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I am trying to solve the following problem:

Let $ABC$ be an equilateral triangle with side $l$.

If $P$ and $Q$ are points respectively in sides $AB$ and $AC$, different from the triangle vertices, prove that $$|BQ| + |PQ| + |CP| > 2l$$

I can see that, as point $P$ tends to $A$, $|CP|+|PQ|$ tends to $|AC|+|AQ|$. If I could prove this, the problem would be solved (the rest follows from the triangle inequality).

However I have no clue on how to do this. I tried to play with triangle inequality and relations between sides and angles but nothing worked.

How can I proceed?

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  • $\begingroup$ I think you have made a typo in your question - let $Q$ approach $B$ and $P$ approach $C$. Then the limit of $BQ$ and $CP$ is 0, and the limit of $PQ$ is $l$ (draw it out!). Therefore I think the inequality should be $|BQ| + |PQ| + |CP| > l$. $\endgroup$
    – Toby Mak
    Commented Jun 24, 2017 at 23:56
  • $\begingroup$ That's true, but if you let $Q$ and $P$ approach $A$, the sum approaches $|AB|+|AC|$. $\endgroup$
    – J. C.
    Commented Jun 25, 2017 at 0:08
  • $\begingroup$ But that is the more inefficient case - the case should hold for any $P$ and $Q$, not just one example that you have described. $\endgroup$
    – Toby Mak
    Commented Jun 25, 2017 at 0:23
  • $\begingroup$ Now I realized you might have mislabeled your sketch; actually $Q$ approaches $C$, not $B$, and the inverse for $P$. $\endgroup$
    – J. C.
    Commented Jun 25, 2017 at 0:38
  • $\begingroup$ If $P$ and $Q$ are the midpoints then the statement is trivial. If either $P$ or $Q$ are far from $A$ (by far I mean farther than midway) the result follows by the first case. So you are left only with the case when both $P$ and $Q$ are close to $A$. $\endgroup$
    – user 1987
    Commented Jun 25, 2017 at 0:43

4 Answers 4

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Upon reflection ...

enter image description here

$$2s = |\overline{AB^\prime}| \leq |\overline{AP}|+|\overline{PQ^\prime}|+|\overline{Q^\prime B^\prime}| = p+q+r$$

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    $\begingroup$ It was something amazing! $\endgroup$ Commented Jun 25, 2017 at 10:13
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    $\begingroup$ Exactly what I was looking for! Fantastic $\endgroup$
    – J. C.
    Commented Jun 25, 2017 at 13:22
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Let $AQ=x$, $AP=y$ and $l=1$.

Thus, $$PQ=\sqrt{x^2-xy+y^2},$$ $$PC=\sqrt{y^2-y+1}$$ and $$BQ=\sqrt{x^2-x+1}$$ and we need to prove that $$\sqrt{x^2-xy+y^2}+\sqrt{x^2-x+1}+\sqrt{y^2-y+1}\geq2.$$ Now, by Minkowwski $$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\sqrt{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}\geq$$ $$\geq\sqrt{\left(x-\frac{1}{2}+y-\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)^2}=\sqrt{(x+y-1)^2+3}$$ and $$\sqrt{x^2-xy+y^2}\geq\frac{x+y}{2}.$$ Let $x+y=2a$.

Hence, $a\leq1$ and we need to prove that $$a+\sqrt{(2a-1)^2+3}\geq2$$ or $$\sqrt{4a^2-4a+4}\geq2-a$$ or $$4a^2-4a+4\geq a^2-4a+4,$$ which is obvious.

Done!

By my solution easy to make a geometric proof.

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  • $\begingroup$ Quite an interesting approach! I was looking for a more geometric argument, but thank you for the feedback. $\endgroup$
    – J. C.
    Commented Jun 25, 2017 at 13:21
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This is not an answer, rather it is a suggested approach to the problem. Here is a Geogebra (link)diagram:

If you go to the Geogebra link you can move points $P$ and $Q$.

The problem is reduced to proving that $|DI|>|IQ|$.

Geometric Diagram

Note from the construction that $|PQ|=|DQ|$ and $|CP|=|BG|=|BF|$ and that $2l=|IH|$.

Therefore

$$ |PQ|+|BQ|+|CP|=|DQ|+|QB|+|BF|=|DF| $$

So to establish the result it must be shown that

$$ |DI|>|FH|=|IQ| $$

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Refer to the graph (using smartphone, will draw at PC later) below:

enter image description here

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