If $f$ is a continuous path in $\Bbb{R}^2$, and $f(0)\in B_1(0,0)$ and $f(1)\in B_1(10,10)$, how to prove that $f$ crosses the line $\{(x,y):x+y=5\}$

Question

$f:[0,1]\to\mathbb{R}^2$ is continuous, $f(0) \in B_{1}(0,0)$ and $f(1) \in B_{1}(10,10)$. Prove there exists $t \in [0,1]$ such that $f(t) \in \{(x,y): x+y=5\}$.

I am thinking we need to use extreme value theorem or intermediate value theorem. Which one and how?

Just for information $B_1$(x,y) is the circle of radius 1 around pt (x,y)

Answer 1

Hint: $g\colon\mathbb R^2\to\mathbb R$, $(x,y)\mapsto x+y-5$ is continuous and so is $g\circ f$.

Second hint: If $(x,y)\in B_r(a,b)$, then $a-r<x<a+r$ and $b-r<y<b+r$.

Answer 2

Here's my hint: consider the change of coordinates

$$u=x+y$$ $$v=x-y$$

Rephrasing the question in these coordinates:

$f:[0,1]\to\mathbb{R}^2$ is continuous, $f(0) \in B_{1}(0,0)$ and $f(1) \in B_{1}(20,0)$. Prove there exists $t \in [0,1]$ such that the first coordinate of $f(t)$ is 5.

Of course, this change of coordinates does nothing conceptually, but I think it is much clearer why the statement is true.