1993 IMO Problems/Problem 4

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For three points $A,B,C$ in the plane, we define $m(ABC)$ to be the smallest length of the three heights of the triangle $ABC$, where in the case $A$, $B$, $C$ are collinear, we set $m(ABC) = 0$. Let $A$, $B$, $C$ be given points in the plane. Prove that for any point $X$ in the plane,

$m(ABC) \leq m(ABX) + m(AXC) + m(XBC)$.


Solution

First we prove the claim for all points $X$ within or on the triangle $ABC$: In this specific case, suppose without loss of generality that $a\ge b\ge c$. Then the length of the smallest height of $ABC$ is that of the height from vertex $A$. This has a value of $\frac{2[ABC]}{a}$. Similarly, for triangles $ABX,BCX,CAX$ we have that the length of the smallest height is twice the area of the respective triangle divided by the longest side of that triangle. It is clear that since $a$ is the longest side of $ABC$, no two points within $ABC$ have distance exceeding $a$. Thus, since $X$ is within $ABC$, the longest side of any of the triangles $ABX,BCX,CAX$ does not exceed $a$. So, we have

$m(ABX)+m(BCX)+m(CAX)\ge\frac{2[ABX]}{a}+\frac{2[BCX]}{a}+\frac{2[CAX]}{a}$

$=\frac{2[ABC]}{a}=m(ABC)$

as desired.

Now on to prove the assertion for $X$ outside the triangle $ABC$. We shall assume without loss of generality that out of the points $A,B,C$ point $A$ is that farthest from $X$.

If $ABCX$ is concave or a degenerate quadrilateral, assume without loss of generality that $B$ inside or on triangle $ACX$. We shall prove that $m(ACX)\ge m(ABC)$ to prove the main claim. If the shortest height of $ACX$ was from vertex $A$ then it is clear that the height of $ABC$ from $A$ is smaller than that since ray $CB$ is closer to $A$ than ray $CX$. The case of the height from $C$ being the smallest of triangle $ACX$ is analogous, so we move on to the case of the height from $X$ is the smallest. If this is the case, then it is clear that the height of $ABC$ from $B$ is smaller than $m(ABC)$ since $\angle XAC\ge \angle BAC$. Thus the claim is proved.

If, instead, $ABCX$ is convex, we can assign the letter $D$ to represent to point of intersection of $AX$ and $BC$. Before proving the main claim, we shall prove that for triangle $ABX$ we have $m(ABX)\ge m(ABD)$. We prove this by considering each of the vertices that the shortest height of $ABX$ is on. If the shortest height is that from $X$, then it is obvious that the height from $D$ to side $AB$ is smaller than that from $X$ since $D$ is closer to $A$ than $X$ is and so $m(ABX)\ge m(ABD)$. If the shortest height of $ABX$ was from $B$, then since the height from $B$ to $XD$ is equal to the height from $B$ to $AD$, we have $m(ABX)\ge m(ABD)$. If instead the shortest height of $ABX$ was from $A$, the it is clear that $\angle B<90$. Thus, the projection of $A$ onto $BD$ is on the same side of $B$ as $C$. Now it is obvious that the height of $ABD$ from $A$ is smaller than that of $ABX$ from $A$ since the ray $BD$ is closer to $A$ than the ray $BX$. Thus, we have $m(ABX)\ge m(ABD)$ in all cases. Notice that the case for triangle $ACX$ is analogous. Therefore, we have $m(ABX)+m(ACX)\ge m(ABD)+m(ACD)$. But, this is the case of $D$ on the triangle $ABC$, and this case was shown in the first part of this proof.

So, we have that $m(ABC) \leq m(ABX) + m(AXC) + m(XBC)$ is true in all cases, as desired.

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