Challenge Math Challenge - July 2021

[fresh_42]

In the proof I gave, I did not assume that $x, y, z$ must obey the triangle inequality. The only assumptions made are what are given in the question. I do not understand why the derivative-based solution will not help find the maximum if $c>z\geq x+y > y \geq x>0$, mentioned above as the more difficult case. Was there an add-on to the original question which is mentioned in some post? Or is a simpler, more intuitive solution expected?

There is no maximum if the boundary is excluded, under the assumption I didn't make a mistake.

$c>z\geq x+y > y \geq x>0$ makes $f(x,y)\leq 0$ and $x=y=z=c/3$ isn't a solution anymore because $z=c/3< 2c/3 =x+y.$

 

[phantomvommand]

Q14 can be solved by dropping perpendiculars from B to s (let it meet at point L) , A to s (meet at point M) and C to s (meet at point N) . One can then see 3 sets of similar triangles. One set will give you AZ/BZ = AM/BL. Applying this 2 more times to the 2 other sets of similar triangles will give you the intended result.

 

[Not anonymous]

There is no maximum if the boundary is excluded, under the assumption I didn't make a mistake.

$c>z\geq x+y > y \geq x>0$ makes $f(x,y)\leq 0$ and $x=y=z=c/3$ isn't a solution anymore because $z=c/3< 2c/3 =x+y.$

The original question did not mention $c>z\geq x+y > y \geq x>0$, so the proof of my earlier attempted answer did not impose that constraint, though it did take into account $x, y, z >0$ and $x + y+ z = c$. I assume that the updated question only adds the constraint $c>z\geq x+y > y \geq x>0$ to the other conditions mentioned in the original question, so the condition $x + y+ z = c$ still holds true. If that is correct, below is my updated attempted solution

Spoiler: Question 13 (Updated version)

$$
f(x,y,z) = 4x^2y^2 - (x^2+y^2-z^2)^2 = (2xy - (x^2+y^2-z^2))(2xy + (x^2+y^2-z^2))
$$
$$
= (z^2 - (x-y)^2)((x+y)^2-z^2) = (z^2-(x-y)^2)(x+y-z)(x+y+z)
$$
$$
= -c(z^2-(x-y)^2)(z-x-y)
$$ (Eq. 1)

Since $z \geq x+y$ and $x+y+z = c$ and $x,y,z>0$, we must have $0 < x,y \leq \dfrac{c}{2} \leq z < c$. Hence, in Eq. 1, $(z - x - y) \geq 0$ and $(z^2-(x-y)^2) > 0$, the latter since $-\dfrac{c}{2} < (x-y) < \dfrac{c}{2} \leq z$. The expression $-c(z^2-(x-y)^2)(z-x-y)$ can therefore never take a positive value under the given constraints.
∴ $f(x,y,z) \leq 0$ under the given conditions. The maximum achievable value is 0, and this is achieved when $z = \dfrac{c}{2}$, since this makes $x+y = c - z = \dfrac{c}{2}$ and hence $(z-x-y)$ in the product expression of (Eq. 1) becomes zero.

 

[Not anonymous]

11. Assume we have put a Cartesian coordinate system on France and got the following positions:
Paris $(0, 0)$, Lyon $(3, -8)$ and Marseille $(4, -12)$. Look up the definitions and calculate the distance between Lyon and Marseille according to

• the Euclidean metric.
• the maximum metric.
• the French railway metric.
• the Manhattan metric.
• the discrete metric.

Spoiler: Question 11

Euclidean metric (a.k.a. Euclidean distance) = $\sqrt{(4-3)^2 + (-12 - (-8))^2} = \sqrt{1+16} \approx 4.123$
Maximum metric (a.k.a. Chebyshev distance) = $\max{|4-3|, |-12 - (-8)|} = 4$
French railway metric = $\|(3, -8)\| + \|(4, -12)\|$ (since these 2 points are not collinear with $(0, 0)$) = $\sqrt{3^2 + (-8)^2} + \sqrt{4^2 + (-12)^2} \approx 21.193$
Manhattan metric = $|(4-3)| + |-12 - (-8)| = 1+4 = 5$
Discrete metric = $1$, since $(3, -8) \neq (4, -12)$