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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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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0 replies
jlacosta
May 1, 2025
0 replies
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Inequality Involving Complex Numbers with Modulus Less Than 1
tom-nowy   0
a minute ago
Let $x,y,z$ be complex numbers such that $|x|<1, |y|<1,$ and $|z|<1$.
Prove that $$ |x+y+z|^2 +3>|xy+yz+zx|^2+3|xyz|^2 .$$
0 replies
tom-nowy
a minute ago
0 replies
J
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Inequality
nguyentlauv   2
N 2 minutes ago by nguyentlauv
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
2 replies
nguyentlauv
May 6, 2025
nguyentlauv
2 minutes ago
J
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japan 2021 mo
parkjungmin   0
2 minutes ago

The square box question

Is there anyone who can release it
0 replies
parkjungmin
2 minutes ago
0 replies
J
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easy sequence
Seungjun_Lee   17
N 7 minutes ago by GreekIdiot
Source: KMO 2023 P1
A sequence of positive reals $\{ a_n \}$ is defined below. $$a_0 = 1, a_1 = 3, a_{n+2} = \frac{a_{n+1}^2+2}{a_n}$$Show that for all nonnegative integer $n$, $a_n$ is a positive integer.
17 replies
Seungjun_Lee
Nov 4, 2023
GreekIdiot
7 minutes ago
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No more topics!
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Prove that DP=DR
WakeUp   8
N Jul 25, 2020 by dchenmathcounts
Source: Polish Second Round 2001
Points $A,B,C$ with $AB<BC$ lie in this order on a line. Let $ABDE$ be a square. The circle with diameter $AC$ intersects the line $DE$ at points $P$ and $Q$ with $P$ between $D$ and $E$. The lines $AQ$ and $BD$ intersect at $R$. Prove that $DP=DR$.
8 replies
WakeUp
Mar 6, 2012
dchenmathcounts
Jul 25, 2020
J
Prove that DP=DR
G H J
Reveal topic tags
geometry proposed
geometry
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G H BBookmark kLocked kLocked NReply
Source: Polish Second Round 2001
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WakeUp
1347 posts
WakeUp
#1 h  Mar 6, 2012, 7:02 PM • 2 Y
Y by Adventure10, Mango247
Points $A,B,C$ with $AB<BC$ lie in this order on a line. Let $ABDE$ be a square. The circle with diameter $AC$ intersects the line $DE$ at points $P$ and $Q$ with $P$ between $D$ and $E$. The lines $AQ$ and $BD$ intersect at $R$. Prove that $DP=DR$.
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WakeUp
1347 posts
WakeUp
#2 h  Mar 6, 2012, 9:13 PM • 1 Y
Y by Adventure10
Since $EA\perp AC$ clearly $EA$ is a tangent to the circle with diameter $AC$. Then $\angle PAE=\angle PQA=\angle QAC$. Clearly this makes $\triangle PAE$ and $\triangle BAR$ congruent since they're already both right-angled and $BA=AE$. Thus $PE=BR$, implying $DP=DR$.
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vanu1996
607 posts
vanu1996
#3 h  Dec 5, 2013, 3:26 AM • 2 Y
Y by Adventure10, Mango247
$\angle PAE=\angle PQA=\angle RAB$,so $\angle DAP=\angle DAR=45-\angle RAB$,also $\angle PDA=\angle RDA=45$,hence $DPA$ and $DRA$ are congruent,so $DP=DR$.
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jayme
9792 posts
jayme
#4 h  Dec 5, 2013, 9:56 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
see
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=468077
and also my proof on
http://perso.orange.fr/jl.ayme vol. 7 Miniatures geometriques p. 87-89.
Sincerely
Jean-Louis
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armpist
527 posts
armpist
#5 h  Dec 5, 2013, 1:34 PM • 2 Y
Y by Adventure10, Mango247
jayme wrote:
Dear Mathlinkers,
see
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=468077
and also my proof on
http://perso.orange.fr/jl.ayme vol. 7 Miniatures geometriques p. 87-89.
Sincerely
Jean-Louis

Dear J-L,

There is smth wrong with your reference: your paper has only 83 pages.

I wrote to you about it before, and hopefully this time you will make
needed corrections.


Friendly,

M.T.
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jayme
9792 posts
jayme
#6 h  Dec 5, 2013, 3:31 PM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
ops, I have not charge the new version... done
Sorry and sincerely
Jean-Louis
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AlastorMoody
2125 posts
AlastorMoody
#7 h  Mar 9, 2019, 1:40 AM • 4 Y
Y by lolmanc123, Adventure10, Mango247, Mango247
$$\angle CAP=x=\angle ACQ=\angle ARB=\angle APE \implies \Delta APE \cong \Delta ARB \Longrightarrow PE=BR \implies DP=DR$$
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WolfusA
1900 posts
WolfusA
#8 h  Jul 17, 2020, 1:01 PM
Y by
Consider the construction on cartesian plane. Let $$A=(0,0),\ B=(1,0),\ D=(1,1),\ E=(0,1),\ C=(2c,0)$$for some $c>1$. Then
$$P=\left(c-\sqrt{c^2-1},1\right),\ Q =\left(c+\sqrt{c^2-1},1\right),\ R =\left(1,\frac{1}{c+\sqrt{c^2-1}}\right)$$and finally
$$|DP|=1-c+\sqrt{c^2-1}= 1-\frac{1}{c+\sqrt{c^2-1}}=|DR|$$QED
This post has been edited 4 times. Last edited by WolfusA, Jul 26, 2020, 8:40 PM
Reason: post #9 from a troll deleted
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dchenmathcounts
2443 posts
dchenmathcounts
#10 h  Jul 25, 2020, 7:50 AM • 3 Y
Y by Mango247, Mango247, Mango247
Boring length bash.

Notice this is equivalent to proving $EP=BR.$ Let $EP=x,$ $BR=y,$ $AC=2r,$ and $AB=h.$

By Power of a Point $EA^2=h^2=EP\cdot EQ=x(x+2\sqrt{r^2-h^2})=(x+\sqrt{r^2-h^2})^2+h^2-r^2,$ implying $r^2=(x+\sqrt{r^2-h^2}^2)$ or $x=r-\sqrt{r^2-h^2}.$

Now note $\frac{AC}{QC}=\frac{r+\sqrt{r^2+h^2}}{h}=\frac{AB}{RB}=\frac{h}{y},$ implying $y=\frac{h^2}{r+\sqrt{r^2+h^2}}=r-\sqrt{r^2-h^2}.$ Thus $x=y.$
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