We are given three non-collinear points $A$, $B$, and $C$. Point $C$ is the midpoint of segment $[BD]$. Let $f$ be an affine transformation defined by $f(A) = A$, $f(B) = D$, and $f(C) = B$. We need to: 1. Determine the set of invariant points under $f$.

GeometryAffine TransformationBarycentric CoordinatesGeometric TransformationsInvariant PointsVectors
2025/5/14

1. Problem Description

We are given three non-collinear points AA, BB, and CC. Point CC is the midpoint of segment [BD][BD].
Let ff be an affine transformation defined by f(A)=Af(A) = A, f(B)=Df(B) = D, and f(C)=Bf(C) = B.
We need to:

1. Determine the set of invariant points under $f$.

2. Deduce that there exists a point $G$ on the line $(BC)$ that is invariant under $f$, and express $\vec{GB}$ in terms of $\vec{GC}$.

3. Prove that $f$ is an affinity and specify its characteristic elements.

2. Solution Steps

1. Invariant points:

Let MM be an invariant point. Then f(M)=Mf(M) = M.
We can write MM as a barycentric combination of AA, BB, and CC: M=αA+βB+γCM = \alpha A + \beta B + \gamma C, where α+β+γ=1\alpha + \beta + \gamma = 1.
Applying ff to MM, we have:
f(M)=αf(A)+βf(B)+γf(C)=αA+βD+γBf(M) = \alpha f(A) + \beta f(B) + \gamma f(C) = \alpha A + \beta D + \gamma B.
Since f(M)=Mf(M) = M, we have:
αA+βD+γB=αA+βB+γC\alpha A + \beta D + \gamma B = \alpha A + \beta B + \gamma C.
βD+γB=βB+γC\beta D + \gamma B = \beta B + \gamma C.
Since CC is the midpoint of [BD][BD], we have 2C=B+D2C = B + D, or D=2CBD = 2C - B.
Substituting this into the equation, we get:
β(2CB)+γB=βB+γC\beta (2C - B) + \gamma B = \beta B + \gamma C.
2βCβB+γB=βB+γC2\beta C - \beta B + \gamma B = \beta B + \gamma C.
2βCβB+γBβBγC=02\beta C - \beta B + \gamma B - \beta B - \gamma C = 0.
2βC2βB+γBγC=02\beta C - 2\beta B + \gamma B - \gamma C = 0.
(2βγ)C+(γ2β)B=0(2\beta - \gamma)C + (\gamma - 2\beta)B = 0.
Since BB and CC are linearly independent (as A,B,CA, B, C are non-collinear), we must have:
2βγ=02\beta - \gamma = 0 and γ2β=0\gamma - 2\beta = 0.
Thus, γ=2β\gamma = 2\beta.
Since α+β+γ=1\alpha + \beta + \gamma = 1, we have α+β+2β=1\alpha + \beta + 2\beta = 1, so α+3β=1\alpha + 3\beta = 1, or α=13β\alpha = 1 - 3\beta.
Therefore, the set of invariant points is the line A+λ(B12C)A + \lambda(B - \frac{1}{2}C), where λ\lambda is a scalar.

2. Invariant point $G$ on line $(BC)$:

Let GG be a point on line (BC)(BC) such that f(G)=Gf(G) = G. We can write G=λB+(1λ)CG = \lambda B + (1 - \lambda)C for some scalar λ\lambda.
f(G)=λf(B)+(1λ)f(C)=λD+(1λ)Bf(G) = \lambda f(B) + (1 - \lambda) f(C) = \lambda D + (1 - \lambda) B.
Since CC is the midpoint of [BD][BD], D=2CBD = 2C - B.
f(G)=λ(2CB)+(1λ)B=2λCλB+BλB=(12λ)B+2λCf(G) = \lambda (2C - B) + (1 - \lambda) B = 2\lambda C - \lambda B + B - \lambda B = (1 - 2\lambda) B + 2\lambda C.
Since f(G)=Gf(G) = G, we have (12λ)B+2λC=λB+(1λ)C(1 - 2\lambda)B + 2\lambda C = \lambda B + (1 - \lambda)C.
Equating coefficients, we get 12λ=λ1 - 2\lambda = \lambda and 2λ=1λ2\lambda = 1 - \lambda.
From either equation, we get 3λ=13\lambda = 1, so λ=13\lambda = \frac{1}{3}.
Thus, G=13B+23CG = \frac{1}{3}B + \frac{2}{3}C. This means that GG is the barycenter of (B,1)(B, 1) and (C,2)(C, 2).
Now, we need to express GB\vec{GB} in terms of GC\vec{GC}.
G=13B+23CG = \frac{1}{3}B + \frac{2}{3}C implies 3G=B+2C3G = B + 2C.
Then, GB=GB=G(3G2C)=2C2G=2(GC)=2CG=2GC\vec{GB} = \vec{G} - \vec{B} = \vec{G} - (3\vec{G} - 2\vec{C}) = 2\vec{C} - 2\vec{G} = -2(\vec{G} - \vec{C}) = -2\vec{CG} = 2\vec{GC}.
So, GB=2GC\vec{GB} = 2\vec{GC}.

3. Affine transformation:

We have f(A)=Af(A) = A, f(B)=Df(B) = D, f(C)=Bf(C) = B, and D=2CBD = 2C - B. Since f(G)=Gf(G) = G and G=13B+23CG=\frac{1}{3}B+\frac{2}{3}C, the line (AG)(AG) is invariant, and (AG)(AG) is the axis. Let's consider the direction BC\vec{BC}. f(BC)=f(BC)=f(B)f(C)=DB\vec{f(BC)} = f(\vec{BC}) = \vec{f(B)f(C)} = \vec{DB}. Since CC is the midpoint of BDBD, we have DB=2BC\vec{DB} = -2\vec{BC}. Thus ff is an affinity with axis (AG)(AG), direction BC\vec{BC} and ratio -

2. Therefore, f is an affinity of axis (AG) and ratio -

2.

3. Final Answer

1. The set of invariant points is the line defined by $A + \lambda(B - \frac{1}{2}C)$, where $\lambda$ is a scalar.

2. $G = \frac{1}{3}B + \frac{2}{3}C$, and $\vec{GB} = 2\vec{GC}$.

3. $f$ is an affinity with axis $(AG)$ and ratio $-2$.

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