I think it is best to change the variables:

X = x - 1/2, Y = y - 1/2, Z = z - 1/2.

You'll then get

(Y+Z)A² + (Z+X)B² + (X+Y)C² - 2ZAC - 2YAC - 2XBC = 0 and

(1/2 - X)(1/2 - Y)(1/2 - Z) = (1/2 + X)(1/2 + Y)(1/2 + Z).

The first one could be rewritten as

(*)	X(B - C)2 + Y(C - A)2 + Z(A - B)2 = 0

that suggests, I think, some degree of freedom in finding X, Y, Z, even if restricted by the second condition.

On the other hand, if the sum

(B - C)² + (C - A)² + (A - B)²

is zero and, assuming the three terms in it are not collinear, then (*) necessarily implies

X = Y = Z = 0.

This could be rewritten as

A(AY + AZ - BZ - CY) +
B(BX + BZ - AZ - CX) +
C(CX + CY - AY - BX) = 0, or

(1)	Am + Bn + Ck = 0.

Note that m + n + k = 0, even if not all of X, Y, Z is 0.

If any of m, n, or k is not 0 then A, B, C are collinear. For example, if, say, m is not 0, then (1) implies

(2)	(B-A)n + (C-A)k = 0,

so that (B-A) and (C-A) are linearly dependent, which could not be.

Thus we must have

(3)	m = n = k = 0,

which is a linear system of three equations in X, Y, Z. The determinant of the system is

0		A-C		A-B
B-C		0		B-A
C-B		C-A		0

which appears to be 0. So there bound to be non zero X, Y, Z that satisfy (3).

Now, whether they satisfy your second condition I can't tell right away.

The above problem is my first cut at Problem 1711 from Vol. 78, No. 1, February 2005 of "Mathematics Magazine" from the MAA.

In triangle ABC, let A' be on BC, B' on CA, and C' on AB, and suppose that the cevians AA', BB', and CC' meet at M. Prove that if triangle ABC is similar to triangle A'B'C', then M is the centroid of triangle ABC.

Does the word "cevians" in the problem statement force the point M to be on the triangle or in its interior?

Cevian: the "line segment" from a vertex of a triangle to a point on the opposite side or its extension.

If A, B, C, A', B', and C' are considered as complex numbers, then

A' = xB + (1-x)C
B' = yC + (1-y)A
C' = zA + (1-z)B

, where 0 <= x,y,z <= 1.

If that last restriction on x, y, and z is not true, then I don't think the conclusion of the problem is true.

No, M could be any point.

>A' = xB + (1-x)C
>B' = yC + (1-y)A
>C' = zA + (1-z)B
>
>, where 0 <= x,y,z <= 1.

Only if the primed points are on the sides of the triangle, not their extensions. On the extensions x, y, x could be greater than 1 or negative.

>If that last restriction on x, y, and z is not true, then I
>don't think the conclusion of the problem is true.

You have an example?

The fact that A'B'||AB means that x = 1/y, etc. So that

xyz = 1.

Hint: interpret the barycentric coordinates of M in terms of x, y, z.

If the point M was on the other side of line BC from vertex A, then the cevian AA' ( by definition ) could not contain M.

>
>>If that last restriction on x, y, and z is not true, then I
>>don't think the conclusion of the problem is true.
>
>You have an example?
>

I think the following is an example:

Let X = -2/5 and Y = Z = 1 with

A = 0, B = 1, and C = (5 + i*sqrt(5))/10

Then

A' = (5 + 3*i*sqrt(5))/20
B' = -(5 + i*sqrt(5))/20
C' = 1/10

Triangles ABC and A'B'C' are similar if

(C-A)/(B-A) = (C'-A')/(B'-A')

The point M in this case is definitely
not the centroid of triangle ABC.

Something is wrong here, because

if Y = 1, then B' = C
if Z = 1, then C' = A

I'm sorry for the confusion. I'm using

A' = xB + (1-x)C
B' = yC + (1-y)A
C' = zA + (1-z)B

with X, Y, and Z solving the following two equations:

(1-2X)(1-2Y)(1-2Z) = (1+2X)(1+2Y)(1+2Z)

X(A-B)^2 + Y(B-C)^2 + Z(C-A)^2 = 0

The change of variables is

x = 1/2 - Y = 1/2 - 1 = -1/2
y = 1/2 - Z = 1/2 - 1 = -1/2
z = 1/2 - X = 1/2 - (-2/5) = 9/10

Hope this clears things up.

But then AA', BB', CC' do not meet.

Also, I do not think that the problem from Math Magazine meant to include complex solutions. As I already mentioned, you must be able to reformulate the problem in terms of barycentric, or trilinear, coordinates.

If M = (a, b, c), where a+b+c=1, then, say,

C' = bA + aB

This must have repercussions on your x, y, z.

When you say they do not meet, do you mean the line segments
do not meet or the lines do not meet? I agree that the line
segments do not meet, but the lines meet at a point outside
the triangle ABC.

>
>Also, I do not think that the problem from Math Magazine
>meant to include complex solutions....
>

The complex numbers are just a way to describe two dimensional
points. If you were to plot the points

A = (0,0)
B = (1,0)
C = (1/2,s/20)
A' = (1/4,3s/20)
B' = (-1/4,-s/20)
C' = (1/10,0)

you would see that A', B', and C' lie on
lines BC, CA, and AB respectively; that
triangles ABC and A'B'C' are similar; that
lines AA', BB', and CC' are concurrent at
a point outside triangle ABC.

No, I mean their extensions do not meet. With x, y, z, as above, this would violate Ceva's theorem.

>The complex numbers are just a way to describe two
>dimensional
>points.

I thought you were looking for complex x, y, z.

If you were to plot the points
>
> A = (0,0)
> B = (1,0)
> C = (1/2,s/20)
>A' = (1/4,3s/20)
>B' = (-1/4,-s/20)
>C' = (1/10,0)
>
>you would see that A', B', and C' lie on
>lines BC, CA, and AB respectively; that
>triangles ABC and A'B'C' are similar; that
>lines AA', BB', and CC' are concurrent at
>a point outside triangle ABC.

No, I do not believe that. A can't lie on the line through C and B', because their y's are symmetric in 0, while their x's are not.

And in any event, if those A', B', C', have anything to do with x, y, z, as you defined them, then Ceva's theorem would not hold.

x, y, z are OK. A', B' C' are not. As I said, for the given points, B' does not even lie on AC.

However, if the points A', B', C' are computed correctly for the given x, y, z, triangle ABC and A'B'C' cease to be similar.

Sorry for that. All the calculations were correct, but I copied
down the wrong value for C. The values are

A = (0,0)
B = (1,0)
C = (1/2,s/10)
A' = (1/4,3s/20)
B' = (-1/4,-s/20)
C' = (1/10,0)

where s = sqrt(5).

That's OK. But, still, If you compute A', B', C' with your x, y, z, they triangles are no longer similar.

A' = xB + (1-x)C = (-1/2)(1) + (1-(-1/2))(C) = -1/2 + (3/2)C

B' = yC + (1-y)A = (-1/2)(C) + (1-y)(0) = (-1/2)C

C' = zA + (1-z)B = (9/10)(0) + (1-9/10)(1) = 1/10

Triangles ABC and A'B'C' are similar if

 C-A     C'-A'
----- = -------.
 B-A     B'-A' 

 C-0       1/10 - (-1/2 + (3/2)C)
----- = ---------------------------
 1-0     (-1/2)C - (-1/2 + (3/2)C)

         3/5 - (3/2)C 
    C = --------------
           1/2 - 2C

(1/2)C - 2C^2 = 3/5 - (3/2)C

0 = 10C^2 - 10C + 3 

     10 +- sqrt(10^2 - 4*10*3)
C = ---------------------------
               2*10

C = 1/2 +- sqrt(-20)/20

C = 1/2 +- 2*i*sqrt(5)/20

C = (1/2,s/10)

This is how I came up with the value of C 
in the first place (I picked the plus sign).

There's no mistaking the fact that ABC is isosceles while A'B'C' is not.

Where am I making the mistake in the following:

|A'-C'|^2 = |B'-C'|^2

(1/4 - 1/10)^2 + (3s/20 - 0)^2 = (-1/4 - 1/10)^2 + (-s/20 - 0)^2

(3/20)^2 + (3s/20)^2 = (7/20)^2 + (s/20)^2

9 + 9s^2 = 49 + s^2

s^2 = 5

The triangles are similar and the point M does not coincide with the center of ABC.

Thanks. I was beginning to think I needed a refresher course
in arithmetic.

So, let ABC and A'B'C' be similar with spiral similarity f (that is, f(A) = A', f(B) = B' and f(C) = C'). Let A' lie on BC, B' on CA and C' on AB. Let AA', BB' and CC' be concurrent at P. Let P' be f(P). In fact, since f is so significant, I shall denote f(X) by X'for a general X. Trivially if P = P', we must have that the angle of rotation of f is pi and so that P is the centroid of ABC. Conversely, the medial triangle and centroid satisfy all the given conditions.

We examine the more interesting case of P' != P. In that case, it is clear that A'PB' = APB = A'P'B' so that A'PP'B' is cyclic. Similarly, B'PP'C' is cyclic and so is C'PP'A'. It follows that both P and P' lie on the circumcircle of A'B'C'. Now let K be the fixed point of f. PKP' = PAP' (Both are the angle of rotation of f) and so K also lies on this circumcircle.

Let S_A be the circle B'C'H', and define S_B and S_C similarly. The reason for doing this is that since B'AC' = C'A'B', A must lie on S_A and similarly for B and C. Then we have that angles H'AC' and C'BH' are fixed, and since AB passes through C', the spiral similarity taking A to B with centre H' is fixed in terms of A'B'C'. Similar comments may be made with respect to the other pairs of vertices. Now let A*B*C* be the anticomplementary triangle of A'B'C'. Note that these comments apply replacing A by A*, B by B* and C by C*. So there is a spiral similarity g centred at H' and taking ABC to A*B*C*.

Now we know that scaling through G' by a factor -1/2 (I will denote this by t) takes A*B*C* to A'B'C'. Let L = g(K), and observe that f must be the composition of t with g, so t(L) = K, so in particular G' lies on KL. Now, H'KL = H'AA* = pi/2, and so K lies on the orthocentroidal circle of A'B'C'. We deduce that the orthocentroidal circle and circumcircle of A'B'C' intersect, and so that A'B'C' must be obtuse angled. Hence ABC must also be obtuse angled.

If ABC is strictly obtuse angled then backtracking through that lot provides exactly 2 triangles A'B'C' satisfying the desired condition other than the medial triangle. These correspond to the two points of intersection of the orthocentroidal circle with the circumcircle. If ABC is right angled, the construction breaks down and there are no other solutions. The above argument shows there are no solutions for an acute angled triangle.

This leads me to suspect that the original problem was posed for an acute angled triangle and that that condition got lost somewhere down the line. I apologise for not including a diagram. I do not have the necessary software to hand.

Thankyou

sfwc
<><

Why to go to such a length?

i² = -1 and also
(-i)² = -1.

So that

i = sqrt(-1) = -i.

Or, without involving complex numbers,

1 = sqrt(1²) = sqrt((-1)²) = -1

Hence

1 = -1.

What's wrong?