Difference between revisions of "2021 MPFG Problem 19"

m (Solution 1)
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==Solution 1==
 
==Solution 1==
Connect O with the 4 vertices of <math>T</math>. Extend the line made by connecting the top vertex of <math>T</math> with <math>O</math>, intersecting at the base/vertex of <math>t</math>.  
+
Connect <math>O</math> with the 4 vertices of <math>T</math>. Extend the line made by connecting the top vertex of <math>T</math> with <math>O</math>, intersecting at the base/vertex of <math>t</math>.  
  
 
<math>S</math> equals to <math>1</math> regular tetrahedron with <math>4</math> protruding tetrahedrons.
 
<math>S</math> equals to <math>1</math> regular tetrahedron with <math>4</math> protruding tetrahedrons.

Revision as of 05:07, 27 August 2025

Problem

Let $T$ be a regular tetrahedron. Let $t$ be the regular tetrahedron whose vertices are the centers of the faces of $T$. Let $O$ be the circumcenter of either tetrahedron. Given a point $P$ different from $O$, let $m(P)$ be the midpoint of the points of intersection of the ray $\overrightarrow{OP}$ with $t$ and $T$. Let $S$ be the set of eight points m(P) where P is a vertex of either $t$ or $T$. What is the volume of the convex hull of $S$ divided by the volume of $t$? Express your answer as a fraction in simplest form.

Solution 1

Connect $O$ with the 4 vertices of $T$. Extend the line made by connecting the top vertex of $T$ with $O$, intersecting at the base/vertex of $t$.

$S$ equals to $1$ regular tetrahedron with $4$ protruding tetrahedrons.

3d.png 2d.png

$S_{tetra} = (\frac{5}{3})^3 = \frac{125}{27}$

$S_{total} = \frac{125}{27} \cdot (1+\frac{4}{5}) = \boxed{\frac{25}{3}}$

~cassphe