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Question:

Question

My solution: Using the mid-point formula, we can easily prove that the coordinates of the point $M$ are $(\frac{a}{2}, \frac{b}{2})$. After this, we can use the distance formula to compute the distance $BM$ & $MA$ as follows (I'm doing so in an equation instead of individually):

$\sqrt{(0 + \frac{a}{2})^2 + (b + \frac{b}{2})^2} = \sqrt{(a + \frac{a}{2})^2 + (0 + \frac{b}{2})^2}$

$\implies (0 + \frac{a}{2})^2 + (b + \frac{b}{2})^2 = (a + \frac{a}{2})^2 + (0 + \frac{b}{2})^2 ... (1)$

$\implies \frac{a^2}{4} + b^2 + \frac{b^2}{4} + b^2 = a^2 + \frac{a^2}{4} + a^2 + \frac{b^2}{4}$

$\implies b^2 = a^2$

$\implies b = a$ (since distance can never be negative)

Putting $b=a$ in $1$, we get $0=0.$ Similarly, we can prove for $C$. Now, is this approach correct or do I need to individually calculate the distances and then show that they are equal?

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    $\begingroup$ You have a mistake in the formula of the distance. For example the distance between $M$ and $B$ is $d(M,B)=\sqrt{\left(0-\frac{a}{2}\right)^2+\left(b-\frac{b}{2}\right)^2}=\sqrt{ \frac{a^2}{4}+\frac{b^2}{4}}$. For the proof, check Roman83's answer $\endgroup$ – Darío G Apr 26 '16 at 14:34
  • $\begingroup$ Oh! I'm so sorry. And I've seen his answer. That's my question. Instead of finding only $d(M, B)$. What if I write something like $d(M, B) = d(M, A)$ which is equation to $0 = 0$. Hence the distances are equal? Will it be all right? $\endgroup$ – MathEnthusiast Apr 26 '16 at 14:44
  • $\begingroup$ It is true that $d(M,B)=d(M,A)$ simply because $M$ is the midpoint of $AB$. What you have to prove is that $d(0M)=d(MB)$, but this is easy since you already know that $M$ is the point with coordinates $\left(\frac{a}{2},\frac{b}{2}\right)$. $\endgroup$ – Darío G Apr 26 '16 at 15:06
  • $\begingroup$ Okay! I got it, thanks so much! $\endgroup$ – MathEnthusiast Apr 26 '16 at 15:16
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The very first line under the radical sign is wrong: it should be $b - \frac{b}{2}$ that gets squares, and similarly for $a - \frac{a}{2}$.

You re hoping to show these two square roots are equal, which you do by squaring them, but that doesn't work. Because even though $-2$ and $2$ have squares that are equal, the numbers themselves are not. YOu need to say something like this:

"To show sqrt1 = sqrt2, we'll show that their squares are equal. Since nonnegative numbers whose squares are equal must be equal, and since sqrts are always nonnegative by definition, this will show the distances are in fact equal."

When you then do out the algebra, you need double-implication (both forward and backward) at each step. Otherwise you're showing that something implies a true statement, which tells you nothing about the original thing. For instance, I could claim that $2 = 1$, and therefore $1 = 2$, and by adding equations, get that $3 = 3$. The truth of that last stemenet tells you nothing about the truth of the first one (which is obviously false!).

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$$MA=\sqrt {\left( \frac a2 -0\right)^2+\left(\frac b2 -b \right)^2}=\sqrt{\frac {a^2}4+\frac{b^2}4}$$ Similarly, $$MB=MC=\sqrt{\frac {a^2}4+\frac{b^2}4}$$

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