projection of an convex polyhedron inside a cube on any face covers it, max V

I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE

1571 replies

rrusczyk

Mar 24, 2025

SmartGroot

Mar 26, 2025

k a March Highlights and 2025 AoPS Online Class Information

jlacosta 0

Mar 2, 2025

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0 replies

jlacosta

Mar 2, 2025

0 replies

IMO 2008, Question 1

orl 153

N 4 minutes ago by QueenArwen

Source: IMO Shortlist 2008, G1

Let $H$ be the orthocenter of an acute-angled triangle $ABC$ . The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing through $H$ intersects the sideline $BC$ at points $A_{1}$ and $A_{2}$ . Similarly, define the points $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ .

Prove that the six points $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia

153 replies

orl

Jul 16, 2008

QueenArwen

4 minutes ago

P(x) doesn't have 2024 distinct real roots

gatnghiep 1

N 7 minutes ago by kiyoras_2001

Given a polynomial $P(x)=x^{2024}+a_{2023}x^{2023}+...+a_1x+1$ with real coefficient. It is known that $|a_{1012}|<2$ and $a_k = a_{2024-k}, \forall k = 1,2,...,2012$ . Prove that $P(x)$ can't have $2024$ distinct real roots.

1 reply

gatnghiep

Oct 2, 2024

kiyoras_2001

7 minutes ago

A lies on the radical axis of BQX and CPX

a_507_bc 35

N 35 minutes ago by jordiejoh

Source: APMO 2024 P1

Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$ . Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$ , respectively, such that both $P$ and $Q$ lie between $B$ and $C$ . Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$ . Prove that the points $A, X$ , and $Y$ are collinear.

35 replies

a_507_bc

Jul 29, 2024

jordiejoh

35 minutes ago

sum(ab/4a^2+b^2) <= 3/5

truongphatt2668 3

N an hour ago by Nguyenhuyen_AG

Source: I remember I read it somewhere

Let $a,b,c>0$ . Prove that:
$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$

3 replies

truongphatt2668

Monday at 1:23 PM

Nguyenhuyen_AG

an hour ago

No more topics!

projection of an convex polyhedron inside a cube on any face covers it, max V

parmenides51 1

N Dec 18, 2019 by greenturtle3141

Source: Quantum - English M29

A cube contains a convex polyhedron whose projection onto any of the cube's faces covers the entire face. Show that the volume of the poiyhedron is not less than $1/3$ that of the cube.

(V. Prasolov)

1 reply

parmenides51

Aug 4, 2018

greenturtle3141

Dec 18, 2019

projection of an convex polyhedron inside a cube on any face covers it, max V

G H J

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Source: Quantum - English M29

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parmenides51

30629 posts

parmenides51

#1 h Aug 4, 2018, 7:59 AM • 2 Y

Y by Adventure10, Mango247

This post has been edited 1 time. Last edited by parmenides51, Aug 29, 2020, 7:37 PM

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greenturtle3141

3541 posts

greenturtle3141

#2 h Dec 18, 2019, 7:56 AM • 1 Y

Y by Adventure10

Solution

First notice that the polyhedron contains a point from every edge. Otherwise, if it does not contain a point from some edge, then the projection on either face perpendicular to that edge will be missing either endpoint of the edge.

Hence, if we choose 12 points, one from every edge, such that each point is inside or on the polyhedron, then the convex hull of those 12 points will be contained within the polyhedron, so the volume of that convex hull will necessarily be less than or equal to that of the polyhedron.

Now assume it is a unit cube. It suffices to show that the convex hull of 12 points, one taken from each edge, has a minimal volume of $1/3$ .

Observe that this convex hull is simply the unit cube with eight right tetrahedrons cut out. We now claim that the minimal case must occur when each of 12 points lies at a vertex (with potential overlap). To see this, isolate a single point on some edge. Let the areas of the opposite faces of the two attached aforementioned tetrahedrons (with this point as a vertex) be $A$ and $B$ . Note that moving this point along the edge only affects the volumes of these two tetrahedrons, and the sum of these volumes is given by:
$\frac13Ax + \frac13B(1-x)$ $= \frac13((A-B)x+B)$ Where $x$ is the positive distance from the point to a chosen vertex of the cube belonging to its edge. For $A>B$ , the volume sum is minimized when $x=0$ , and for $A \leq B$ , the sum is minimized when $x=1$ . i.e. the minima occurs at the extremes, hence proven.

Now it suffices to minimize the convex hull of a subset of the vertices of the cube, such that this subset contains a vertex from every edge. Color the vertices alternating black and white.

We take cases: Suppose that the subset does not contain all four vertices of some color. Then the subset lacks at least a certain black vertex and a certain white vertex. These vertices cannot be adjacent, lest their edge has no vertices in the subset. Hence the vertices are opposite corners of the cube. Note that the convex hull of the other six vertices will then have volume $2/3$ (compute volume of the complement to see this), and $2/3 > 1/3$ , so this case can be eliminated.

Otherwise, the subset contains four vertices, all of some same color. The convex hull of such a subset will have volume $1/3$ (computed once again by considering the complement), and hence this is the minimal case. We are done.

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