A Potpourri of Math Problems
by djmathman, Aug 30, 2013, 3:39 AM
from three different subjects. That's what I dealt with today.
Also combo isn't on here because combo is evil
Also combo isn't on here because combo is evil
MOSP 1999 wrote:
Prove that for any positive real numbers 
![\[\dfrac1{a+b}+\dfrac1{b+c}+\dfrac1{c+a}+\dfrac1{2\sqrt[3]{abc}}\geq\dfrac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}.\]](//latex.artofproblemsolving.com/3/9/f/39f12ad42239ffd89a5f390ba67b83304cdf77ec.png)
![\[\dfrac1{a+b}+\dfrac1{b+c}+\dfrac1{c+a}+\dfrac1{2\sqrt[3]{abc}}\geq\dfrac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}.\]](http://latex.artofproblemsolving.com/3/9/f/39f12ad42239ffd89a5f390ba67b83304cdf77ec.png)
![\[\dfrac{c^2}{c^2(a+b)}+\dfrac{a^2}{a^2(b+c)}+\dfrac{b^2}{b^2(c+a)}+\dfrac{(\sqrt[3]{abc})^2}{2abc},\]](http://latex.artofproblemsolving.com/e/d/e/edea28d55d50d4f187bf2cbd4cab2759452062f7.png)
![\[\dfrac{(a+b+c+\sqrt[3]{abc})^2}{a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc}=\dfrac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)},\]](http://latex.artofproblemsolving.com/9/c/6/9c66941c795b164a72de57df8fecaef6702763eb.png)
Panafrican 2013.1 wrote:
Let 
be a convex quadrilateral with 
parallel to 
. Let 
and 
be the midpoints of 
and 
, respectively. Prove that if 
, then 
.
be a convex quadrilateral with 
parallel to 
. Let 
and 
be the midpoints of 
and 
, respectively. Prove that if 
, then 
.
,
be the feet of the altitudes from 
to 
respectively. Note that by the angle equality we have that 
and 
are isogonal conjugates of each other, but since 
is a median of 
it follows that 
-
.
,
,
.
is the 
-
and 
as desired. Bulgaria 1998, PEN A17 wrote:
Let 
and 
be natural numbers such that
![\[A=\dfrac{(m+3)^n+1}{3m}\]](//latex.artofproblemsolving.com/c/c/6/cc6ea4853ef98b5c073d7e292dfbfa0773d19e8d.png)
is an integer. Prove that
is odd.
and 
be natural numbers such that![\[A=\dfrac{(m+3)^n+1}{3m}\]](http://latex.artofproblemsolving.com/c/c/6/cc6ea4853ef98b5c073d7e292dfbfa0773d19e8d.png)
is an integer. Prove that
is odd.
and 
are both odd. If 
gives 
is odd. Therefore, since 
we have by LTE![\[v_2(A)=v_2[(m+3)^n+1]-v_2(3m)=v_2(m+4)+v_2(n)-v_2(m)=v_2(m+4)-v_2(m).\]](http://latex.artofproblemsolving.com/9/d/4/9d4b032598b3a414be71e21651bde8fca96e6415.png)
then 
and 
,
and 
terms are divisible by 
.
is to be even, since 
we must have 
.
then by bounding we can easily see that 
,
,
.This post has been edited 11 times. Last edited by djmathman, Sep 5, 2013, 2:30 PM
does equal 
for 
odd since![\[x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1})\]](http://latex.artofproblemsolving.com/5/3/e/53e2cc3815cfc4a0999f377887d8b9c25ceaa5d6.png)
April 2025
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