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David E Speyer
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From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

enter image description here

A quick proof: Let $ABC$ be equilateral and $P$ arbitrary. Construct $BPQ$ equilateral. Let $AB=AC=BC=w$, $AP=x$, $BP=y$ and $CP=z$. Then $BP=PQ=BQ=y$ by construction, $CP=z$ and $CB=x$$CB=w$ obviously, so it remains to check that $CQ=x$. Now note that triangle $CBQ$ is the $60^\circ$ rotation of $ABP$ around $B$.

From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

enter image description here

A quick proof: Let $ABC$ be equilateral and $P$ arbitrary. Construct $BPQ$ equilateral. Let $AB=AC=BC=w$, $AP=x$, $BP=y$ and $CP=z$. Then $BP=PQ=BQ=y$ by construction, $CP=z$ and $CB=x$ obviously, so it remains to check that $CQ=x$. Now note that triangle $CBQ$ is the $60^\circ$ rotation of $ABP$ around $B$.

From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

enter image description here

A quick proof: Let $ABC$ be equilateral and $P$ arbitrary. Construct $BPQ$ equilateral. Let $AB=AC=BC=w$, $AP=x$, $BP=y$ and $CP=z$. Then $BP=PQ=BQ=y$ by construction, $CP=z$ and $CB=w$ obviously, so it remains to check that $CQ=x$. Now note that triangle $CBQ$ is the $60^\circ$ rotation of $ABP$ around $B$.

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David E Speyer
  • 156.3k
  • 14
  • 421
  • 763

From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

enter image description here

A quick proof: Let $ABC$ be equilateral and $P$ arbitrary. Construct $BPQ$ equilateral. Let $AB=AC=BC=w$, $AP=x$, $BP=y$ and $CP=z$. Then $BP=PQ=BQ=y$ by construction, $CP=z$ and $CB=x$ obviously, so it remains to check that $CQ=x$. Now note that triangle $CBQ$ is the $60^\circ$ rotation of $ABP$ around $B$.

From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

enter image description here

A quick proof: Let $ABC$ be equilateral and $P$ arbitrary. Construct $BPQ$ equilateral. Let $AB=AC=BC=w$, $AP=x$, $BP=y$ and $CP=z$. Then $BP=PQ=BQ=y$ by construction, $CP=z$ and $CB=x$ obviously, so it remains to check that $CQ=x$. Now note that triangle $CBQ$ is the $60^\circ$ rotation of $ABP$ around $B$.

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Edgardo
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From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z"

Post Made Community Wiki by Edgardo