Putnam Problem of the Day

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10520885]

[math]
\text{Given a point }(a,b)\text{ with }0<b<a\text{, determine the minimum perimeter of a}
\\
\text{triangle with one vertex at }(a,b)\text{, one on the }x\text{-axis, and one on the line }y=x\text{.}
\\
\text{You may assume that a triangle of minimum perimeter exists.}
[/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10520889]

Previous Thread >>10518553

[Anonymous No.10520916]

>>10520885
1/∞
The triangle can be as small as you can fucken draw it.

[Anonymous No.10520920]

>>10520916
It's perimeter, not area retard.

[Anonymous No.10520937]

>>10520920
I meant you can make the sides infinitismally small you double retard.

[Anonymous No.10520952]

>>10520937
Of course you can't do that. Are you really retarded?

[Anonymous No.10520965]

>>10520937
Explain to me, if you fix a and b, how can you draw a triangle with infinitesimal small perimeter?
For a simple counter argument, pick (a,b) = (3,1), try to pick one point on y = x, and one point on y = 0 so that it has 1/inf perimeter.
Fucking retard doesn't know what he says.

[Anonymous No.10520989]

>>10520965
Nice strawman you triple retard.
(a,b) = (1/∞,1/∞)
The point on y = x is at 1/∞ and at the x axis point y = 1/∞.

[Anonymous No.10521011]

>>10520885
Thinly veiled homework thread of the day

[Anonymous No.10521330]

>>10520885

isn't this just a convex optimization problem?

||(a,b) - (c,0)|| + ||(c,0) - (d,d)|| + ||(d,d) - (a,b)||

[Anonymous No.10521355]

>>10521330
cont.

i think the solution turns out to be (a,0) and ((b-a)/2, (b-a)/2)

[Anonymous No.10521380]

>>10520885
Classic
Any triangle constructed this way can be unfolded into 3 connected line segments: the line segment connecting (a,b) to the x=y line is mirrored across said line, the line segment connecting the x=y line to the x axis is unchanged, and the line segment connecting (a,b) to the x axis is mirrored across the x axis. So, these are 3 line segments connected in series from (b,a) to (a,-b). These connected line segments will produce cusps when the perimeter is not minimal, but will produce a straight line from (b,a) to (a,-b) when the perimeter is minimal (because a straight line is the shortest possible path between two points). So, the perimiter is sqrt(2)*(the distance from 0 to (a,b)).

[Anonymous No.10521382]

>>10521330
cont.

pretty sure you can just solve it analytically for the two variables c and d, but i don't feel like working it all out.

[Anonymous No.10522212]

>>10520965
Can't you just pick the points [math](\varepsilon_1, \varepsilon_1)[/math] and [math](\varepsilon_2, 0)[/math], where [math]0 < \varepsilon_1 \neq \varepsilon_2[/math]?

Then the triangle would have perimeter twice the distance from the origin to (a,b). Although this is under the erroneous assumption that such a triangle exists. But if we, as the problem states, make that assumption, this would be correct (I believe). I may have misunderstood the question however.

[Anonymous No.10522216]

can we get the prismrivers on the next one, please?

[Anonymous No.10522222]

>>10521380
This is pretty clever.

[Anonymous No.10522233]

>>10522212
Nevermind, just realized this is dumb.

[Anonymous No.10522256]

>>10521382
Solving it analytically is absolutely awful (I tried).

[Anonymous No.10522365]

>>10522256

how? it a linear system with two equations and two unknowns, right?

[Anonymous No.10522392]

>>10522365
No.
Distance involves square roots.
Each partial derivative has two nasty terms.

[Anonymous No.10522551]

>>10522392

doh. you're right. it should still be doable though.

[Anonymous No.10522774]

Well you have one line that guaranteed goes from the x-axis to the line x=y, which is just a vertical line.
This line will be of length x, from (x, 0) to (x, x). There will be two other line segments of lengths sqrt((a - x)^2 + b^2) and of sqrt((a - x)^2 + (b - x)^2) going from the x-axis to (a, b) and from line x=y to (a, b) respectively. This gives us:
P = x + sqrt((a - x)^2 + b^2) + sqrt((a - x)^2 + (b - x)^2)
I'm not putting this in latex nor I am solving it any further for you, you lazy fuck

[Anonymous No.10522784]

>>10522774
Nevermind, just realized that the line doesn't have to be vertical, disregard this

[Anonymous No.10522785]

>>10522216
please respond

[Anonymous No.10522944]

>>10521380
nice
I made an interactive graph to visualize:
https://www.desmos.com/calculator/vubu3gckxt

[Anonymous No.10523215]

>>10520885
sqrt(1/2)(2 sqrt((b^2 (a^2 + b^2))/(a + b)^2) + sqrt(b^4/a^2 + a^2 + (2 b^3)/a - 2 a b + 2 b^2) + sqrt((a^6 + 3 a^4 b^2 - 4 a^3 b^3 + 3 a^2 b^4 + 4 a b^5 + b^6)/(a^2 (a + b)^2)))

[Anonymous No.10523240]

the minimum triangle perimeter is "2b". The point (a,b) is under the line y=x(for obvious reasons) and, as you can make the triangle as small as you want, the point the x-axis you should choose to reduce the total area to 0 is x=a. Doing that, you have one vertical straight line connecting all three dots, making a triangle with an null area.