square on a square tile (2017 AESC MSU Internet Olympiad 7.4)

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0 replies

jlacosta

Apr 2, 2025

0 replies

Eazy equation clap

giangtruong13 0

an hour ago

Find all $x,y,z$ satisfy that: $\frac{x}{y+z}=2x-1; \frac{y}{x+z}=3y-1;\frac{z}{x+y}=5x-1$

0 replies

giangtruong13

an hour ago

0 replies

inequality

revol_ufiaw 3

N 2 hours ago by MS_asdfgzxcvb

Prove that that for any real $x \ge 0$ and natural number $n$ ,
$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$

3 replies

revol_ufiaw

3 hours ago

MS_asdfgzxcvb

2 hours ago

How to judge a number is prime or not?

mingzhehu 0

2 hours ago

A=（10X1+1）(10X+1），X1，X∈N+
B=（10 X1+3）（10X+7），X∈N，X1∈N
C=(10 X1+9)(10X+9), X∈N，X1∈N
D=(10 X1+1)(10X+3), X1∈N+，X∈N
E=（10 X1+7）(10X+9），X∈N，X1∈N
F=（10 X1+1）（10X+7），X1∈N+，X∈N
G=（10 X1+3）（10X+9），X∈N，X1∈N
H=（10 X1+1）10X+9），X1∈N+，X∈N
I=（10 X1+3）（10X+3），X1∈N，X∈N
J=( 10X1+7)(10X+7),X∈N，X1∈N

For any natural number P∈{P=10N+1，n∈N}，make P=A or B or C
If P can make the roots of function group(ABC) without any root group completely made up of integer, P will be a prime
For any natural number P∈{P=10N+3，n∈N}，make P=D or E
If P can make the roots of function group(DE) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+7，n∈N}，make P=F or G
If P can make the roots of function group(FG) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+9，n∈N}，make P=H or I or J
If P can make the roots of function group(GIJ) without any root group completely made up
of integer, P will be a prime

0 replies

mingzhehu

2 hours ago

0 replies

What is an isogonal conjugate and why is it useful?

EaZ_Shadow 6

N 2 hours ago by maxamc

What is an isogonal conjugate and why is it useful? People use them in Olympiad geometry proofs but I don’t understand why and what is the purpose, as it complicates me because of me not understanding it.

6 replies

EaZ_Shadow

Dec 28, 2024

maxamc

2 hours ago

No more topics!

square on a square tile (2017 AESC MSU Internet Olympiad 7.4)

parmenides51 4

N Feb 19, 2021 by natmath

A square sheet of $ABCD$ paper was dropped onto a square tile so that the corner $A$ of the sheet fell into the corner of the tile, side $BC$ went through the corner of the tile $E$ , and the corner $C$ went to the edge of the tile. Find the angle $BAE$ .
IMAGE

4 replies

parmenides51

Feb 19, 2021

natmath

Feb 19, 2021

square on a square tile (2017 AESC MSU Internet Olympiad 7.4)

G H J

Reveal topic tags

geometry

square

G H BBookmark kLocked kLocked NReply

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parmenides51

30629 posts

parmenides51

#1 h Feb 19, 2021, 4:50 PM • 1 Y

Y by samrocksnature

https://cdn.artofproblemsolving.com/attachments/a/e/b18541c0b0c7ea6cdacc2bcab7a68d263ed131.png

This post has been edited 2 times. Last edited by parmenides51, Feb 19, 2021, 5:18 PM

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parmenides51

30629 posts

parmenides51

#2 h Feb 19, 2021, 4:51 PM • 1 Y

Y by samrocksnature

posted for the image link

Attachments:

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natmath

8219 posts

natmath

#3 h Feb 19, 2021, 6:27 PM • 1 Y

Y by samrocksnature

solution

Define $x=AB$ , $y=\frac{AE}{2}$ , and $\theta=\angle BAE$ .
We have
$2y\cos\theta=x$ $x\sqrt{2}\sin (45-\theta)=y$ Multiplying the 2 equations gives
$\cos\theta (\cos\theta-\sin\theta)=\frac{1}{2}$ $-\cos\theta\sin\theta=-\frac{1}{2}+\cos^2\theta$ Define $u=\sin\theta$ ,
$2u\sqrt{1-u^2}=1-2u^2$ $4u^2(1-u^2)=4u^4-4u^2+1$ $0=8u^4-8u^2+1$ $u^2=\frac{2 \pm \sqrt{2}}{4}$ Since $u$ is positive, we require that $1-2u^2>0$ , i.e. $u^2<\frac{1}{2}$ . This means the appropriate solution is $u^2=\frac{2-\sqrt{2}}{4}$
Since $\sin^2\theta=\frac{1-\cos 2\theta}{2}$ , we have
$1-\cos 2\theta=\frac{2-\sqrt{2}}{2}$ $-\cos 2\theta =-\frac{\sqrt{2}}{2}$ $2\theta=45^\circ$ $\theta=\boxed{22.5^\circ}$

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donian9265

541 posts

donian9265

#4 h Feb 19, 2021, 9:54 PM • 1 Y

Y by natmath

another solution

Let $\angle{BAE}=\theta$ and let $AE=2x$ . This means $AB=2x\cos(\theta)$ and $BE=2x\sin(\theta)$ . Furthermore, if we call the intersection of the edges of tiles where $E$ and $C$ are on $F$ , then it is clear $EF=x$ and $\angle{XEC}=\theta$ . Therefore, $EC=\frac{x}{\cos(\theta)}$ . From this, we know $2x\sin(\theta)+\frac{x}{\cos(\theta)}=2x\cos(\theta)$ $2\sin(\theta)\cos(\theta)=\cos^2(\theta)-1$ $\sin(2\theta)=\cos(2\theta)$ This implies $2\theta=\frac{\pi}{4}$ or $\theta=\angle{}BAE=\boxed{\frac{\pi}{8}}$ . Note that all other values for $\theta$ that satisfy $\sin(2\theta)=\cos(2\theta)$ would lead to the sum of the angles of $\triangle{BAE}$ exceeding $\pi$ .

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natmath

8219 posts

natmath

#5 h Feb 19, 2021, 10:09 PM

Y by

Oh lol. I kept on making arithmetic errors and had to go back to fix stuff. As a result, I missed that really nice equation. Nice solution @above.

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