When is EF longer than AC? (a generalization)

Question

ABC is an isosceles right triangle, M is on AC, and EMF is a straight line.

When is EF longer than AC?

]1

Note: This is a generalization of the following problem, which has M in the center of AC:

Prove that EF is longer than AC

In that case, I showed that EF is always longer than AC.

I can show the following:

If M is closer to C than A, then EF is longer.

If M is closer to A than C, then EF can be shorter, but it will be longer when EA is long enough.

If $|EB| \ge \sqrt{2}$, then EF is always longer no matter where M is.

If there are no answers in two days, I will post mine.

Answer 1

We may suppose that $$A(0,1),B(0,0),C(1,0),M(m,1-m),E(0,1-m-am),F\left(m+\frac{m-1}{a},0\right)$$ where $0\lt m\lt 1$ is the $x$ coordinate of $M$ and $a\lt -1$ is the slope of the line $EF$.

Then, we have $$\begin{align}&|EF|\gt |AC|\\\\&\iff |EF|^2\gt |AC|^2\\\\&\iff \left(m+\frac{m-1}{a}\right)^2+(1-m-am)^2\gt 2\\\\&\iff m^2+\frac{2m(m-1)}{a}+\frac{(m-1)^2}{a^2}+1+m^2+a^2m^2-2m-2am+2am^2-2\gt 0\\\\&\iff (a+1)^2(a^2+1)m^2-2(a+1)(a^2+1)m+(a+1)(1-a)\gt 0\\\\&\iff (a+1)(a^2+1)m^2-2(a^2+1)m+1-a\lt 0\\\\&\iff \color{red}{\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt m\lt 1}\tag1\end{align}$$

Here note that $$0\lt f(a)=\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \color{blue}{\frac 12}$$ for $a\lt -1$ :
$$\begin{align}f(a)&=\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\cdot\frac{-a^2-1-\sqrt{2a^2(a^2+1)}}{-a^2-1-\sqrt{2a^2(a^2+1)}}\\&=\frac{a-1}{-a^2-1-\sqrt{2a^2(a^2+1)}}\\&=\frac{1-\frac 1a}{-a-\frac 1a+\sqrt{2(a^2+1)}}\end{align}$$ is increasing for $a\lt -1$ with $$\lim_{a\to -\infty}f(a)=0,\qquad\lim_{a\to -1^-}f(a)=\frac 12.$$

Added : Now I'm going to check if my answer agrees with what the OP wrote.

If M is closer to C than A, then EF is longer.

If $m\ge\frac 12$, then $(1)$ holds for any $(a,m)$.

If M is closer to A than C, then EF can be shorter, but it will be longer when EA is long enough.

If $m\lt\frac 12$, then EF can be shorter, but it will be longer when $a$ is small enough.

If $|EB|\ge \sqrt 2$, then EF is always longer no matter where M is.

If $1-m-am\ge \sqrt 2$, i.e. $m\ge\frac{1-\sqrt 2}{a+1}$, then $(1)$ holds for any $(a,m)$ because we have $$\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \frac{1-\sqrt 2}{a+1}.$$

Answer 2

Edit: Adding the comment by the OP.

Let N be the midpoint of AC and M is a point somewhere between N and C.

The picture clearly shows $EF > AC$ because $EF = UV + VC = UV + AC$.

Thus, we only need to study the case when M is somewhere between A and N. See below.

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The picture shows the following idea:- As $E$ slides along the $y$-axis (and $F$ moves correspondingly), there is a critical position $E’$ ($F’$ correspondingly) such that $AC = E’F’$. If $AE < AE’$, $EF$ will be shorter than $AC$.

To make the two length comparable, I translate $E’F’$ to $XC$ such that $E'F'CX$ is a //gm. $CX$ is extended to cut the $y$-axis at $Y$. The circle drawn with $C$ as center and $CA$ as radius shows more clearly of the effect.

$E’A = E’B – AB = XZ – 1 = \sqrt(2) \cos \beta – 1$; where $\beta$ is the angle that EF inclines to the y-axis.

Answer 3

In my solution to the original problem, I used algebra and analytic geometry.

I tried to do the same thing for this generalization, but the final simplification does not occur here.

Here is what I've got.

Suppose $AB = BC = 1$.

Then $M =(a, 1-a) $, where $0 < a < 1$.

Let $E = (0, 1+v) $, where $v > 0 $.

Then line $EM$ is $\frac{y-(1+v)}{x} =\frac{(1-a)-(1+v)}{a} =\frac{-a-v}{a} =-1-\frac{v}{a} $ or $y =1+v-x(1+\frac{v}{a}) $.

Putting $y=0$, $x =\frac{1+v}{1+\frac{v}{a}} $, so $F =(\frac{1+v}{1+\frac{v}{a}}, 0) $.

$|AC|^2 =2 $

$\begin{array}\\ |EF|^2 &=(\frac{1+v}{1+v/a})^2+(1+v)^2\\ &=(1+v)^2(1+\frac{1}{(1+v/a)^2})\\ &=(1+v)^2(\frac{1+(1+v/a)^2}{(1+v/a)^2})\\ &=(1+v)^2(\frac{a^2+(a+v)^2}{(a+v)^2})\\ &=(1+v)^2(\frac{2a^2+2av+v^2}{(a+v)^2})\\ \end{array} $

According to Wolfy,

$\begin{array}\\ |EF|^2-|AC|^2 &=(1+v)^2 (\frac{2a^2+2av+v^2}{(a+v)^2}) -2\\ &=v \frac{2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v}{(a+v)^2}\\ \end{array} $

Therefore, when

$0 \lt 2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v\\ = a^2(2v+4)+a(2v^2+4v-2)+v^3+2 v^2-v\\ $

for $0 < a < 1$ and $v > 0 $, |EM| > |AC|.

Again according to Wolfy, if $a = 1-c$, then

$f(v, a)\\ =2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v\\ =2 c^2 (v+2)-2 c (v^2+3 v+1)+v^3+3 v^2+3 v/2\\ $

If $c \le 0$ (i.e., $a \ge \frac12$), all the terms are positive, so |EM| > |AC| for all $v > 0$.

However, if $c > 0$ (i.e., $a < \frac12$), for small $v$ this is about $4c^2-2c =2c(2c-1) \lt 0 $, so there are values of $v$ for which the expression is negative so that |EM| < |AC|.

Again according to Wolfy, this occurs at the following values of $v$:

$ a=1/3: v< 0.19795\\ a=1/4: v< 0.27832\\ a=1/10: v< 0.38509\\ a=1/100: v< 0.41382\\ a=1/10000: v<0.41421\\ $

Looking at these values, it seems that the $v$ bound is approaching $\sqrt{2}-1$. To show this is true, Wolfy gets $f(\sqrt{2}-1, a) =2(1+\sqrt{2})a^2 \gt 0 $ for $a > 0$.