If two triangles $ABC$ and $DEF$ are such that the perpendicular from $A,B,C$ to $EF,FD,DE$ are congruent, prove that the perpendiculars from $D,E,F$ to $BC,CA,AB$ are concurrent.  Source: Challenge and Thrills of pre-college mathematics.   Thanks in advance!
123785
Let $ABC$ and $PQR$ be any two triangles in the same plane. Assume that the perpendiculars from the points $A,B,C$ to the sides $QR,RP,PQ$, respectively, are concurrent. Prove that the perpendiculars from $P,Q,R$ to $BC,CA,AB$ respectively are also concurrent.   I was suggested to prove this using vectors. But I have no idea how to use them...
116465
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$ then proving $\triangle$ is equilateral.  $\bf{My\; Try::}$ We can write it as $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1$$ So $$\left(\frac{a+b-c}{a}\right)^{a}\cdot \left(\frac{b+c-a}{b}\right)^{b}\cdot \left(\frac{c+a-b}{c}\right)^{c}\geq 1$$ So $$(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$$ Now how can i solve after that, Help required, Thanks
10093
Self-studying “Basic Mathematics” by Serge Lang, got stuck doing final problem of chapter 5.    I took a sample triangle and labeled two lines and points.    As L1 is the perpendicular bisector of segment PQ, I know KP = KQ. Points K, Q and M form a right triangle.  That’s the information I gathered, do not know how to proceed from there. Any help would be appreciated.
34932
I need some help proving this, I've seen it proven in the other direction (prove the formula if it is an equilateral) but cant figure out how to prove it this way around.     Given three complex numbers $z_1, z_2, z_3$ prove that the points $z_1, z_2, z_3$ are vertices of an equilateral triangle in $\Bbb C$, if $$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3$$
96285
Can anyone show some hints to this? If $\gcd(a,b)=1$, then $a^{\phi(b)}+b^{\phi(a)}=1 \pmod{ab}$. I know that $a^{\phi(b)}=1 \pmod{b}$, and similarly, $b^{\phi(a)}=1 \pmod{a}$, but then how do I combine the work so that I get the result I need? Thanks!
58621
I've got a question concerning how to proof the Pythagoras theorem using the following assumption:  $x$ is perpendicular to $y$ (if and only if) $||x+y||^2 = ||x||^2 + ||y||^2$, where $x$ and $y$ are vectors.  I have a basic understanding of linear algebra, however I'm a beginner with this. The question provides hints how to prove the above mentioned equation.  Namely that I should use the properties of the dot product and the definition of the norm of a vector. Those being symmetry, scaling and distributivity as the dot product properties and the norm of a vector being the squared root of the dot product between the same vector.  I was thinking about using the fact that if a vector is perpendicular to another vector the dot product between those vectors should be 0. But that is not provided as a hint so I'm not sure. I know the under lying thought behind it is the cosine rule for vectors, that being:  $$x\cdot y = ||x||\,||y|| \cos(\theta)$$  If the angle between the two vectors is perpendicular you should use $\cos(\pi/2)$ which is $0$ and $||x||\cdot 0 = 0$ and $||y||\cdot 0 = 0$ with the vectors not necessarily being $0$. Thus $x\cdot y = 0$. How would I apply this to the equation I first mentioned to prove the Pythagoras theorem?  I have a few more thoughts on how I could prove this but I'm not sure if they're correct.  I hope someone could point me in the right direction.
10195
If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that: $$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$  Here is my try: $$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} &amp; \le \frac{1}{\sqrt{(a+1)^2 \cdot (b^2+1)}}\\ &amp; = \frac{1}{(a+1) \cdot \sqrt{b^2+1}}\\ &amp; \le \frac{1}{(a+1) \cdot b}\\ &amp; \le \frac{1}{ab} \end{align}$$  So, $$S \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{1}$$  Can you help me to continue it, please? Thanks!
113901
Prove the following: If $a \mid bc$, then $a  \mid \gcd(a, b)c$.   I tried to set $\gcd(a, b)$ to $b$ and used the fundamental theorem of arithmetic to prove that it is divisible by $a$, but I can't prove that $a \mid bc$, if and only if $a\mid b$ and $a\mid c$. Please help. Thanks.
63211
I am pretty sure that this is such a stupid, stupid question, but how do you prove that $d_2(x,y)={|x-y|\over {1+|x-y|}}$ satisfying the third condition to be a metric, which is the triangle inequality.   I know that I should really be able to do this, but I just can't, I keep seeing a contradiction. Please help me.
14425
I've figured out the following statement is true but I was wondering how you actually go about proving something like this?   $A \times (B \triangle C) = (A \times B) \triangle (A \times C)$
47380
If $x$ is positive integer, prove that for all integers $a$, $(a+1)(a+2)\cdots(a+x)$ is congruent to $0\!\!\!\mod x$.  Any hints? What are the useful concepts that may help me solve this problem?
76682
Using the definition of congruence mod n, prove that if a ≡ b (mod n) and c ≡ d (mod n), then a + c ≡ b + d (mod n). I've seen a couple of different examples of how to do this, mostly using s and t, but I am really struggling to understand the proof, and how to write it.
36455
I'm having some trouble figuring out how to give a direct proof using the definitions of mod m and congruence modulo m, without using any theorem involving mod or congruence.  if $a, b, c$ &amp; $m$ are integers such that $m \geq 2$, $c &gt; 0$ and $a \equiv b \pmod{m}$, then $ac \equiv bc\pmod{mc}$
74811
Inside an equilateral triangle $ABC$,an arbitrary point $P$ is taken from which the perpendiculars $PD,PE$ and $PF$ are dropped onto the sides $BC,CA$ and $AB$,respectively.Show that the ratio $\dfrac{PD+PE+PF}{BD+CE+AF}$ does not depend upon the choice of the point $P$ and find its value.    I have no idea how to proceed. I could not do anything beyond calculating the area of $PDB,PCE,PFA$ but here I get nothing. Please help!
68989
The question is as follows:      Let $A'$, $B'$ and $C'$ be the midpoint triangle of triangle $ABC$. In other words, $A'$, $B'$ and $C'$ are the midpoints of segments $BC$, $CA$ and $AB$, respectively. Show that triangles $A'B'C'$ and $ABC$ have the same centroid.    My reasoning was through the means of the Midline Theorem. After drawing the diagram, we can see, for example, $C'A'$ is a midline of $\triangle ABC$, which means that $C'A'$ is parallel to the base $AC$ and half the length of it as well. Therefore, the median $B'B$ would pass through the midpoint of $C'A'$, including the midpoint of $AC$.   Is my reasoning valid or, at least, is it on the right track? Any advice on solving this problem will be greatly appreciated.
78418
Have some  non-null $\vec{a}$ and $\vec{b}$.      I am trying to prove this to no avail:  $$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}|| \iff \vec{a} \perp \vec{b}$$    If we start with  $$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}|| \implies \vec{a} \perp \vec{b}$$  Our hypothesis is  $$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}||$$  Which tells us that both horizontal sides of this triangle have the same length, so we got an isosceles triangle. Not sure what to make out of that though.  Anyway, the hypothesis is equivalent to  $$\sqrt{(\vec{a} + \vec{b})\cdot (\vec{a} + \vec{b})} = \sqrt{(\vec{a} - \vec{b})\cdot (\vec{a} - \vec{b})}$$  I can't make much out of that.    A hint to begin tackling this problem would be appreciated.
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In the following figure, $ABCD$ is a parallelogram, and $O$ is any point. Parallelograms $OAEB, OBFC, OCGD, ODHA$ are completed. Prove that $EFGH$ is a parallelogram.   We can obtain a fairly trivial proof using affine geometry. As $OAEB, OBFC, OCGD, ODHA$ are parallelograms,   $$\vec{A} + \vec{B} = \vec{O} + \vec{E} \implies \vec{A}  + \vec{B} - \vec{O} = \vec{E} \tag1$$  $$\vec{C} + \vec{B} - \vec{O} = \vec{F}\tag2$$  $$\vec{D} + \vec{C} - \vec{O} = \vec{G} \tag3$$  $$\vec{D} + \vec{A} - \vec{O} = \vec{H}\tag4$$  Adding $(1)$ to $(3)$ and $(2)$ to $(4)$, we get,   $$\vec{E} + \vec{G} = \vec{A} + \vec{B} + \vec{C} + \vec{D} - 2\vec{O} \tag5$$  $$\vec{F} + \vec{H} =  \vec{A} + \vec{B} + \vec{C} + \vec{D} - 2\vec{O} \tag6$$  Clearly, $(5)$ and $(6)$ are equal, therefore,   $$\vec{E} + \vec{G} = \vec{F} + \vec{H}$$  Therefore, $EFGH$ is a parallelogram. Can somebody give an elementary proof using Euclidean geometry? Also, I noticed that in my proof, nowhere did I use the fact that $ABCD$ is a parallelogram, but constructing an example, it was quickly clear that the result stated does not generalize to all quadrilaterals. How come? Is my proof incorrect?