USAMO 1985 #3

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

March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
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0 replies

jlacosta

Mar 2, 2025

0 replies

inequalities

Cobedangiu 4

N an hour ago by sqing

Source: own

$a,b>0$ and $a+b=1$ . Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$

4 replies

Cobedangiu

Yesterday at 6:10 PM

sqing

an hour ago

Interesting inequalities

sqing 1

N an hour ago by sqing

Source: Own

Let $a,b,c> 0$ and $a+b+c=3$ . Prove that
$\frac{a^2}{a^2+b+c+ \frac{3}{2}}+\frac{b^2}{b^2+c+a+\frac{3}{2}}+\frac{c^2}{c^2+a+b+\frac{3}{2}} \leq \frac{6}{7}$ Equality holds when $(a,b,c)=(0,\frac{3}{2},\frac{3}{2})$ or $(a,b,c)=(0,0,3) .$

1 reply

1 viewing

sqing

an hour ago

sqing

an hour ago

Problem 4: ISL 2008/G3 Constructed Four Times

ike.chen 25

N an hour ago by Yiyj1

Source: USEMO 2022/4

Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points $P, Q, R, S$ lie in the interiors of segments $AB, BC, CD, DA,$ respectively, such that $\angle PDA = \angle PCB, \text{ } \angle QAB = \angle QDC, \text{ } \angle RBC = \angle RAD, \text{ and } \angle SCD = \angle SBA.$ Let $AQ$ intersect $BS$ at $X$ , and $DQ$ intersect $CS$ at $Y$ . Prove that lines $PR$ and $XY$ are either parallel or coincide.

Tilek Askerbekov

25 replies

ike.chen

Oct 23, 2022

Yiyj1

an hour ago

Inspired by old results

sqing 8

N an hour ago by sqing

Source: Own

Let $a,b,c> 0$ and $abc=1$ . Prove that
$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$ Where $0<k \leq 1.$

8 replies

sqing

Monday at 1:42 PM

sqing

an hour ago

No more topics!

USAMO 1985 #3

Mrdavid445 1

N Apr 24, 2015 by Jettywang828

Let $A,B,C,D$ denote four points in space such that at most one of the distances $AB,AC,AD,BC,BD,CD$ is greater than $1$ . Determine the maximum value of the sum of the six distances.

1 reply

Mrdavid445

Jul 26, 2011

Jettywang828

Apr 24, 2015

USAMO 1985 #3

G H J

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AMC

USA(J)MO

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combinatorics unsolved

combinatorics

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Mrdavid445

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Mrdavid445

#1 h Jul 26, 2011, 7:10 PM • 2 Y

Y by Adventure10, Mango247

Let $A,B,C,D$ denote four points in space such that at most one of the distances $AB,AC,AD,BC,BD,CD$ is greater than $1$ . Determine the maximum value of the sum of the six distances.

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Jettywang828

79 posts

Jettywang828

#2 h Apr 24, 2015, 10:45 PM • 1 Y

Y by Adventure10

Answer

$5+\sqrt{3}$

Solution

Call the sum of the six distances $S$ .

First, in order to achieve the desired maximum, there MUST be one distance greater than one; otherwise the best possible configuration is simply the four points being the four vertices of a regular tetrahedron.

Say $AB$ is the largest distance of the six; then $1<AB<2$ , the latter inequality because of the triangle inequality; if $AB = 2$ , then $C$ and $D$ would have to be the midpoint of $AB$ , and then $S$ will only be $6$ .

Draw 2 spheres each of radius $1$ with centers at $A$ and $B$ . Observe that points $C$ and $D$ must lie in the interior of both spheres, because their distances from $A$ and $B$ cannot be greater than $1$ . Observe further that in order to maximize $S$ , $C$ and $D$ must actually be on the circle that is the intersection of the spheres, and must be opposite each other on this circle, forming a diameter. This will make $AC$ , $AD$ , $BC$ , and $BD$ , all equal to $1$ , and will also maximize $CD$ within the given space.

Now we know that the four points are coplanar and form a rhombus of side length 1, and the only remaining unknown is the length of $AB$ itself.
We are trying to make $AB+CD$ as large as possible, so we notice that they are diagonals our rhombus, and that $\frac{AB}{2}$ and $\frac{CD}{2}$ are the legs of a right triangle with hypotenuse $1$ . Sum of the lengths of legs of a right triangle with fixed length is maximized when the triangle is isosceles, but our triangle can't be isosceles, as that would make $CD>1$ . Then we want the triangle as close to isosceles as possible, which happens when $CD=1$ . Wait- so $\triangle ACD$ and $\triangle BCD$ are equilateral! We have two equilateral triangles of side $1$ which share a side to form a rhombus, and so $AB$ is just two times the height of one of these, or $\sqrt{3}$ .

This brings us to our final answer of $S = \boxed{5+\sqrt{3}}$ , since all of the other lengths are $1$ .

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