_B+S_C\right)S_\phi+3S_\phi^2=0$ . Using the formulas $S_BS_C+S_CS_A+S_AS_B=S^2$ and $S_A+S_B+S_C=S_\omega$ , this rewrites as $S^2+2S_\omega S_\phi+3S_\phi^2=0$ . This is a quadratic equation for $S_\phi$ , and the two solutions are

$S_\phi=\frac{-S_\omega\pm\sqrt{S_\omega^2-3S^2}}{3}$ .

Since $S_\theta=S\cdot\cot\theta$ for any angle $\theta$ , this yields $\cot\phi=\frac{-\cot\omega\pm\sqrt{\cot^2\omega-3}}{3}$ . Using $\cot\omega=\frac{a^2+b^2+c^2}{4\Delta}$ , where $\Delta$ is the area of triangle ABC, and applying Heron's formula for $\Delta$ , sooner or later we arrive at

$\cot\phi=\frac{-\left(a^2+b^2+c^2\right)\pm 2\sqrt{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}}{12\Delta}$ .

This is the required value of $\phi$ . Do you call this a good olympiad problem?

References

[1] Floor van Lamoen and Paul Yiu, The Kiepert Pencil of Kiepert Hyperbolas, Forum Geometricorum 1 (2001) p. 125-132.

Darij

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a