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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Combinatorics problem
lgx57   1
N 4 minutes ago by alexheinis
There are $100$ positive integer arrays $\{a_{1,n}\},\{a_{2,n}\},\cdots,\{a_{100,n}\}$, and they are all infinity arrays.
Prove that: There exists $ m<k$, that for every integer $1 \le i \le 100, a_{i,m}<a_{i,k}$
1 reply
lgx57
Apr 14, 2025
alexheinis
4 minutes ago
Span to the infinity??
dotscom26   1
N 4 minutes ago by P0tat0b0y
The equation \[
\sqrt[3]{\sqrt[3]{x - \frac{3}{8}} - \frac{3}{8}} = x^3 + \frac{3}{8}
\]has exactly two real positive solutions \( r \) and \( s \). Compute \( r + s \).
1 reply
dotscom26
an hour ago
P0tat0b0y
4 minutes ago
Inequalities
sqing   19
N 2 hours ago by sqing
Let $ a,b,c> 0 $ and $ ab+bc+ca\leq  3abc . $ Prove that
$$ a+ b^2+c\leq a^2+ b^3+c^2 $$$$ a+ b^{11}+c\leq a^2+ b^{12}+c^2 $$
19 replies
sqing
Tuesday at 1:54 PM
sqing
2 hours ago
circumcenter, excenter and vertex collinear (Singapore Junior 2012)
parmenides51   6
N 4 hours ago by lightsynth123
In $\vartriangle ABC$, the external bisectors of $\angle A$ and $\angle B$ meet at a point $D$. Prove that the circumcentre of $\vartriangle ABD$ and the points $C, D$ lie on the same straight line.
6 replies
parmenides51
Jul 11, 2019
lightsynth123
4 hours ago
No more topics!
Trig + Triangle Laws: How come?
hastapasta   4
N Mar 28, 2023 by Carlos Jacob Rubio Barrio
My question is after the problem and my solution.
Problem:
In the quadrilateral $ABCD$, $\angle{ADC}=90^\circ$, $AB=2$, $BD=5$.
(1) Find $\cos{\angle{ADB}}$.
(2) If $DC=2\sqrt{2}$, find $BC$. (2018 China Gaokao Syllabus I #17)

Solution:
Draw a diagram (see the attachment):
(1) First, $\frac{5}{\sin{A}}=\frac{2}{\sin{\angle{ADB}}}$. Since $\sin{A}=\frac{\sqrt{2}}{2}$, we can plug it back and find that $\sin{\angle{ADB}}=\frac{\sqrt{2}}{5}$. Therefore, through the Pythagorean Identities, we find that $\cos{\angle{ADB}}=\boxed{\frac{\sqrt{23}}{5}}$.

(2) Notice that since $\angle{ADC}=90^\circ$, its sine is $1$. Therefore, using the fact that $\angle{ADC}=\angle{ADB}+\angle{BDC}$, we can say: $\sin{\angle{ADC}}=\sin{(\angle{ADB}+\angle{BDC})}=1$.
Expand using sum of sines: $\frac{\sqrt{2}}{5}\cos{\angle{BDC}}+\frac{\sqrt{23}}{5}\sin{\angle{BDC}}=1$.
Using $(\sin{\angle{BDC}})^2+(\cos{\angle{BDC}})^2=1$, call $\cos{\angle{BDC}}$ equal to $x$ (notice that since $\angle{BDC}$ is acute, its sine must be positive, so that's why I used the cosine to set the value to avoid problems), $\sin{\angle{BDC}}=\sqrt{1-x^2}$. Plug it back in and solve the quadratic equation, we find that $c=\frac{\sqrt{2}}{5}$. Now using the law of cosines, we find that $BC=\sqrt{5^2+(2\sqrt{2})^2-2*5*2\sqrt{2}*\frac{\sqrt{2}}{5}}=\boxed{5}$.

Wow! Such a long bash! Thanks for reading the solution. Now, here's the question:
In the answer key to part (2) of this problem, it said:

Because of the problem's given information and part (1) [its solution is identical to mine, just differently worded], it is obvious that $\cos{\angle{BDC}}=\sin{\angle{ADB}}$.



How come? I bashed all of this out to obtain that fact. How come it's "obvious?" Please explain. Notice that if the answer key omitted these details, it wouldn't be an answer key because the bashing is vital to this problem it looks like!

Thanks!
4 replies
hastapasta
Mar 29, 2022
Carlos Jacob Rubio Barrio
Mar 28, 2023
Trig + Triangle Laws: How come?
G H J
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hastapasta
131 posts
#1
Y by
My question is after the problem and my solution.
Problem:
In the quadrilateral $ABCD$, $\angle{ADC}=90^\circ$, $AB=2$, $BD=5$.
(1) Find $\cos{\angle{ADB}}$.
(2) If $DC=2\sqrt{2}$, find $BC$. (2018 China Gaokao Syllabus I #17)

Solution:
Draw a diagram (see the attachment):
(1) First, $\frac{5}{\sin{A}}=\frac{2}{\sin{\angle{ADB}}}$. Since $\sin{A}=\frac{\sqrt{2}}{2}$, we can plug it back and find that $\sin{\angle{ADB}}=\frac{\sqrt{2}}{5}$. Therefore, through the Pythagorean Identities, we find that $\cos{\angle{ADB}}=\boxed{\frac{\sqrt{23}}{5}}$.

(2) Notice that since $\angle{ADC}=90^\circ$, its sine is $1$. Therefore, using the fact that $\angle{ADC}=\angle{ADB}+\angle{BDC}$, we can say: $\sin{\angle{ADC}}=\sin{(\angle{ADB}+\angle{BDC})}=1$.
Expand using sum of sines: $\frac{\sqrt{2}}{5}\cos{\angle{BDC}}+\frac{\sqrt{23}}{5}\sin{\angle{BDC}}=1$.
Using $(\sin{\angle{BDC}})^2+(\cos{\angle{BDC}})^2=1$, call $\cos{\angle{BDC}}$ equal to $x$ (notice that since $\angle{BDC}$ is acute, its sine must be positive, so that's why I used the cosine to set the value to avoid problems), $\sin{\angle{BDC}}=\sqrt{1-x^2}$. Plug it back in and solve the quadratic equation, we find that $c=\frac{\sqrt{2}}{5}$. Now using the law of cosines, we find that $BC=\sqrt{5^2+(2\sqrt{2})^2-2*5*2\sqrt{2}*\frac{\sqrt{2}}{5}}=\boxed{5}$.

Wow! Such a long bash! Thanks for reading the solution. Now, here's the question:
In the answer key to part (2) of this problem, it said:

Because of the problem's given information and part (1) [its solution is identical to mine, just differently worded], it is obvious that $\cos{\angle{BDC}}=\sin{\angle{ADB}}$.



How come? I bashed all of this out to obtain that fact. How come it's "obvious?" Please explain. Notice that if the answer key omitted these details, it wouldn't be an answer key because the bashing is vital to this problem it looks like!

Thanks!
Attachments:
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hastapasta
131 posts
#2
Y by
I used GeoGebra since I didn't know how to create an image using LaTeX. Can someone give me help on that? Sorry. I read the wiki and tutorials and I'm still very stuck on how to create images in LaTeX.

Thanks!
This post has been edited 1 time. Last edited by hastapasta, Mar 29, 2022, 8:46 PM
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Lamboreghini
6486 posts
#3
Y by
Notice that $\angle BDC+\angle ADB=90^\circ.$ So the fact that $\cos\angle BDC=\sin \angle ADB$ is a direct result from the fact that $\cos\theta=\sin\left(90^\circ-\theta\right).$ The proof of this isn't too hard. Just consider the unit circle. Notice that rotating the point $(1,0)$ by an angle of $\theta$ counterclockwise takes it to the point $(\cos\theta,\sin\theta),$ and rotating it $\theta$ clockwise sends it to $(\cos(90^\circ-\theta),\sin(90^\circ-\theta)).$ Also notice that these two points are symmetric about $y=x,$ so the points $\cos(90^\circ-\theta),\sin(90^\circ-\theta))$ and $(\sin\theta,\cos\theta)$ are the same, so $\sin\theta=\cos(90^\circ-\theta)$ and $\cos\theta=\sin(90^\circ-\theta).$
Z K Y
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hastapasta
131 posts
#4
Y by
Oh, yeah! Didn't realize that! Thanks!
Z K Y
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Carlos Jacob Rubio Barrio
7 posts
#5
Y by
Why sin A=\sqrt{2}/2???
Z K Y
N Quick Reply
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