Let $M$ be a point in the interior of triangle $ABC$ in the plane. Prove $AM+BM+CM<AB+BC+CA$.

The above question was posed to someone I know who is taking high-school Euclidean geometry. I'm not sure what theorems she can rely on in her proof (though they all follow from Euclid's axioms anyway), but I do know that she does not use trigonometry at all. She turned to me (a mathematician by training) for help; and I can't seem to prove it. So I turn to you all for a proof (using facts from high-school geometry only).

One thing I can prove is that $\sup(AM,BM,CM)<\sup(AB,BC,CA)$. Indeed, say the longest of the interior segments is $\overline{AM}$. Drop an altitude (perpendicular) from $A$ to the point $D\in\overline{BC}$, and consider the side — $\overline{AB}$ or $\overline{AC}$ — such that $\overline{AM}$ lies between $\overline{AD}$ and that side. (If $\overline{AM}\subset\overline{AD}$, consider either $\overline{AB}$ or $\overline{AC}$.) Say it's $\overline{AB}$. Then examining right triangle $ADB$ shows easily that $AM<AB$. However, I can't seem to prove that each of the three sides can be used in turn for one of the interior segments in that proof — which would suffice for the problem above.

Another idea I had was to prove that angle $AMB$ is strictly larger than angle $ACB$ (and likewise for the other two angles) and to use that to prove the claim. But I can't seem to do either: neither to prove the inequality of angle measures, nor, assuming that inequality, to prove the claim sought.

Any help would be much appreciated — again, using high-school geometry only. I suspect there's a simple proof I'm not seeing.

  • 1
    $\begingroup$ The claim is relatively straightforward to prove for $M$ on the perimeter. Then the fact that the sum is convex as $M$ moves on any line from the perimeter to the opposite vertex shows that the claim also holds in the interior. Thus it would suffice if you could formulate and prove this convexity in geometrical terms. $\endgroup$ – joriki Jun 7 '13 at 6:38

enter image description here

The method is to prove $AC+BC>AM+BM$:

Extend $AM$, let $ME=MB \implies \angle MBE=\angle MEB$, $AM$ cross $BC$ at $F$ (because $M$ is inside of $\triangle ABC$).

If $E$ is on $MF$, then $AF\ge AE=AM+ME$. $BC>CF \implies BC+AC> FC+AC>AF \ge AE=AM+BM$,

If $E$ is on the extension $MF$, $\angle MBE> \angle CBE, \angle BEC>\angle MEB =\angle MBE > \angle CBE$, so in $\triangle CEB$, $BC>CE, \implies BC+AC>AC+CE>AE=AM+BM$.

So we've proven $AC+BC>AM+BM$. For the same reason, $BC+AB>AM+MC$, $AB+AC>BM+MC$.

Finally, we have $AC+BC+AB>AM+BM+CM$.

edit1: more simple way:

$AC +CF> AF =AM +MF, MF + BF > BM \\ \implies AC+CF+BF+MF > AM+BM+MF \\ \implies AC+BC> AM+BM$

  • $\begingroup$ my pleasure if it helps! $\endgroup$ – chenbai Jun 7 '13 at 12:13

The distance from a fixed point is a convex function, and the sum of convex functions gives a convex function. Any triangle is a convex set, and convex functions on convex sets attain their maxima on the boundary. Through this simple principle, if $R$ ranges over a triangle $ABC$ we have

$$ RA+RB+RC \leq AB+AC+BC-\min(AB,AC,BC).$$

  • $\begingroup$ Nice proof, but I don't think it'd be accessible to a typical high-school geometry student. $\endgroup$ – msh210 Dec 16 '17 at 23:57

I would like to share a proof that relies on 2 theorems/axioms that should be known to someone taking high school Euclidean geometry.

Triangle Inequality (TI):

The sum of the lengths of any two sides of a triangle is greater or equal than the length of the remaining side, where the inequality is strict if the triangle is non-degenerate (meaning it has a non-zero area).

Triangle Convexity (TC):

If a straight line is drawn "through" a triangle, which means at least one interior point of the triangle belongs to the line, then the line intersects the boundary of the triangle exactly twice.

In what follows any triangle is non-degenerate.

As pointed out by the excellent answer from chenbai, it is sufficient to prove the following:

Theorem 1:

Given a triangle $ABC$ and an interior point M of the triangle, we have $$AM + MB < AC + CB$$ enter image description here

First we prove a special case of Theorem 1:

Theorem 2:

Given a triangle $ABC$ and a point $M$ lying on any of the sides of the triangle, then $$AM + MB < AC + CB$$


The case in which $M$ coincides with any of the 3 vertices $A,B,C$ follows immediately from (TI) applied to triangle $ABC$.

Suppose then that $M$ does not coincide with any of the vertices of the triangle; we need to consider 3 remaining cases.

Case 1: $M$ lies on side $AB$

enter image description here

This is trivial as $AM+MB=AB$ and by (TI) applied to triangle $ABC$ we have $AM+MB=AB<AC+CB$

Case 2: $M$ lies on side $BC$

enter image description here

Considering triangle $ACM$ and applying (TI) we get $$AM<AC+CM$$ Adding $MB$ to both sides of the inequality we obtain $AM+MB<AC+CM+MB$, but $CM+MB=CB$ hence we have shown that $$AM+MB<AC+CB$$

Case 3: $M$ lies on side $AC$.

It simply follows by symmetry from Case 2 above.

We can finally prove Theorem 1.

First let's extend the segment $AM$ such that it crosses the side $CB$ at point $M'$; this follows by the (TC) property since $M$ is an interior point of the triangle.

enter image description here

By Theorem 2, we have

$$AM'+M'B < AC + CB \tag 1$$

By (TI) applied to triangle $BMM'$ we have $$MB < MM' +M'B$$ Adding $AM$ to both side of the inequality we obtain:

$$AM+MB < AM + MM' +M'B \tag 2 $$

Since $AM + MM' = AM'$, $(2)$ becomes $$AM + MB < AM' + M'B \tag 3 $$

Now from $(1)$ and $(3)$ by transitivity we get $$AM + MB < AM'+M'B < AC + CB$$ as we wanted to prove.


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