Arithmetic

May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies

jlacosta

May 1, 2025

0 replies

Number Theory problem

StoicFTW 0

5 minutes ago

Determine the number of solutions $(i, j, k)$ which satisfy the simultaneous equations
$i j=6 i+6 j+6 k$ and
$k^2=i^2+j^2$ where $i, j$ and $k$ are integers such that $k^2>i^2$ and $k^2>j^2$ .

0 replies

StoicFTW

5 minutes ago

0 replies

Inequalities from SXTX

sqing 24

N 6 minutes ago by sqing

T702. Let $a,b,c>0$ and $a+2b+3c=\sqrt{13}.$ Prove that $\sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$ S

$\sqrt{a^2+1} \geq \frac{ \sqrt{13} a +6}{7}$

T703. Let $a,b$ be real numbers such that $a+b\neq 0.$ . Find the minimum of $a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $a,b,c>0$ and $a+b+c=3.$ Prove that $\frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)} \geq 6$ S

$\frac{a^2+7}{(a+3)^2} + \frac{b^2+7}{(b+3)^2} +\frac{c^2+7}{(c+3)^2} \geq \frac{3}{2}$

24 replies

+1 w

sqing

Feb 18, 2025

sqing

6 minutes ago

a tst 2013 test

Math2030 9

N an hour ago by Shan3t

Given the sequence $(a_n): a_1=1, a_2=11$ and $a_{n+2}=a_{n+1}+5a_{n}, n \geq 1$
. Prove that $a_n$ not is a perfect square for all $n > 3$ .

9 replies

Math2030

Saturday at 5:26 AM

Shan3t

an hour ago

Challenge Problem in Similar Triangles

yes45 1

N an hour ago by LilKirb

$\triangle ABC$ has a right angle at $B$ . $AB = 9$ and $AC = 15$ . Point $D$ lies on $AB$ such that $AD = 2$ and point $E$ lies on $AC$ such that $AE$ = $\frac{12}{5}$ . Find the perimeter of $BCED$ .
Answer Confirmation

$\frac{168}{5}$

Solution

The perimeter of $BCED$ = $BC + CE + ED + BD$ .

$BC = 12$ .

$CE = 15 - \frac{12}{5} = \frac{63}{5}$ .

$BD = 9 - 2 = 7$ .

Now, to get the perimeter, we only need to solve for $ED$ .

Draw the altitude of $\triangle ADE$ from base $AE$ , and you will get two right triangles. Let point F be the point where that altitude intersects with $AE$ . $\triangle AFD$ shares angles $\angle{A}$ with $\triangle ABC$ and also has a right angle, so we can prove that $\triangle AFD \sim \triangle ABC$ through AA Postulate. $\frac{AB}{AC} = \frac {AF}{AD} = \frac{3}{5}$ , so $FA = 2 \cdot \frac{3}{5} = \frac{6}{5}$ and $FE = \frac{12}{5} - \frac{6}{5} = \frac{6}{5}$ . Since $FE \cong AF$ , $\angle{EFD} \cong \angle{DFA}$ (as both are right angles), and $DF \cong DF$ , we can prove $\triangle AFD \cong \triangle EFD$ through SAS postulate. $ED$ corresponds with $AD$ , so $ED = AD = 2$ .

$12 + \frac{63}{5} + 7 + 2 = \boxed{\frac{168}{5}}$ .

1 reply

yes45

Yesterday at 5:26 PM

LilKirb

an hour ago

Computer Science

rrusczyk 0

Nov 10, 2009

I've frequently ranted about how much easier it was for people in my generation to get started with programming than it is for students now. This post that levans pointed me to not only gives a vivid example, but also explains that making it harder for kids to get started with computing puts restrictions on their educational development in other areas, too. One of these days . . . .

0 replies

rrusczyk

Nov 10, 2009

0 replies

Master the Hard Stuff: CS Edition

rrusczyk 6

N Aug 19, 2009 by pascal_1588

Just last night I was recalling (not fondly) the days of malloc and calloc and worrying about memory when coding in C. Then today, a parent pointed me at this article, which gives, I think, a pretty good defense of why it was a good thing for me to have to deal with. I think there's something to his argument that is a lot like the argument I make when I argue that rapid acceleration within mathematics to fancier and fancier tools is not the best route for a gifted young student to take.

6 replies

rrusczyk

Aug 15, 2009

pascal_1588

Aug 19, 2009

Project Euler

rrusczyk 7

N Aug 18, 2009 by jhangil

Do any of you have experience with Project Euler? If so, what do you think?

7 replies

rrusczyk

Aug 13, 2009

jhangil

Aug 18, 2009

No more topics!

Arithmetic

Rounak_iitr 3

N Apr 12, 2025 by evt917

p2. In a room of $20$ people, $65\%$ are wearing a jacket and $40\%$ are wearing a hat. If $4$ people are wearing neither, how many people are wearing both $?$

3 replies

Rounak_iitr

Dec 4, 2023

evt917

Apr 12, 2025

Arithmetic

G H J

Reveal topic tags

arithmetic

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Rounak_iitr

456 posts

Rounak_iitr

#1 h Dec 4, 2023, 4:15 PM

Y by

p2. In a room of $20$ people, $65\%$ are wearing a jacket and $40\%$ are wearing a hat. If $4$ people are wearing neither, how many people are wearing both $?$

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MC413551

2228 posts

MC413551

#2 h Dec 4, 2023, 5:56 PM

Y by

13 have a jacket and 8 have a hat and 4 have neither so 20-4=16 and 13+8-x=16 so there are 5 people wearing both

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imnotgoodatmathsorry

82 posts

imnotgoodatmathsorry

#3 h Apr 12, 2025, 1:52 AM

Y by

Rounak_iitr wrote:

p2. In a room of $20$ people, $65\%$ are wearing a jacket and $40\%$ are wearing a hat. If $4$ people are wearing neither, how many people are wearing both $?$

THIS IS VENN DIAGRAM!
There are $13$ peoples who wears jacket and $8$ peoples who wears hat is $16$ people (That are wearing each one of two or both)
So, there will be $(13+8)-16=5$ people that are wearing both

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evt917

2424 posts

evt917

#4 h Apr 12, 2025, 1:57 AM

Y by

nahh how is this hsm

this should be in msm

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