1993 IMO Problems/Problem 4

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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1993 IMO Problems/Problem 4

Solution