The coloring... again

Geometry Level 4

A line drawn through the centroid of a triangle intersects two sides internally, forming a yellow triangle and a green quadrilateral, as shown below. What is the minimum possible ratio of the yellow area to the green area?

The answer is 0.8.

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5 solutions

Mark Hennings
Nov 12, 2017

Suppose that the origin lies at the triangle vertex that is one of the three vertices of the yellow triangle, and let the other two triangle vertices have position vectors a \mathbf{a} and b \mathbf{b} with respect to the origin. If the endpoints of the line that defines the yellow triangle have position vectors λ a \lambda\mathbf{a} and μ b \mu\mathbf{b} for some 0 < λ , μ < 1 0 < \lambda,\mu < 1 , then the yellow triangle has area 1 2 λ μ a × b \tfrac12\lambda\mu|\mathbf{a}\times\mathbf{b}| , while the whole triangle has area 1 2 a × b \tfrac12|\mathbf{a}\times\mathbf{b}| . Thus the ratio of the yellow and green areas is λ μ 1 λ μ \frac{\lambda\mu}{1-\lambda\mu} .

The centroid of the triangle has position vector 1 3 ( a + b ) \tfrac13(\mathbf{a}+\mathbf{b}) , and the three points with position vectors λ a \lambda\mathbf{a} , 1 3 ( a + b ) \tfrac13(\mathbf{a}+\mathbf{b}) and μ b \mu\mathbf{b} must be collinear, which implies that μ = λ 3 λ 1 \mu = \frac{\lambda}{3\lambda - 1} . For this to be consistent with the fact that both λ \lambda and μ \mu lie between 0 0 and 1 1 , we deduce that in fact 1 2 < λ , μ < 1 \tfrac12 < \lambda,\mu < 1 .

Thus we need to minimize λ μ 1 λ μ = λ 2 3 λ 1 λ 2 \frac{\lambda\mu}{1 - \lambda\mu} = \frac{\lambda^2}{3\lambda - 1 - \lambda^2} over the range 1 2 < λ < 1 \tfrac12 < \lambda < 1 . This happens when λ = μ = 2 3 \lambda = \mu = \tfrac23 . The minimum ratio is 4 5 \boxed{\tfrac45} .

How was mu=lambda/(3lambda-1) determined?

Sean Chen - 3 years, 6 months ago

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We want 1 3 ( a + b ) λ a \tfrac13(\mathbf{a} + \mathbf{b}) - \lambda \mathbf{a} and μ b λ a \mu\mathbf{b} - \lambda\mathbf{a} to be parallel, so that one is a multiple of the other. Thus 1 3 λ = ξ λ 1 3 = ξ μ \tfrac13 - \lambda \; = \; -\xi\lambda \hspace{2cm} \tfrac13 \; = \; \xi\mu Eliminating ξ \xi from these equations gives us the answer.

Mark Hennings - 3 years, 6 months ago

How did you minimize in the last step?!

Mohamed Osama - 3 years, 6 months ago

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Look for a turning point by differentiating the function of λ \lambda .

Mark Hennings - 3 years, 6 months ago
David Vreken
Nov 26, 2017

If the line division is not parallel to the third side, then the yellow section below the parallel and the green section above the parallel can be switched by a 180° rotation (as pictured below), giving a yellow area larger than a yellow area of a parallel line division.
Therefore, a minimum yellow area (and a minimum yellow to green ratio) must occur when the line through the centroid is parallel to the third side of the triangle.

Since all centroids divide a median in a 2:1 proportion, a second parallel line can be drawn through the yellow section halfway up, and 1 of its 2 parts would be equal to 1 of the 4 parts along the parallel line through the centroid, and 1 of the 6 parts along the base. The triangle can then be subdivided into congruent parallelograms using the base and the median though the base as guidelines (as pictured below), and the each of triangular parts can be matched with a matching triangular part through a 180° rotation to form more congruent parallelograms (also pictured below). After all the transformations, there are 9 congruent parallelograms – 4 yellow and 5 green, for a ratio of 4/5 or 0.8.

i think your solution was the best !!

Relue Tamref - 3 years, 6 months ago
Kevin Tong
Nov 22, 2017

This is a brute force method using calculus. Assume we have a right triangle A B C ABC with coordinates A ( 0 , 0 ) A(0,0) , B ( 0 , 3 ) B(0,3) , and C ( 3 , 0 ) C(3,0) . This way, the centroid is at point 1 , 1 1,1 . Now, assume our line has y intercept α \alpha and crosses through the centroid to make a triangle with the x and y axis. The equation for the line would then be y = ( 1 α ) x + α y=(1-\alpha)x+\alpha . To find the x intercept, we set y = 0 y=0 to get x = α α 1 x=\frac{\alpha}{\alpha -1} . Then the formula for the triangle, which we'll generalize to a rectangle, would be A = α α α 1 A=\alpha\cdot\frac{\alpha}{\alpha -1} . Then the ratio of the yellow triangle to the quadrilateral would then be A 3 3 A \frac{A}{3\cdot3 - A} . We can now take the derivative and solve for α \alpha , giving us our answer. d d α α 2 α 1 9 α 2 α 1 = d d α α 2 α 1 9 α 9 α 2 α 1 = d d α α 2 9 α 9 α 2 18 α 2 18 α 2 α 3 9 α 2 + 2 α 3 ( 9 α 9 α 2 ) 2 = 9 α 2 18 α ( 9 α 9 α 2 ) 2 = 0 9 α 2 18 α = 0 α = 2 \begin{aligned}&\frac{d}{d\alpha}\frac{\frac{\alpha^2}{\alpha -1}}{9-\frac{\alpha^2}{\alpha -1}}= \frac{d}{d\alpha}\frac{\frac{\alpha^2}{\alpha -1}}{\frac{9\alpha-9-\alpha^2}{\alpha -1}}=\frac{d}{d\alpha}\frac{\alpha^2}{9\alpha-9-\alpha^2}\\ &\frac{18\alpha^2-18\alpha-2\alpha^3-9\alpha^2+2\alpha^3}{(9\alpha-9-\alpha^2)^2} = \frac{9\alpha^2-18\alpha}{(9\alpha-9-\alpha^2)^2}=0 \\ &9\alpha^2-18\alpha=0 \implies \alpha=2 \end{aligned} Putting this back in the ratio function, we get 2 2 9 ( 2 ) 9 2 2 = 4 5 \frac{2^2}{9(2)-9-2^2}=\boxed{\frac{4}{5}}

But why should we assume that the triangle is right-angled?

Mark Hennings - 3 years, 6 months ago

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I did it for simplicity, since the x and y axis intersect at a right angle, I let that be one of the angles in my triangle. In general, it does not have to be right angled, but the ratio would remain the same. Imagine you have a right triangle and that line, creating the minimum ratio. Now, imagine putting this into some photo editing software, and dragging one of the vertices where ever, forming a completely different triangle. The lines move to adjust to that change, and as a result, keep the ratio.

Kevin Tong - 3 years, 6 months ago

How area of triangle generalize to a rectangle explain plz

Roy Satyam - 3 years, 6 months ago

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If we were to use the area of a triangle, it would 1/2 of that of a rectangle of the same base times height. Since we're finding the ratio of the triangle to itself again, the 1/2 term would cancel out, thus removing the need to have it in there.

Kevin Tong - 3 years, 6 months ago

It must be 1/2*x intercept *y intercept

Roy Satyam - 3 years, 6 months ago

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In the problem, I stated that I would generalize it to a rectangle so that I would not have to multiply it by 1/2. You can do it that way, but in the end, the 1/2 would cross out in the ratio.

Kevin Tong - 3 years, 6 months ago
Jeremy Galvagni
Nov 25, 2017

Looking at the first two samples, the ratios won't be changed if the top point is moved. So the triangle may as well be equilateral. As the line slants left or right towards a corner, the ratio approaches 1. So the ratio is minimized when the line is horizontal.

The centroid is 1/3 of the way up. So the horizontal line makes a yellow triangle 2/3 of size, or 4/9 the area of the whole triangle. So the quadrilateral is the other 5/9. The ratio between them is 4/5 = 0.8

Puneet Sharma
Nov 23, 2017

PFA the solution!

That isn't a solution.

Parth Sindhu - 3 years, 6 months ago

Actually this can be turned into a solution.

First notice that there is a linear area preserving transformation which sends your triangle to an equilateral triangle. To do so either use a ( infinite number of) shear transformations or write out explicitly the constraints on 2x2 matrices with determinant one. (You can assume one vertex is at the origin, you have two constraints to bring the vertices where you want and one constraint for the fact the the determinant is one).

The symmetry argument is also mostly fine. As the line which splits the triangle turns around the center there are moments where the triangle is split into two equal triangles. The ratio is then 1 which is the only possible [aboslute] maximum. Somewhere in between there has to be a [absolute] minimum (unless the ratio is constant, which the actual calculation show it isn't). Since there is a minimum, this has to happen when the drawn line is parallel to the bottom side (by symmetry).

So the only catch is that there could be 2 [absolute] minima between the "obvious" maxima. I don't see an easy argument for that yet, but this seems very implausible.

Antoine G - 3 years, 6 months ago

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