Math Challenge - June 2023

[The Fez]

What am I missing on Question 1? It would be easy to construct a line through the centroids of both shapes, which should bisect each one. Centroid of a rectangle is the intersection of the diagonals, and triangle is intersection of the three medians.

 

[mathwonk]

@The Fez: see posts 28, 29.

 

[mathwonk]

Thanks Fresh and Infrared for insight on prob. 7. I finally see it. If the problem were true, the matrix would have to commute with a matrix of negative determinant, but the 2x2 real matrix for multiplication by i, (i.e. 90 degree rotation), only commutes, by definition of complex linearity, with matrices of multiplication by complex numbers, and the matrix for multiplication by a+bi has determinant a^2+b^2.

 

[Throwaway_for_June]

What am I missing on Question 1? It would be easy to construct a line through the centroids of both shapes, which should bisect each one. Centroid of a rectangle is the intersection of the diagonals, and triangle is intersection of the three medians.

Lines through the centroid of the triangle don't always cut the area of the triangle in half. Consider, for example, a triangle with vertices at $(0,0)$, $(2,0)$ and $(0,2)$ in a plane. Then the centroid is at $(\frac{2}{2},\frac{2}{3})$, but the line $y=\frac{2}{3}$ doesn't divide the area of the triangle in half.

There may be other ways to think about it, but, effectively, problem 1 is asking us to find the tangent to a hyperbola that goes through a particular point. I can describe a method for constructing the line, but it's basically "solve this quadratic equation that was developed using differential calculus and analytic geometry" which is pretty unsatisfying as a "compass and straightedge" kind of thing.

 

[mathwonk]

nonetheless if you feel like posting it, i would surely learn something enjoyable.

 

[Throwaway_for_June]

nonetheless if you feel like posting it, i would surely learn something enjoyable.

My capabilities with GeoGebra weren't quite up to making nice illustrations.

Start by constructing the centroid of the rectangle $C_R$ by crossing the diagonals of the rectangle. We know that lines divide the area of the rectangle in two if and only if they go through the centroid. So our eventual solution is going to have to go through that point.

Construct the midpoints of the edges of the triangle and the centroid of the triangle $C_T$. Draw a ray from $C_T$ through $C_R$. If this ray happens to cross the triangle at a vertex or at the midpoint of an edge, then extending it to a line it bisects both the triangle and the rectangle and we're done. Otherwise, the ray crosses exactly one of the edges of the triangle, and one of the vertices on that edge is further from the intersection than the other. Label that vertex $O$. Label the edge that is crossed by the ray $Y$, and the other edge that goes through $O$, $X$.

Now we know that there's a line from $C_R$ through $X$ and $Y$ which divides the area of hte triangle in half.

Construct an altitude from one of the sides of the triangle as the base, and then take the geometric mean of that altitude with half the base. We'll be using this geometric mean as our unit length. (A square with this as the side length will have the same area as the triangle.)

At this point I'm going to assume that $C_R$ is outside the triangle. It doesn't actually make things more difficult, but without oriented distances, it ends up being a case with a sign change.

Extend the edges $X$ and $Y$ to lines. Construct a line parallel to $Y$ through $C_R$ and label the distance along that line from $C_R$ to the extended $X$ edge $D_Y$. Construct the perpendicular distance from $C_R$ to the line $Y$ and label that distance $D_X$.

Using well-known constructions for multiplication, division, addition, and the square root, construct the length $\frac{ 2D_Y } {1 + \sqrt{1+4D_Y D_X}}$. Mark that distance from $O$ along the edge $Y$ and call the point $T$. The line from $C_R$ through $T$ will divide the areas of the rectangle and the triangle in half.

 

[mathwonk]

very nice! but it seems to use only similarity of triangles. i.e. when Dx = Dy = 1, just bh = 1 and b(1+h) = h. From your summary earlier, I was expecting to find a hyperbolic envelope, or some calculus. ?? ( I also thought I had made an error, since I got different answers solving for b and for h, until I realized that [sqrt(5)-1]/2 is the reciprocal of [sqrt(5)+1]/2.)

 

[Throwaway_for_June]

very nice! but it seems to use only similarity of triangles .... From your summary earlier, I was expecting to find a hyperbolic envelope, or some calculus. ??

It was me thinking that everything can work as a nail when I've got a hammer in my hand kind of thing.

I was in a "compass and straightedge" kind of mindset, so I went looking to draw a tangent to some kind of circle that also went through the centroid of the rectangle. Then I remembered (or realized) that the tangents to a particular hyperbola all form triangles with the same area when intersected with the hyperbola's asymptotes. Compass and straightedge don't allow for constructing a hyperbola to draw the tangent against, but I can do that algebraically using analytic geometry and calculus which got me something that worked.

If I had started by thinking in terms of similar triangles then I think I would have ended up with a slightly more elegant construction since we don't actually need the $D_x$ and $D_y$ to be measured along perpendicular axes.

... ( I also thought I had made an error, since I got different answers solving for b and for h, until I realized that [sqrt(5)-1]/2 is the reciprocal of [sqrt(5)+1]/2.)

Yeah. There are lots of subtle identities involving square roots and fractions. Though it's also totally plausible that I made some algebra errors last night.

 

[mathwonk]

actually this sentence puzzled me:

"Now we know that there's a line from C_R
CRthrough XX and Y and Ywhich divides the area of the triangle in half."

and I thought maybe calculus motivated it. I.e. I did not see how the areas cut by the lines through the centroid C_T varied as they rotated. But I didn't draw a picture. (...... Ok I drew a picture and see it now.)

anyway, this is beautiful. thanks for posting!

 

[Throwaway_for_June]

actually this sentence puzzled me:

"Now we know that there's a line from C_R
CRthrough XX and Y and Ywhich divides the area of the triangle in half."

and I thought maybe calculus motivated it. I.e. I did not see how the areas cut by the lines through the centroid C_T varied as they rotated. But I didn't draw a picture.

anyway, this is beautiful. thanks for posting!

Here's a drawing:

N is the centroid of triangle ABC.

If M is the centroid of the rectangle, then triangle ARC has more area than triangle ALC which has half the area of triangle ABC, and triangle IQC has less area than triangle IBC which has half the area of triangle ABC.

So there's some line that goes from M through the segments RQ and AI that produces a triangle that has a vertex at C with the desired area.

The closest this gets to calculus is the idea of continuously sweeping the ray from M.

 

[mathwonk]

thank you! and in reference to the arithmetic identities you remarked, if you do the calculation backwards as I did, you seem to get [sqrt(1+4D_xD_y)-1]/2D_x, which I believe equals your 2D_y/[1 + sqrt(1+4D_xD_y)]. So division is required either way. In particular I believe your arithmetic is correct.