Brocard angle

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Rushil

#1 h Sep 23, 2005, 6:51 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Three circles touch each other externally and all these cirlces also touch a fixed straight line. Let $A,B,C$ be the mutual points of contact of these circles. If $\omega$ denotes the Brocard angle of the triangle $ABC$ , prove that $\cot{\omega}$ = 2.

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yetti

#2 h Oct 14, 2005, 3:03 AM • 1 Y

Y by Adventure10

At first, let $\triangle ABC$ be arbitrary. The Brocard point $Z_1$ is the concurrency point of 3 circles $(O_A), (O_B), (O_C)$ . The circle $(O_A)$ is tangent to the line AB at A and passes through the opposite vertex C, the circle $(O_B)$ is tangent to the line BC at B and passes through the opposite vertex A, the circle $(O_C)$ is tangent to the line CA at C and passes through the opposite vertex B. Then $\omega = \angle Z_1BC = \angle Z_1CA = \angle Z_1AB$ is the Brocard angle of the triangle $\triangle ABC$ . Similarly, the circles $(P_B), (P_C), (P_A)$ were tangent to the lines AB, BC, CA at the other vertices B, C, A, respectively, concur at the other Brocard point $Z_2$ and $\omega = \angle Z_2CB = \angle Z_1CA = \angle Z_2BA$ is the same Brocard angle of the triangle $\triangle ABC$ . We consider the 1st case of $Z_1$ .

Lemma 1: The triangle $\triangle O_AO_BO_C$ is spirally similar to the triangle $\triangle ABC$ with the similarity center $Z_1$ and rotation angle $\phi = \omega - 90^\circ$ .

The angle $\angle AZ_1B$ s spanning the arc $\widehat {AZ_1B}$ of the circle $O_B$ is equal to $180^\circ = \angle ABC$ , where $\angle ABC = \angle B$ is the angle formed by the chord AB and tangent BC of this circle at the triangle vertex B. The Brocard rays $AZ_1, BZ_1$ are the radical axes of the circle pairs $(O_B), (O_A)$ and $(O_B), (O_C)$ . Hence, their center lines $O_BO_A, O_BO_C$ are perpendicular to these Brocard rays and they form the angle $\angle O_B = \angle O_AO_BO_C = 180^\circ - \angle AZ_1B = 180^\circ - (180^\circ - \angle B) = \angle B$ . Similarly, the remaining 2 angles $\angle O_C, \angle O_A$ of the triangle $\triangle O_AO_BO_C$ are equal to $\angle O_C = \angle C, \angle O_A = \angle A$ . Thus the triangles $\triangle O_AO_BO_C \sim \triangle ABC$ are similar. The angle $\angle Z_1O_BB$ is the central angle of the Brocard angle $\angle Z_1AB = \omega$ spanning the arc $\widehat {AZ_1B}$ of the circle $(O_B)$ , hence $\angle Z_1O_BB = 2 \omega$ . Likewise, $\angle Z_1O_CC = \angle Z_1O_AA = 2\omega$ . Consequently, the isosceles triangles $\triangle Z_1O_BB \sim \triangle Z_1O_CC \sim \triangle Z_1O_AA$ are all similar and

$\frac{Z_1O_B}{Z_1B} = \frac{Z_1O_C}{Z_1C} = \frac{Z_1O_A}{Z_1A}$ ,

$\angle O_BZ_1B = \angle O_CZ_1C = \angle O_AZ_1A = \frac{180^\circ - 2\omega}{2} = 90^\circ - \omega$ ,

which implies that the Brocard point $Z_1$ is the spiral similarity center of the triangles $\triangle O_AO_BO_C \sim \triangle ABC$ . Since the oriented angle $\measuredangle O_BZ_1B$ ic clockwise for a triangle $\triangle ABC$ labeled counter-clockwise, the rotation angle is $\phi = - \angle O_BZ_1B = \omega - 90^\circ$ .

Lemma 2: Let the normals to the Brocard rays $AZ_1, BZ_1, CZ_1$ at the point $Z_1$ meet the circles $(O_A), (O_B), (O_C)$ at points D, E, F, respectively, different from $Z_1$ . The triangle $\triangle DEF$ is spirally similar to the triangle $\triangle ABC$ with the similarity center $Z_1$ and rotation angle $-90^\circ$ . The similarity coefficient of these 2 triangles is equal to $\cot \omega$ .

Since the angle $\angle AZ_1D = 90^\circ$ inscribed in the circle $(O_A)$ is right by definition, $AD$ is a diameter of this circle, its center $O_A$ is the midpoint of of this diameter and the points $A, O_A, D$ are collinear. Similarly, BE ia a diameter of the circle $(O_B)$ and the points $B, O_B, E$ are also collinear. Thus the angle $\angle BAE = 90^\circ$ is right. Since the circle $(O_A)$ is tangent to the line AB at the vertex A, the angle $\angle BAD = 90^\circ$ is also right. As a result, the points $D, O_A, A, E$ are all collinear and similarly, the points $E, O_B, B, F$ are collinear and the points $F, O_C, C, D$ are collinear. The angle $\angle DEF \equiv \angle AEB$ is spanning the arc $\widehat {AEB}$ of the circle $(O_B)$ opposite to the arc $\overarc{AZ1B}$ , hence, $\angle DEF = 180^\circ - \angle AZ_1B = 180^\circ - (180^\circ - \angle B) = \angle B$ (see the proof of lemma 1). Likewise, $\angle EFD = \angle C, \angle FDE = \angle A$ and consequently, the triangles $\triangle DEF \sim \triangle ABC$ are similar. The angles $\angle Z_1EB = \angle Z_1AB = \omega$ spanning the same arc $Z_1B$ of the circle $(O_B)$ are equal. Likewise, $\angle Z_1FC = \angle Z_1BC = \omega$ and $\angle Z_1DA = \angle Z_1CA = \omega$ . As a result, the isosceles triangles $\triangle Z_1O_BE \sim \triangle Z_1O_CF \sim \triangle Z_1O_AD$ are all similar and

$\frac{Z_1O_B}{Z_1E} = \frac{Z_1O_C}{Z_1F} = \frac{Z_1O_A}{Z_1D}$ ,

which implies that the Brocard point $Z_1$ is the common spiral similarity center of the triangles $\triangle DEF \sim \triangle O_AO_BO_C \sim \triangle ABC$ . By definition, the rotation angle of the triangle $\triangle DEF$ with respect to the triangle $\triangle ABC$ is $90^\circ$ clockwise, i.e., $-90^\circ$ counter-clockwise. The similarity coefficient of the triangles $\triangle DEF \sim \triangle ABC$ is $\frac{Z_1E}{Z_1B}$ . Since the angle $\angle BZ_1E = 90^\circ$ is right by definition and since we have shown that the angle $\angle Z_1EB = \omega$ is equal to the Brocard angle, $\frac{Z_1E}{Z_1B} = \cot \omega$ .

To prove the problem proposition, we will use only lemma 1. (I got carried away before realizing that the problem was simpler that it seemed.)

(to be continued)

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yetti

#3 h Oct 14, 2005, 3:04 AM • 1 Y

Y by Adventure10

Let $r_1, r_2$ be the radii of the 2 given externally tangent circles $(O_1), (O_2)$ . Let A be the tangency point of these 2 circles and let KL be one of their common external tangents with the tangency points $K \in (O_1), L \in (O_2)$ . We follow the usual construction of the circle $(O_3)$ tangent to the line KL and to the circles $(O_1), (O_2)$ at points B, C, i.e., we invert the circles $(O_1), (O_2)$ and the line KL in a circle (A) centered at their tangency point A lying on the center line $O_1O_2$ . We choose the radius of the inversion circle (A) to be $r = \sqrt{r_1r_2}$ . (We could also choose $r = \sqrt{2r_1r_2}$ or $r = 2 \sqrt{r_1r_2}$ , but we will stick with the original choice. This is a known trick for problems related to Archimedes' arbelos.) Since the circles $(O_1), (O_2)$ both pass through the inversion center, they are carried into straight lines $o_1, o_2$ both perpendicular to the center line $O_1O_2$ . Let the 2 circles intersect the center line $O_1O_2$ at points P, Q different from their tangency point A. Since $AP = 2r_1,\ AQ = 2r_2$ , the points P, Q are carried into the points $P', Q' \in O_1O_2$ , such that

$AP' = \frac{r^2}{AP} = \frac{r_1r_2}{2r_1} = \frac{r_2}{2}$

$AQ' = \frac{r^2}{AQ} = \frac{r_1r_2}{2r_2} = \frac{r_1}{2}$

The lines $o_1, o_2$ , i.e., the inverted circles $(O_1), (O_2)$ , are the normals to the center line $O_1O_2$ at the points P', Q', recpectively. Since the triangles $\triangle PKA \sim \triangle ALQ$ are similar and right, the angle $\angle KAL = 90^\circ$ is also right. The normal to the center line $O_1O_2$ at the tangency point A is the radical axis of the circles $(O_1), (O_2)$ , hence, it cuts the tangent segment KL at its midpoint R and consequently, AR is the median of the right angle triangle $\triangle KAL$ equal to half of its hypotenuse $KL = 2 \sqrt{r_1r_2}$ (the external tangent distance of 2 externally tangent circles). Thus $AR = \frac{KL}{2} = \sqrt{r_1r_2} = r$ . As a result, the inversion circle (A) with radius r cuts the common external tangent KL at the midpoint R (and at some other point S). Our inversion carries the line $t \equiv KL$ into a circle (T) passing through the inversion center A, through the points R, S, and tangent to the 2 parallel lines $o_1 \parallel o_2$ separated by the distance $P'Q' = AP' + AQ' = \frac{r_1 + r_2}{2}$ . Thus the radius of the circle (T) is $\frac{P'Q'}{2} = \frac{r_1 + r_2}{4}$ . The unknown circle $(O_3)$ tangent to the circles $(O_1), (O_2)$ at points B, C is carried into a circle $(O_3')$ also tangent to the parallel lines $o_1 \parallel o_2$ at points B', C' and to the circle (T). The tangency points B, C of the circle $(O_3)$ are the intersections of the rays AB', AC' with the circles $(O_1), (O_2)$ . Thus we obtained the defined triangle $\triangle ABC$ , which is all we need. Anyway, the tangency point U of the circle $(O_3)$ with the line $t \equiv KL$ is the intersection of the ray AU' with the line KL and $(O_3)$ is the circumcircle of the triangle $\triangle BCU$ . Or, the center of the circle $(O_3)$ is the intersection of the perpendicular bisector of the segment BC with the ray $AO_3'$ (because a circle and its inverted image are centrally similar with the homothety center at the inversion center).

Because of the basic properties of inversion, the triangles $\triangle AC'B' \sim \triangle ABC$ are similar and consequently, they have the same Brocard angle. Thus we do not have to worry about the triangle $\triangle ABC$ at all and we can find the Brocard angle of the inverted triangle $\triangle AB'C'$ instead. Obviously, the side B'C' of this triangle is perpendicular to the parallel lines $o_1 \parallel o_2$ and from the rectangle P'Q'C'B', $B'C' = P'Q' = \frac{r_1 + r_2}{2}$ . The other 2 sides of this rectangle are equal to

$P'B' = Q'C' = \frac{AR}{2} + TU' + U'O_3' = \frac{r_1 + r_2 + \sqrt{r_1r_2}}{2}$

because $TU' = U'O_3' = \frac{r_1 + r_2}{4}$ are radii of the congruent circles $(T), (O_3')$ and the circle (T) is centered on the perpendicular bisector of the segment $AR = \sqrt{r_1r_2}$ (see above). By Pythagorean theorem for the right angle triangles $\triangle AP'B', \triangle AQ'C'$ their hypotenuses AB', AC' are

$AB'^2 = AP'^2 + P'B'^2 = \frac{r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4}$

$AC'^2 = AQ'^2 + Q'C'^2 = \frac{r_1^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4}$

Let $(O_B)$ be a circle tangent to the side B'C' of the triangle $\triangle AB'C'$ at the vertex B' an passing through the opposite vertex A. Since $B'C' \perp P'B'$ , the circle $(O_B)$ is centered on the line $o_1 \equiv P'B'$ , on its intersection with the perpendicular bisector of the segment AB'. Let $(O_C)$ be a circle tangent to the side AC' at the vertex C' and passing through the opposite vertex B'. The circle $(O_C)$ is centered on the perpendicular bisector of the segment B'C', which is parallel to the lines $o_1 \equiv P'B', o_2 \equiv Q'C'$ . Similarly, we could define circle $(O_A)$ , but we do not need it. According to lemma 1, the side B'C' of the triangle $AB'C'$ forms the angle $90^\circ - \omega$ with the side $O_BO_C$ of the triangle $\triangle O_AO_BO_C$ ( $\omega$ is the Brocard angle). Therefore, the line $O_BO_C$ forms the angle $\omega$ with the perpendicular $o_1 \equiv P'B'$ to the line B'C'. Since $O_B \in P'B'$ , the angle $\angle O_CO_BB' = \omega$ is the Brocard angle of the triangles $\triangle AB'C' \sim \triangle O_AO_BO_C$ . Let a normal to the line $o_1 \equiv P'B' \equiv O_BB'$ through the point $O_C$ intersects this line at a point V. Then from the right angle triangle $\triangle O_BO_CV$ , $\cot \omega = \frac{O_BV}{OC_V}$ . We know one leg $O_CV = O_3'B' = \frac{r_1 + r_2}{4}$ of this triangle. Now we have to calculate the other leg.

Let M, N be the midpoints of the sides AB', B'C'. (Obviously, $N \equiv O_3$ is the center of the circle $(O_3')$ .) The right angle triangles $\triangle O_BMB' \sim \triangle AP'B'$ are similar, having the angle at the vertex B' common. Hence,

$\frac{O_BB'}{MB'} = \frac{AB'}{P'B'}$

$O_BB' = \frac{MB' \cdot AB'}{P'B'} = \frac{AB'^2}{2P'B'} = \frac{r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4\left(r_1 + r_2 + \sqrt{r_1r_2}\right)}$

Likewise, the right angle triangles $\triangle C'NO_C \sim \triangle AQ'C'$ are similar, because the angle $\angle AC'O_C = 90^\circ$ is right due to the tangency of the circle $(O_C)$ to the line B'C' at the point C'. Therefore, the angles $\angle AC'N + \angle NC'O_C = 90^\circ$ add up to a right angle and $\angle NC'O_C = 90^\circ - \angle AC'N = 90^\circ - \angle C'AP' = \angle AC'P'$ . Hence,

$\frac{O_CN}{NC'} = \frac{AQ'}{Q'C'}$

$O_CN = \frac{NC' \cdot AQ'}{Q'C'} = \frac{(r_1 + r_2)r_1}{4\left(r_1 + r_2 + \sqrt{r_1r_2}\right)}$

The other leg of the right angle triangle triangle $\triangle O_BO_CV$ and $\cot \omega$ are then

$O_BV = O_BB' + B'V = O_BB' + O_CN = \frac{(r_1 + r_2)r_1 + r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4\left(r_1 + r_2 + \sqrt{r_1r_2}\right)}$

$\cot \omega = \frac{O_BV}{O_CV} = \frac{(r_1 + r_2)r_1 + r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{(r_1 + r_2)\left(r_1 + r_2 + \sqrt{r_1r_2}\right)} =$

(factoring out $r_1 + r_2$ )

$= \frac{(r_1 + r_2)r_1 + r_2^2 + (r_1 + r_2)^2 + 2(r_1 + r_2) \sqrt{r_1r_2} + r_1r_2}{(r_1 + r_2)(r_1 + r_2 + \sqrt{r_1r_2})} =$

$= \frac{(r_1 + r_2)\left(r_1 + r_1 + r_2 + 2 \sqrt{r_1r_2}\right) + r_2(r_1 + r_2)}{(r_1 + r_2)\left(r_1 + r_2 + \sqrt{r_1r_2}\right)} =$

$= \frac{2r_1 + 2r_2 + 2 \sqrt{r_1r_2}}{r_1 + r_2 + \sqrt{r_1r_2}} = 2$

Q.E.D.

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Rushil

#4 h Oct 14, 2005, 8:45 AM • 2 Y

Y by Adventure10, Mango247

Excellent proofs!!!!1

Can you tell me which program you used for the wonderful diagrams and where I can sownload them???

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yetti

#5 h Oct 15, 2005, 2:15 PM • 2 Y

Y by Adventure10, Mango247

Tx. I use The Geometer's Sketchpad to make the drawings. This program is not available for a free download, but it is very nice and very cheap ( $\$$ 40). See the last message in Geometrical problem for more info. The label "AGif UNREGISTERED" stands just for the Active GIF Creator, which is a graphics converter. Unregistered version (not purchased, just downloaded) puts this label into the GIF images. Search the web, I got many hits for "Active GIF Creator". Normally, I use Adobe Photoshop to export the Sketchpad drawings into the transparent GIF format, but I did not have it on the computer, which I used at this time.

Yetti.

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