Question

If $A(\sin t, \cos t), B(\frac{1-e^{\cos^2 t}}{\sin t}, \frac{1-e^{\sin^2 t}}{\cos t})$ and $C (\csc t, \sec t)$ are three points, then which of following can't be true?

• A. $AB + BC > AC$
• B. $AB + BC = AC$
• C. $AC + BC = AB$
• D. $BA + AC = BC$

Answer 1

Answer: (B)

$AB + BC = AC$

Answer 2

The given points are $A(\sin t, \cos t)$, $B(\frac{1-e^{\cos^2 t}}{\sin t}, \frac{1-e^{\sin^2 t}}{\cos t})$ and $C (\csc t, \sec t)$. For the coordinates to be well-defined, $\sin t \ne 0$ and $\cos t \ne 0$, which means $t \ne k\pi/2$ for any integer $k$.

Let $x_A = \sin t$, $y_A = \cos t$. Then $x_C = \csc t = 1/\sin t = 1/x_A$ and $y_C = \sec t = 1/\cos t = 1/y_A$.

We need to determine if the points A, B, C are collinear. Three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ are collinear if the slope of the line segment connecting the first two points is equal to the slope of the line segment connecting the first and third points. That is, $\frac{y_B - y_A}{x_B - x_A} = \frac{y_C - y_A}{x_C - x_A}$. This condition can be rewritten as $(y_B - y_A)(x_C - x_A) = (y_C - y_A)(x_B - x_A)$.

Let's calculate the differences: $x_B - x_A = \frac{1-e^{\cos^2 t}}{\sin t} - \sin t = \frac{1-e^{\cos^2 t} - \sin^2 t}{\sin t}$ $y_B - y_A = \frac{1-e^{\sin^2 t}}{\cos t} - \cos t = \frac{1-e^{\sin^2 t} - \cos^2 t}{\cos t}$

$x_C - x_A = \frac{1}{\sin t} - \sin t = \frac{1-\sin^2 t}{\sin t} = \frac{\cos^2 t}{\sin t}$ $y_C - y_A = \frac{1}{\cos t} - \cos t = \frac{1-\cos^2 t}{\cos t} = \frac{\sin^2 t}{\cos t}$

For collinearity, we require: $\frac{\frac{1-e^{\cos^2 t} - \sin^2 t}{\sin t}}{\frac{\cos^2 t}{\sin t}} = \frac{\frac{1-e^{\sin^2 t} - \cos^2 t}{\cos t}}{\frac{\sin^2 t}{\cos t}}$ $\frac{1-e^{\cos^2 t} - \sin^2 t}{\cos^2 t} = \frac{1-e^{\sin^2 t} - \cos^2 t}{\sin^2 t}$

Let $u = \sin^2 t$ and $v = \cos^2 t$. We know that $u+v=1$. Substitute $u$ and $v$ into the equation: $\frac{1-e^v - u}{v} = \frac{1-e^u - v}{u}$ Since $u+v=1$, we have $u=1-v$ and $v=1-u$. $\frac{1-e^v - (1-v)}{v} = \frac{1-e^u - (1-u)}{u}$ $\frac{v-e^v}{v} = \frac{u-e^u}{u}$ $1 - \frac{e^v}{v} = 1 - \frac{e^u}{u}$ $\frac{e^v}{v} = \frac{e^u}{u}$

Let $f(x) = \frac{e^x}{x}$. We need to find when $f(v) = f(u)$. The domain for $u = \sin^2 t$ and $v = \cos^2 t$ (since $t \ne k\pi/2$) is $(0, 1)$. Let's find the derivative of $f(x)$: $f'(x) = \frac{x e^x - e^x}{x^2} = \frac{e^x(x-1)}{x^2}$. For $x \in (0, 1)$, $x-1 < 0$, $e^x > 0$, and $x^2 > 0$. Therefore, $f'(x) < 0$ for $x \in (0, 1)$. This implies that $f(x)$ is a strictly decreasing function on the interval $(0, 1)$.

Since $f(x)$ is strictly decreasing, $f(v) = f(u)$ implies $u=v$. $u=v \implies \sin^2 t = \cos^2 t \implies \tan^2 t = 1 \implies \tan t = \pm 1$. This means $t = \pi/4 + k\pi/2$ for integer $k$.

Thus, the points A, B, C are collinear if and only if $t = \pi/4 + k\pi/2$. If $t \ne \pi/4 + k\pi/2$ (and $t \ne k\pi/2$), the points A, B, C are non-collinear.

The triangle inequality states that for any three points A, B, C: $AB + BC \ge AC$ $AC + BC \ge AB$ $BA + AC \ge BC$

The equality holds if and only if the points are collinear and the middle point lies on the segment formed by the other two. The strict inequality holds if the points are non-collinear.

Let's analyze the given options: A. $AB + BC > AC$: This statement is true if A, B, C are non-collinear. It is false if A, B, C are collinear. Since A, B, C are non-collinear for many values of $t$ (e.g., $t=\pi/6$), this statement can be true. B. $AB + BC = AC$: This statement is true if A, B, C are collinear and B lies between A and C. It is false if A, B, C are non-collinear, or if they are collinear but B is not between A and C. Since A, B, C can be non-collinear, this statement can be false. C. $AC + BC = AB$: This statement is true if A, B, C are collinear and C lies between A and B. It is false otherwise. This statement can be false. D. $BA + AC = BC$: This statement is true if A, B, C are collinear and A lies between B and C. It is false otherwise. This statement can be false.

The question asks "which of following can't be true?". This means which statement is never true for any valid value of $t$.

Consider the case when A, B, C are non-collinear. In this case, $AB+BC > AC$, $AC+BC > AB$, and $BA+AC > BC$ are all true. This means that options B, C, D are false. So, for non-collinear points, $AB+BC=AC$ is false. Thus, option B can be false. For non-collinear points, $AC+BC=AB$ is false. Thus, option C can be false. For non-collinear points, $BA+AC=BC$ is false. Thus, option D can be false.

Consider the case when A, B, C are collinear (i.e., $t = \pi/4 + k\pi/2$). Let's take $t=\pi/4$. $A = (\sin(\pi/4), \cos(\pi/4)) = (1/\sqrt{2}, 1/\sqrt{2}) \approx (0.707, 0.707)$ $C = (\csc(\pi/4), \sec(\pi/4)) = (\sqrt{2}, \sqrt{2}) \approx (1.414, 1.414)$ $B = (\frac{1-e^{1/2}}{1/\sqrt{2}}, \frac{1-e^{1/2}}{1/\sqrt{2}}) = (\sqrt{2}(1-\sqrt{e}), \sqrt{2}(1-\sqrt{e})) \approx (1.414(1-1.648), 1.414(1-1.648)) \approx (-0.916, -0.916)$

All three points lie on the line $y=x$. The order of the x-coordinates (and y-coordinates) is $x_B < x_A < x_C$. So, B is approximately $(-0.916)$, A is approximately $(0.707)$, C is approximately $(1.414)$. This means point A lies between B and C.

When A lies between B and C, the collinearity condition implies $BA + AC = BC$. Therefore, for $t=\pi/4$, option D ($BA+AC=BC$) is true. Since option D can be true, it is not the answer.

If $BA+AC=BC$ is true, then $AB+BC=AC$ must be false (unless A, B, C are the same point, which is not the case here). Also $AC+BC=AB$ must be false. And $AB+BC > AC$ must be false (since they are collinear).

So, for $t=\pi/4$: A. $AB + BC > AC$ is FALSE. B. $AB + BC = AC$ is FALSE. C. $AC + BC = AB$ is FALSE. D. $BA + AC = BC$ is TRUE.

The question asks which of the following can't be true. This implies there is one statement that is always false for any $t$. However, based on our analysis, for $t=\pi/4$, options A, B, C are false, and option D is true. For $t=\pi/6$ (non-collinear case), option A is true, and options B, C, D are false.

This means each of the options A, B, C, D can be true for some $t$, or can be false for some $t$. Let's re-read the question carefully: "which of following can't be true?". This phrasing implies that one of the options represents an impossible scenario.

In geometry, for any three points A, B, C, the triangle inequality states $AB+BC \ge AC$, $AC+BC \ge AB$, $BA+AC \ge BC$. If any of these conditions are violated, it means the configuration is impossible. For example, if $AB+BC < AC$, then that configuration is impossible.

Let's look at the options as conditions: A. $AB + BC > AC$: This is true for non-collinear points. So it can be true. B. $AB + BC = AC$: This is true for collinear points where B is between A and C. So it can be true. C. $AC + BC = AB$: This is true for collinear points where C is between A and B. So it can be true. D. $BA + AC = BC$: This is true for collinear points where A is between B and C. We found this is true for $t=\pi/4$. So it can be true.

It seems there might be a misunderstanding of the question or the options. If the question is implicitly assuming A, B, C form a triangle (i.e., are non-collinear), then B, C, D would be false. But we showed they can be collinear.

Let's re-examine the condition $f(x) = e^x/x$. We proved that A, B, C are collinear if and only if $u=v$, i.e., $\sin^2 t = \cos^2 t$. This means that the points are either non-collinear or they are collinear with A being the middle point (between B and C) or C being the middle point (between A and B) or B being the middle point (between A and C).

Let's check the ordering of points for $t=\pi/4$ again. $A = (1/\sqrt{2}, 1/\sqrt{2})$ $B = (\sqrt{2}(1-\sqrt{e}), \sqrt{2}(1-\sqrt{e}))$ $C = (\sqrt{2}, \sqrt{2})$ The x-coordinates are $x_B \approx -0.916$, $x_A \approx 0.707$, $x_C \approx 1.414$. Since $x_B < x_A < x_C$, the point A lies between B and C. In this case, $BA + AC = BC$ holds true. (Option D)

What if the question implies that the points must form a triangle (i.e. non-collinear)? If A, B, C are non-collinear, then $AB+BC > AC$, $AC+BC > AB$, and $BA+AC > BC$ are all true. In this scenario, options B, C, D would be false. But the question asks which can't be true. Since B, C, D can be true (when collinear), this interpretation doesn't work.

Let's consider the standard form of triangle inequality $a+b \ge c$. The options are given in the form of sums of distances. A. $AB+BC > AC$ (strict triangle inequality) B. $AB+BC = AC$ (collinearity, B in middle) C. $AC+BC = AB$ (collinearity, C in middle) D. $BA+AC = BC$ (collinearity, A in middle)

For any three points A, B, C, at least one of these conditions must be true:

• If non-collinear: A is true. B, C, D are false.
• If collinear: One of B, C, D is true. A is false.

The question asks "which of following can't be true?". This means "which statement is never satisfied by these points for any $t$?".

Since A, B, C can be non-collinear (e.g., $t=\pi/6$), option A can be true. Since A, B, C can be collinear (e.g., $t=\pi/4$), and we found A is between B and C for $t=\pi/4$, option D can be true. What about options B and C? Can B be between A and C, or C between A and B?

Let's check the order of points for other values of $t$ where they are collinear. If $t = 3\pi/4$: $A = (\sin(3\pi/4), \cos(3\pi/4)) = (1/\sqrt{2}, -1/\sqrt{2})$ $C = (\csc(3\pi/4), \sec(3\pi/4)) = (\sqrt{2}, -\sqrt{2})$ $B = (\sqrt{2}(1-\sqrt{e}), -\sqrt{2}(1-\sqrt{e})) \approx (-0.916, 0.916)$ These points lie on the line $y=-x$. $x_B \approx -0.916$, $x_A \approx 0.707$, $x_C \approx 1.414$. Again, $x_B < x_A < x_C$. So A is between B and C. This still supports D.

It seems that for all $t$ where they are collinear ($t = \pi/4 + k\pi/2$), A always lies between B and C. Let's verify this. $x_A = \sin t$ $x_C = 1/\sin t$ $x_B = \frac{1-e^{\cos^2 t}}{\sin t}$ For $t = \pi/4 + k\pi/2$, we have $\sin^2 t = \cos^2 t = 1/2$. So $x_B = \frac{1-e^{1/2}}{\sin t} = \frac{1-\sqrt{e}}{\sin t}$. Since $\sqrt{e} \approx 1.648$, $1-\sqrt{e}$ is negative. So $x_B$ has the opposite sign of $\sin t$. $x_A$ has the same sign as $\sin t$. $x_C$ has the same sign as $\sin t$.

If $\sin t > 0$ (e.g., $t=\pi/4$), then $x_B < 0$, $x_A > 0$, $x_C > 0$. Also, $x_A = \sin t$. $x_C = 1/\sin t$. Since $\sin t \in (0,1)$ for $t \in (0, \pi/2)$, $x_A < x_C$. So, for $t=\pi/4$, $x_B < 0 < x_A < x_C$. This means B is to the left of A, which is to the left of C. Thus, A is between B and C. So $BA + AC = BC$ is true. (Option D)

If $\sin t < 0$ (e.g., $t=5\pi/4$), then $x_B > 0$, $x_A < 0$, $x_C < 0$. Also, $x_A = \sin t$. $x_C = 1/\sin t$. Since $\sin t \in (-1,0)$ for $t \in (\pi, 3\pi/2)$, $x_A > x_C$. So, for $t=5\pi/4$, $x_C < x_A < 0 < x_B$. This means C is to the left of A, which is to the left of B. Thus, A is between C and B. This implies $CA + AB = CB$, which is $AC + BA = BC$. (Option D again!)

It seems that for all values of $t$ where A, B, C are collinear, A is always the middle point. If A is always the middle point when they are collinear, then:

• $BA + AC = BC$ (Option D) is true when collinear.
• $AB + BC = AC$ (Option B) is never true (because B is not the middle point).
• $AC + BC = AB$ (Option C) is never true (because C is not the middle point).

So, if A is always the middle point when collinear, then options B and C can't be true (in the sense of "never true for any $t$ where collinear"). However, the question asks which can't be true for any $t$. If A, B, C are non-collinear, then $AB+BC > AC$, $AC+BC > AB$, $BA+AC > BC$ are all true. This means that for non-collinear points, $AB+BC=AC$ is false. So Option B can be false. Similarly, Option C can be false.

This implies that the question is asking for a condition that is always false. If A is always the middle point when collinear, then B and C are never the middle point. So $AB+BC=AC$ is false when collinear. And $AC+BC=AB$ is false when collinear. And $AB+BC>AC$ is true when non-collinear, false when collinear. And $BA+AC=BC$ is true when collinear, false when non-collinear.

The only options that are always false are those that contradict the triangle inequality. For any three points A, B, C, it is always true that $AB+BC \ge AC$, $AC+BC \ge AB$, $BA+AC \ge BC$. None of the options contradict these fundamental inequalities. The options are statements about the equality or strict inequality in the triangle inequality.

Let's assume the question means "Which of the following statements is always false for any valid value of t?". We have shown that:

• Option A ($AB+BC > AC$) is true when A, B, C are non-collinear. Since they can be non-collinear, A can be true.
• Option D ($BA+AC = BC$) is true when A, B, C are collinear. Since they can be collinear, D can be true.
• Options B ($AB+BC = AC$) and C ($AC+BC = AB$) are true if B or C respectively is the middle point when collinear. We found that A is always the middle point when collinear. Therefore, B and C can never be the middle point. So, $AB+BC = AC$ is never true. And $AC+BC = AB$ is never true.

This means that both B and C can't be true. If this is a single choice question, there might be a subtle difference.

Let's reconfirm if B or C can ever be the middle point. The coordinates are $x_B = \frac{1-\sqrt{e}}{\sin t}$, $x_A = \sin t$, $x_C = \frac{1}{\sin t}$. If $\sin t > 0$: $1-\sqrt{e} < 0$. So $x_B < 0$. $x_A > 0$. $x_C > 0$. The order of x-coordinates is $x_B < 0 < x_A < x_C$. So B is the leftmost point, C is the rightmost point, and A is in the middle. This means $BA + AC = BC$ is true.

If $\sin t < 0$: $1-\sqrt{e} < 0$. So $x_B > 0$. $x_A < 0$. $x_C < 0$. The order of absolute values of x-coordinates is $|x_A| < |x_C|$ for $|\sin t| < 1$. $x_C = 1/\sin t$. $x_A = \sin t$. Since $\sin t \in (-1, 0)$, $1/\sin t < \sin t < 0$. So $x_C < x_A < 0$. And $x_B > 0$. The order of x-coordinates is $x_C < x_A < 0 < x_B$. So C is the leftmost point, B is the rightmost point, and A is in the middle. This means $CA + AB = CB$, which is $AC + BA = BC$ is true.

In both cases where A, B, C are collinear, A is always the middle point. Therefore, the scenarios where B is the middle point ($AB+BC=AC$) or C is the middle point ($AC+BC=AB$) are never possible. Thus, both B and C can't be true. Since it is a single choice question, there might be a convention or a specific choice to be made. Often, if multiple choices are mathematically impossible, any one of them is considered a correct answer.

Let's pick B as the answer.

Final check:

1. Are A, B, C always collinear? No, only when $\sin^2 t = \cos^2 t$.
2. When A, B, C are non-collinear, $AB+BC>AC$ (Option A) is true. So A can be true.
3. When A, B, C are collinear, we found A is always the middle point. So $BA+AC=BC$ (Option D) is true. So D can be true.
4. Since A is always the middle point when collinear, B is never the middle point and C is never the middle point. Therefore, $AB+BC=AC$ (Option B) is never true. And $AC+BC=AB$ (Option C) is never true.

Both B and C are impossible scenarios. If it's a single choice question, either B or C would be the answer. Usually, the first impossible option encountered is chosen.

The final answer is $\boxed{B}$