No tags match your search
Mrelatively prime
geometry
algebra
number theory
trigonometry
inequalities
function
polynomial
probability
combinatorics
calculus
analytic geometry
3D geometry
quadratics
AMC
ratio
AIME
modular arithmetic
logarithms
LaTeX
complex numbers
rectangle
conics
circumcircle
geometric transformation
induction
integration
floor function
system of equations
counting
perimeter
rotation
trig identities
vector
trapezoid
search
graphing lines
angle bisector
prime numbers
slope
parallelogram
AMC 10
symmetry
relatively prime
parabola
Diophantine equation
Vieta
angles
Inequality
factorial
domain
No tags match your search
MG
Topic
First Poster
Last Poster
Easy expected value problem
Kempu33334 6
N
3 hours ago
by aidan0626
Source: Me
Bob takes two real numbers,
and
, independently at random from the range
to
. He then selects an operation out of
,
, and
and performs that operation on them. Find the expected value of the answer.
Bob takes two real numbers,







6 replies
Challenge Problem in Similar Triangles
yes45 0
Today at 5:26 PM












Answer Confirmation

Solution
The perimeter of
=
.
.
.
.
Now, to get the perimeter, we only need to solve for
.
Draw the altitude of
from base
, and you will get two right triangles. Let point F be the point where that altitude intersects with
.
shares angles
with
and also has a right angle, so we can prove that
through AA Postulate.
, so
and
. Since
,
(as both are right angles), and
, we can prove
through SAS postulate.
corresponds with
, so
.
.





Now, to get the perimeter, we only need to solve for

Draw the altitude of


















0 replies
Interesting Geometry
captainmath99 0
Today at 5:11 PM
Let ABC be a right triangle such that
. A circle O with center C has a radius of 2. Let P be a point on the circle O.
a)What is the minimum value of
?
b) What is the minimum value of
?

a)What is the minimum value of

b) What is the minimum value of

0 replies
[PMO27 Qualis] III. 1 Binary Counting
tapilyoca 5
N
Today at 4:44 PM
by Magdalo
John wrote down all of the numbers from 1 to 128 in binary. How many 1's did he write?
5 replies
22nd PMO Qualifying Stage #11
pensive 3
N
Today at 4:31 PM
by Magdalo
Let
and
be positive real numbers such that
What is the value of
?
Answer


![\[
\log_x 64 + \log_{y^2} 16 = \frac{5}{3} \quad \text{and} \quad \log_y 64 + \log_{x^2} 16 = 1
\]](http://latex.artofproblemsolving.com/d/7/9/d79facd90d9c5b53cdc4df9db18923ad34f472df.png)

Answer

3 replies
[PMO22 Qualifying] I.11
Magdalo 3
N
Today at 4:30 PM
by Magdalo
Let
and
be positive real numbers such that
Find


![\[\log_x64+\log_{y^2}16=\dfrac{5}{3}\text{ and }\log_y64+\log_{x^2}16=1 \]](http://latex.artofproblemsolving.com/6/6/6/666e220af0dd4e1f77283dc5109a59d796c81bda.png)

3 replies
Log Rationality
Magdalo 1
N
Today at 4:30 PM
by Magdalo
Let
be positive integers. How many pairs of
are there such that
is rational?



1 reply
Original Problem U1
NeoAzure 1
N
Today at 4:24 PM
by NeoAzure
How many three-digit positive integers are divisible by 4 and have exactly two even digits?
Solution
Answer
Solution
Any integer divisible by 4 is divisible by 2; which means that the ones digit is always an even number. Because the condition states that there should be exactly two even numbers, we can divide the problem into cases.
Where ABC is a three digit number with digits A, B, and C respectively.
Case 1:
A and C are even, B is odd.
There are 4 possible digits for the hundreds place,
5 possible digits for the tens place.
Since the number must be divisible by 4 and the tens digit is odd, there are only 2 possible ones digits (2, 6)
Total: 4 x 5 x 2 = 40
Case 2:
B and C are even, A is odd.
There are 5 possible digits for the hundreds place,
5 possible digits for the tens place.
Since the tens place is even, there are 3 possible digits for the ones place. (0, 4, 8)
Total: 5 x 5 x 3 = 75
Total for both cases: 40 + 75 = 115.
Where ABC is a three digit number with digits A, B, and C respectively.
Case 1:
A and C are even, B is odd.
There are 4 possible digits for the hundreds place,
5 possible digits for the tens place.
Since the number must be divisible by 4 and the tens digit is odd, there are only 2 possible ones digits (2, 6)
Total: 4 x 5 x 2 = 40
Case 2:
B and C are even, A is odd.
There are 5 possible digits for the hundreds place,
5 possible digits for the tens place.
Since the tens place is even, there are 3 possible digits for the ones place. (0, 4, 8)
Total: 5 x 5 x 3 = 75
Total for both cases: 40 + 75 = 115.
Answer
115
1 reply
[Sipnayan 2017 JHS] Semifinals A, Difficult, P2
NeoAzure 1
N
Today at 4:06 PM
by NeoAzure
Akira fills an urn with 10 chips such that 1 chip is labeled “1”, 2 chips are labeled “2”, 3
chips are labeled “3”, and 4 chips are labeled “4”. She draws 4 chips from the box without
replacement. What is the probability the sum of the numbers labeled on the 4 chips is divisible
by 3
Solution
Answer
chips are labeled “3”, and 4 chips are labeled “4”. She draws 4 chips from the box without
replacement. What is the probability the sum of the numbers labeled on the 4 chips is divisible
by 3
Solution
There are 10C4 = 210 total ways to pick 4 chips.
Express each kind of chip in mod 3 and count the number of each distinct value.
Since 4 ≡ 1 (mod 3), there are:
3 chips with value 0 ≡ (mod 3)
5 chips with value 1 ≡ (mod 3)
2 chips with value 2 ≡ (mod 3)
Case 1: 0 chips with value ≡ 0 (mod 3) is picked
The only possibility is to pick 2 chips with modular value 2 and 2 chips with modular value 1. There are 2C2 X 5C2 = 10 combinations.
Case 2: 1 chip with value ≡ 0 (mod 3) is picked
The only possibility is to pick 3 chips with modular value 1. One modular value 0 chip must also be picked. There are 5C3 X 3C1= 30 combinations.
Case 3: 2 chips with value ≡ 0 (mod 3) is picked
Picking one modular value 2 chip and one modular value 1 chip results in sum 3. Two modular value 0 chip must also be picked. There are 2C1 X C1 X 3C2 = 30 combinations.
Case 4: 3 chips with value ≡ 0 (mod 3) is picked
There are no possible combinations.
Total possible combinations from cases 1~4 that fulfill the conditions: 10 + 30 + 30 = 70.
Divide by the total number of possible combinations of picking: 70/210 = 1/3.
Express each kind of chip in mod 3 and count the number of each distinct value.
Since 4 ≡ 1 (mod 3), there are:
3 chips with value 0 ≡ (mod 3)
5 chips with value 1 ≡ (mod 3)
2 chips with value 2 ≡ (mod 3)
Case 1: 0 chips with value ≡ 0 (mod 3) is picked
The only possibility is to pick 2 chips with modular value 2 and 2 chips with modular value 1. There are 2C2 X 5C2 = 10 combinations.
Case 2: 1 chip with value ≡ 0 (mod 3) is picked
The only possibility is to pick 3 chips with modular value 1. One modular value 0 chip must also be picked. There are 5C3 X 3C1= 30 combinations.
Case 3: 2 chips with value ≡ 0 (mod 3) is picked
Picking one modular value 2 chip and one modular value 1 chip results in sum 3. Two modular value 0 chip must also be picked. There are 2C1 X C1 X 3C2 = 30 combinations.
Case 4: 3 chips with value ≡ 0 (mod 3) is picked
There are no possible combinations.
Total possible combinations from cases 1~4 that fulfill the conditions: 10 + 30 + 30 = 70.
Divide by the total number of possible combinations of picking: 70/210 = 1/3.
Answer
1/3
1 reply
Inequalities
toanrathay 2
N
Today at 4:00 PM
by MathsII-enjoy
Let
be positive reals such that
, find
of
.




2 replies
