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3 of 3
Recomputed so that area = 3.
Joseph O'Rourke
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Not an answer, just an illustration to accompany the question. $K$ is an isosceles triangle with base $2$ and altitude $3$ (and so area $3$). First, I mistakenly computed the ellipse $E$ of any area with the smallest area symmetric difference with $K$. It has area about $2.4$:


            [![SymDiff13not][1]][1]
After Gerhard's comment, I recomputed constraining $E$ to have area $3$. Then its center is $(0,1)$ and its axes are roughly $0.74$ and $1.29$:
            [![SymDiff13area3][2]][2]
Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958