MathLabs

Problem 4

For three points A,B,CA,B,C in the plane, let m(ABC)m(ABC) be the smallest length of the three heights of triangle ABCABC, and set m(ABC)=0m(ABC)=0 when the points are collinear. Given A,B,CA,B,C, prove that for every point XX in the plane, m(ABC)≤m(ABX)+m(AXC)+m(XBC)m(ABC)\le m(ABX)+m(AXC)+m(XBC).
Step 2 of 5: Handle X inside ABC
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In plain words

The three small triangle areas tile ABC.

m(ABX)+m(BCX)+m(CAX)≥2([ABX]+[BCX]+[CAX])am(ABX)+m(BCX)+m(CAX)\ge\frac{2([ABX]+[BCX]+[CAX])}{a}
Detailed analysis

If X lies in or on ABC, every distance among A,B,C,X is at most a. Thus the longest side in each of ABX, BCX and CAX is at most a. Each minimum height is therefore at least twice its area divided by a.