geometry
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area of a triangle
geometry theorems

The [b]Generalized Brahmagupta formula[/b] states:

If ABCD is an arbitrary convex quadrilateral with the sidelengths a = AB, b = BC, c = CD, d = DA, then the area F of the quadrilateral ABCD is

$\displaystyle F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) -abcd\cos ^{2}\frac{B+D}{2}}$,

where $s=\frac{1}{2}\left( a+b+c+d\right)$ is the semiperimeter of the quadrilateral ABCD, and $B=\measuredangle ABC$ and $D=\measuredangle CDA$ are the angles at the vertices B and D.

[i]Proof.[/i] (A buffalo proof, sorry, I haven't seen any other.)

The Cosine Law for triangle ABC yields $AC^{2}=a^{2}+b^{2}-2ab\cos B$, and the Cosine Law for triangle CDA yields $AC^{2}=c^{2}+d^{2}-2cd\cos D$. Thus, $a^{2}+b^{2}-2ab\cos B=c^{2}+d^{2}-2cd\cos D$ and hence $a^{2}+b^{2}-c^{2}-d^{2}=2ab\cos B-2cd\cos D$. Therefore,

$\left( a^{2}+b^{2}-c^{2}-d^{2}\right) ^{2}=\left( 2ab\cos B-2cd\cos D\right) ^{2}$
$=4a^{2}b^{2}\cos ^{2}B+4c^{2}d^{2}\cos ^{2}D-8abcd\cos B\cos D$.          [b](1)[/b]

On the other hand, the area F of the quadrilateral ABCD equals F = [ABC] + [CDA], where $\left[ P_{1}P_{2}P_{3}\right] $ is an abbreviation for the area of a triangle $P_{1}P_{2}P_{3}$. But $\left[ ABC\right] =\frac{1}{2}ab\sin B$ and $\left[ CDA\right] =\frac{1}{2}cd\sin D$, thus $F=\frac{1}{2}ab\sin B+\frac{1}{2}cd\sin D$, and hence

$16F^{2}=\left( 4F\right) ^{2}=\left( 4\cdot \left( \frac{1}{2}ab\sin B+\frac{1}{2}cd\sin D\right) \right) ^{2}=\left( 2ab\sin B+2cd\sin D\right) ^{2}$
$=4a^{2}b^{2}\sin ^{2}B+4c^{2}d^{2}\sin ^{2}D+8abcd\sin B\sin D$.

Adding this equation to [b](1)[/b] results in

$\left( a^{2}+b^{2}-c^{2}-d^{2}\right) ^{2}+16F^{2}$
$=\left( 4a^{2}b^{2}\cos ^{2}B+4c^{2}d^{2}\cos ^{2}D-8abcd\cos B\cos D\right) $
$+\left( 4a^{2}b^{2}\sin ^{2}B+4c^{2}d^{2}\sin ^{2}D+8abcd\sin B\sin D\right) $
$=4a^{2}b^{2}\left( \cos ^{2}B+\sin ^{2}B\right) +4c^{2}d^{2}\left( \cos ^{2}D+\sin ^{2}D\right) -8abcd\left( \cos B\cos D-\sin B\sin D\right) $
$=4a^{2}b^{2}+4c^{2}d^{2}-8abcd\cos \left( B+D\right) $
$=\left( 4a^{2}b^{2}+4c^{2}d^{2}+8abcd\right) -8abcd-8abcd\cos \left( B+D\right) $
$=\left( 2ab+2cd\right) ^{2}-8abcd\left( 1+\cos \left( B+D\right) \right) $
$\displaystyle =\left( 2ab+2cd\right) ^{2}-8abcd\cdot 2\cos ^{2}\frac{B+D}{2}=\left( 2ab+2cd\right) ^{2}-16abcd\cos ^{2}\frac{B+D}{2}$,

thus

$\displaystyle 16F^{2}=\left( \left( 2ab+2cd\right) ^{2}-16abcd\cos ^{2}\frac{B+D}{2}\right) -\left( a^{2}+b^{2}-c^{2}-d^{2}\right) ^{2}$
$\displaystyle =\left( \left( 2ab+2cd\right) ^{2}-\left( a^{2}+b^{2}-c^{2}-d^{2}\right) ^{2}\right) -16abcd\cos ^{2}\frac{B+D}{2}$.

But

$\left( 2ab+2cd\right) ^{2}-\left( a^{2}+b^{2}-c^{2}-d^{2}\right) ^{2}$
$=\left( \left( 2ab+2cd\right) +\left( a^{2}+b^{2}-c^{2}-d^{2}\right) \right) \left( \left( 2ab+2cd\right) -\left( a^{2}+b^{2}-c^{2}-d^{2}\right) \right) $
$=\left( 2ab+2cd+a^{2}+b^{2}-c^{2}-d^{2}\right) \left( 2ab+2cd-a^{2}-b^{2}+c^{2}+d^{2}\right) $
$=\left( \left( a^{2}+b^{2}+2ab\right) -\left( c^{2}+d^{2}-2cd\right) \right) \left( \left( c^{2}+d^{2}+2cd\right) -\left( a^{2}+b^{2}-2ab\right) \right) $
$=\left( \left( a+b\right) ^{2}-\left( c-d\right) ^{2}\right) \left( \left( c+d\right) ^{2}-\left( a-b\right) ^{2}\right) $
$=\left( \left( a+b\right) +\left( c-d\right) \right) \left( \left( a+b\right) -\left( c-d\right) \right) \left( \left( c+d\right) +\left( a-b\right) \right) \left( \left( c+d\right) -\left( a-b\right) \right) $
$=\left( a+b+c-d\right) \left( a+b-c+d\right) \left( a-b+c+d\right) \left( -a+b+c+d\right) $
$=\left( \left( a+b+c+d\right) -2d\right) \left( \left( a+b+c+d\right) -2c\right) \left( \left( a+b+c+d\right) -2b\right) \left( \left( a+b+c+d\right) -2a\right) $
$=\left( 2s-2d\right) \left( 2s-2c\right) \left( 2s-2b\right) \left( 2s-2a\right) =16\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) $,

and hence

$\displaystyle 16F^{2}=16\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) -16abcd\cos ^{2}\frac{B+D}{2}$

and

$\displaystyle F^{2}=\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) -abcd\cos ^{2}\frac{B+D}{2}$,

what completes the proof of the generalized Brahmagupta formula.

A special case is the (special, "non-generalized") [b]Brahmagupta formula[/b], which says that for a [i]cyclic[/i] quadrilateral ABCD, with the above notations, the area equals

$F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) }$.

In fact, if the quadrilateral ABCD is cyclic, we have B + D = 180, so that $\displaystyle \frac{B+D}{2}=90^{\circ}$ and $\displaystyle \cos \frac{B+D}{2}=0$, so that the generalized Brahmagupta formula

$\displaystyle F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) -abcd\cos ^{2}\frac{B+D}{2}}$

simplifies to

$\displaystyle F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right)}$.

You can also consider the Heron formula for a triangle as a degenerate case of the Brahmagupta formula for a cyclic quadrilateral. In fact, if we consider a triangle ABC as a cyclic quadrilateral ABCD with coinciding vertices D and A, then the side d = DA is zero, hence $s=\frac{1}{2}\left( a+b+c+0\right) =\frac{1}{2}\left( a+b+c\right) $, and we get the formula

  $F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-0\right) }=\sqrt{s\left( s-a\right) \left( s-b\right) \left( s-c\right) }$

for the area of our triangle ABC. This is the well-known [b]Heron formula[/b].

[b]PS.[/b] If you wonder why I wrote the proof above in such a detailed way - I already had it in my notes, typed down for a mathematical olympiad.

  Darij

Can you explain to me why $\displaystyle 1+\cos (B+D)= 2 \cos^2 \frac{B+D}{2}$

That's just the double angle formula: $\cos 2 \theta = 2 \cos^2 \theta - 1$ for all angles $\theta$.

Also note that since $abcd\cos^2\tfrac{B+D}{2}\ge0$, this formula proves that of all quadrilaterals with four given sides of length $a$, $b$, $c$, and $d$, the area is maximized when the quadrilateral is cyclic.

Isnt that just one Brahmagupta'stheorem? I mean there is another totally unrelated one is nt there?

Yes, there is. (The other one is pretty useless and trivial though.)

You can prove that using vectors and cross products though, it really isn't that bad.

Is there a general formula which gives the area of cyclic polygons?

[quote="dule_00"]Can anybody prove this without trigonometry?[/quote]
Of course no! The formula itself has already included a trigonometric relation!  :D

Sorry i meant for formula if $ABCD$ is cyclic.

Let ABCD be a cyclic quadrilateral with the sidelengths a = AB, b = BC, c = CD, d = DA.  Brahmaguptas formula
$ F=\sqrt{\left( s-a\right)\left( s-b\right)\left( s-c\right)\left( s-d\right)} $
is obvious if both pairs of opposite sides are parallel.
Let us assume that AB and CD intersect at point X and that x = XD, y = XB. We have
$ 4F=\sqrt{\left( y+b+x+c \right)\left( -y+b+x+c \right)\left( y-b+x+c \right)\left( y+b-x-c\right)} -\sqrt{\left( x+d+y-a \right)\left( -x+d+y-a \right)\left( x-d+y-a \right)\left( x+d+a-y\right)} $ .
Note that due to the similarity of XAD and XBC this expression of the form
$ 4F=\sqrt{ A B C D } -\sqrt{ \lambda A \lambda B \lambda C \lambda D }
= \sqrt{ \left( A-\lambda A \right) \left( B+ \lambda B \right) \left( C- \lambda C \right) \left( D+\lambda D \right) }$ .
Applying this formula, x and y vanish and Brahmagupta appears.
(cf. my communication in Forum Geom., 12, 191-192)


algebra
polynomial
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algebra solved

... some are even straightforward. It's my intention not to post the following one into "Number Theory".

A sequence of integers $a_1$, $a_2$, $a_3$, ... is given by the conditions $a_1=1$, $a_2=12$, $a_3=20$, and

$a_{n+3} = 2a_{n+2} + 2a_{n+1} - a_n$ for every $n\geq1$.

Prove that for every natural number n, the number $1+4a_na_{n+1}$ is a perfect square.

[b]PS.[/b] I was asked to post some Kvant problems here. Sorry if they were already posted. For those speaking Russian, the solution can be found here:

http://kvant.mccme.ru/1989/12/resheniya_zadachnika_kvanta_ma.htm

  Darij

If we put $(b_1,b_2,b_3)=(7,31,71)$ and $b_{n+3}=2b_{n+2}+2b_{n+1}-b_n$ (the same recurrence relation as $\{a_n\}$), then we can show that $1+4a_na_{n+1}=b_n^2$.

This is easy because we can write $a_n,b_n$ in terms of the roots of $x^3-2x^2-2x+1$: $a_n=\alpha(-1)^n+\beta u^n+\gamma \bar u^n$ ($\bar u$ isn't the usual complex conjugate, but rather the conjugate over $\mathbb Q$), and $b_n=x(-1)^n+yu^n+z\bar u^n$. We only need to check the identity for $5$ values of $n$ (at most), because it's an identity involving some constant coefficients multiplied by $1^n,(u^2)^n,(\bar u^2)^n,(-u)^n,(-\bar u)^n$, and these regarded as sequences indexed by $n$ satisfy some linear recurrence determined by the polynomial which has all of them as roots.

The numbers involved aren't that large :).

Looks very nice to me. Note that your last step, namely checking the identity for 5 values of n, can be simplified a bit by doing the recursion backwards and defining the sequences $a_n$ and $b_n$ for n < 1. In fact, this gives $a_0=6$, $a_{-1}=2$, $b_0=5$ and $b_{-1}=-7$; these numbers are much more handy for checking the identity than $a_4=63$, $a_5=154$, $b_4=197$ and $b_5=505$.

Let me record the Binet expressions:

$a_{n}=3\cdot \left( -1\right) ^{n}+\frac{15-\sqrt{5}}{10}\left( \frac{3+\sqrt{5}}{2}\right) ^{n}+\frac{15+\sqrt{5}}{10}\left( \frac{3-\sqrt{5}}{2}\right) ^{n}$;

$b_{n}=3\cdot \left( -1\right) ^{n}+\frac{5+7\sqrt{5}}{5}\left( \frac{3+\sqrt{5}}{2}\right) ^{n}+\frac{5-7\sqrt{5}}{5}\left( \frac{3-\sqrt{5}}{2}\right) ^{n}$.

The proposed solution uses a slightly different method: It defines the sequence $b_n$ by the equation $b_n = a_{n+2} - a_{n+1} - a_{n}$, and then shows that $1 + 4 a_n a_{n+1} = b_n^2$ by induction after n: In fact, for n = 1, we can directly verify this, and now, assuming that for some n, we have $1 + 4 a_{n-1} a_n = b_{n-1}^2$, we can prove $1 + 4 a_n a_{n+1} = b_n^2$ as follows: Since

$b_{n}=\underbrace{a_{n+2}}_{=2a_{n+1}+2a_{n}-a_{n-1}}-a_{n+1}-a_{n}=\left( 2a_{n+1}+2a_{n}-a_{n-1}\right) -a_{n+1}-a_{n}$
$=a_{n+1}+a_{n}-a_{n-1}=\underbrace{\left( a_{n+1}-a_{n}-a_{n-1}\right)}_{=b_{n-1}} +2a_{n}=b_{n-1}+2a_{n}$,

we have $b_{n-1}=b_{n}-2a_{n}$ and thus

$1+4a_{n-1}a_{n}=\underbrace{b_{n-1}}_{=b_{n}-2a_{n}}\ ^{2}=\left( b_{n}-2a_{n}\right) ^{2}=b_{n}^{2}-4a_{n}\underbrace{b_{n}}_{=a_{n+1}+a_{n}-a_{n-1}}+4a_{n}^{2}$
$=b_{n}^{2}-4a_{n}\left( a_{n+1}+a_{n}-a_{n-1}\right) +4a_{n}^{2}$
$=b_{n}^{2}-4a_{n}a_{n+1}-4a_{n}^{2}+4a_{n}a_{n-1}+4a_{n}^{2}=b_{n}^{2}-4a_{n}a_{n+1}+4a_{n}a_{n-1}$.

Subtracting $4a_n a_{n-1}$, we get

$1=b_{n}^{2}-4a_{n}a_{n+1}$,

and thus $1 + 4 a_n a_{n+1} = b_n^2$, as intended. This completes the proof.

By the way, the formula $b_n = a_{n+2} - a_{n+1} - a_{n}$ needs not be guessed. It can be found by a systematic reasoning. Indeed, as soon as one sees that the sequence ($b_n$) fulfills the same recurrent equation as the sequence ($a_n$) - namely, the equation $b_{n+3} = 2b_{n+2}+2b_{n+1}-b_n$ -, it becomes clear that the sequences ($a_n$) and ($b_n$) belong to the same vector space of sequences fulfilling this recurrent equation; and now, instead of expressing the sequence ($b_n$) as a linear combination of geometric sequences, one expresses it as a linear combination of the sequences ($a_n$), ($a_{n+1}$), ($a_{n+2}$) and gets the formula above.

  Darij


geometry
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Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Weird solution:

We will have $\angle ABQ = \angle QBC = x$

Let R lie on AC s.t. RAB is equilateral.  Let T lie on AB extended past B such that BT = BP, and let U lie on AC such that UAT is equilateral.  Since AU = AT = AB + BP = AQ + QB, we have QB = QU.  As a result, we calculate $\angle QBU = 90 - \dfrac{60 + \angle ABQ}{2} = 60 - \dfrac{x}{2}$.  Meanwhile, $\angle QBR = 60 - x$, so we have $\angle RBU = \angle QBU - \angle QBR = \dfrac{x}{2}$.

Then suppose the bisector of $\angle BTP$ intersects BR at X.  Then since $\angle BRT = \angle RBU = \dfrac{x}{2} = \angle BTX$, we have similar triangles, and by equal ratios $BT^2 = BX \cdot BR$.  Equivalently, $BP^2 = BX \cdot BR$, so triangles BXP and BPR are similar; in particular, BX = XP.

Since TX is the bisector of $\angle BTP$, we have then that T, B, X, and P are concyclic.  Then $\dfrac{x}{2} = \angle XTP = \angle XBP = \angle ABC - \angle ABR = 2x - 60$.  Solving for x, x = 40.  Then $\angle ABC = 80 \implies \angle ACB = 40$.

[hide="Alternate Solution"]
Consider points $ X$ and $ Y$ on $ AB$ and $ AC$ respectively such that $ BY\equal{}BP$ and $ BQ\equal{}QX$.  Also, let $ AP$ meet $ YX$ at $ Z$ and $ BX$ at $ W$.  Assume that $ C\ne X$.  Notice that $ AY\equal{}AB\plus{}BY\equal{}AB\plus{}BP\equal{}AQ\plus{}BQ\equal{}AQ\plus{}QX\equal{}AX$ and $ \angle XAY\equal{}60$, so $ \triangle AXY$ is equilateral.  Now, let $ \angle ABQ\equal{}\alpha\equal{}\angle QBC$.  Thus, $ \angle AQB\equal{}180\minus{}\angle BAC\minus{}\angle ABQ\equal{}120\minus{}\alpha$.  Also, $ \angle QBX\equal{}\angle QXB\equal{}\frac{\angle AQB}{2}\equal{}60\minus{}\frac{\alpha}{2}$.  Hence, $ \angle BXY\equal{}\angle AXY\minus{}\angle AXB\equal{}60\minus{}(60\minus{}\frac{\alpha}{2})\equal{}\frac{\alpha}{2}$.  Now, since $ BP\equal{}BY$, we have that $ \angle BPY\equal{}\angle BYP\equal{}\frac{\angle ABC}{2}\equal{}\alpha$.  Thus, $ \angle PYZ\equal{}\angle AYX\minus{}\angle AYP\equal{}60\minus{}\alpha$.  Since $ AZ$ is a bisector of equilateral $ \triangle AXY$, we see that $ AZ$ is the perpindicular bisector of $ YX$, which implies that $ \angle PXY\equal{}\angle PYZ\equal{}60\minus{}\alpha\implies \angle PXW\equal{}\angle PXY\minus{}\angle WXY\equal{}60\minus{}\frac{3\alpha}{2}$.  Notice that $ \angle CBX\equal{}\angle QBX\minus{}\angle QBC\equal{}60\minus{}\frac{3\alpha}{2}$.  Therefore, $ \angle CBW\equal{}\angle PXW\equal{}\angle PYW$ because of the fact that $ PW$ is the perpindicular bisector of $ YX$.  This means that $ PWYB$ is cyclic.  Therefore, $ \angle XWZ\equal{}\angle ZWY\equal{}\angle PBY\equal{}180\minus{}\angle ABC\equal{}180\minus{}2\alpha$.  Since $ \angle ZWX\equal{}90\minus{}\angle WXZ\equal{}90\minus{}\frac{\alpha}{2}$, we see that $ 90\minus{}\frac{\alpha}{2}\equal{}180\minus{}2\alpha$, so $ 90\equal{}\frac{3\alpha}{2}\implies \alpha\equal{}60$.  This means that $ \angle ABC\equal{}2\alpha\equal{}120$, so $ \angle ACB\equal{}180\minus{}\angle BAC\minus{}\angle ABC\equal{}180\minus{}60\minus{}120\equal{}0$, which is clearly impossible.  This means that we have a contradiction, so $ C\equal{}X$.  Hence, $ QC\equal{}QB$, so $ \alpha\equal{}180\minus{}\angle BAC\minus{}\angle ABC\equal{}120\minus{}2\alpha\implies \alpha\equal{}40$.  Therefore, $ \angle ABC\equal{}2\alpha\equal{}80$ and $ \angle ACB\equal{}\alpha\equal{}40$. [/hide]

[color=darkred]Let $ ABC$ be a triangle with $ A = 60^{\circ}$ . Let $ AP$ bisect $ \widehat { BAC}$ and let $ BQ$ bisect  $ \widehat {ABC}$ , where

$ P\in BC$ and $ Q\in AC$ . If $ AB + BP = AQ + QB$ , then what are the angles of the triangle ?[/color]

[color=darkblue][b][u]Proof I (metric).[/u][/b] Denote $ l_b = BQ$ . Thus, $ l_b = \frac {2ac}{a + c}\cdot\cos\frac B2$ . Therefore, $ AB + BP = AQ + QB$ $ \Longleftrightarrow$

$ c + \frac {ac}{b + c} = \frac {bc}{a + c} + l_b$ $ \Longleftrightarrow$ $ c(c + a)(c + b) + ac(a + c) = bc(b + c) + 2ac(b + c)\cdot\cos\frac B2$ $ \Longleftrightarrow$

$ ab + (a + c)^2 - b^2 = 2a(b + c)\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ ab + 4p(p - b) = 2a(b + c)\cdot\cos\frac B2$ $ \Longleftrightarrow$

$ ab + 4ac\cdot\cos^2\frac B2 = 2a(b + c)\cdot \cos\frac B2$ $ \Longleftrightarrow$ $ b\left(1 - 2\cdot\cos\frac B2\right) = 2c\cdot\cos\frac B2\cdot\left(1 - 2\cdot\cos\frac B2\right)$ $ \Longleftrightarrow$

$ b = 2c\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ \sin B = 2\cdot\sin C\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ 2\sin\frac B2\cos\frac B2 = 2\sin C\cos\frac B2$ $ \Longleftrightarrow$

$ \sin\frac B2 = \sin C$ $ \Longleftrightarrow$ $ B = 2C$ $ \Longleftrightarrow$ $ C = 40^{\circ}$ , $ B = 80^{\circ}$ because $ A = 60^{\circ}$ .

[b][u]Proof II (trigonometric)[/u][/b] Denote ${ x=m(\widehat ABQ})$ . I"ll apply the [b]Sinus' theorem[/b] in the triangles $ ABP$ , $ ABQ$ :

$ AB+BP=AQ+QB$ $ \Longleftrightarrow$ $ 1+\frac {BP}{BA}=\frac {AQ+QB}{AB}$ $ \Longleftrightarrow$  $ 1+\frac {\sin 30}{\sin (30+2x)}=\frac {\sin x+\sin 60}{\sin (60+x)}$ $ \Longleftrightarrow$

$ \sin (60+x)\left[\sin (30+2x)+\frac 12\right]=\sin (30+2x)(\sin x+\sin 60)$ $ \Longleftrightarrow$

$ \cos (x-30)-\cos (90+3x)+\cos (30-x)=$ $ \cos (x+30)-\cos (30+3x)+\cos (2x-30)-\cos (90+2x)$ $ \Longleftrightarrow$

$ \underline {\cos (x-30)}+\underline {\underline {\sin 3x}}+\underline {\underline {\underline {\cos (x-30)}}}=$ $ \underline {\underline {\underline {\cos (x+30)}}}-\underline {\underline {\cos (30+3x)}}+\cos (2x-30)+\underline {\sin 2x}$ $ \Longleftrightarrow$

$ [\cos (x-30)-\cos (90-2x)]+[\sin 3x+\sin (60-3x)]=$ $ [\cos (x+30)-\cos (x-30)]+\cos (2x-30)$ $ \Longleftrightarrow$

$ 2\sin \left(30-\frac x2\right)\sin\left(60-\frac {3x}{2}\right)+2\sin 30\cos (3x-30)=\cos (2x-30)-2\sin 30\sin x$ $ \Longleftrightarrow$

$ \cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)=\cos (2x-30)-\cos (90-x)$ $ \Longleftrightarrow$

$ \cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)=2\sin \left(30+\frac x2\right)\sin\left(60-\frac {3x}{2}\right)$ $ \Longleftrightarrow$

$ \sin (120-3x)=2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $ \Longleftrightarrow$

$ 2\sin\left(60-\frac {3x}{2}\right)\cos\left(60-\frac {3x}{2}\right)=2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $ \Longleftrightarrow$

$ \sin\left(60-\frac {3x}{2}\right)=0\ \ \vee\ \ \cos\left(60-\frac {3x}{2}\right)=\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)$ . Thus, $ \boxed {\ x=40\ }$ or

$ \sin\left(30+\frac {3x}{2}\right)+\sin\left(30-\frac x2\right)=\sin\left(30+\frac x2\right)$ $ \Longleftrightarrow$ $ 2\sin\left(30+\frac x2\right)\cos x=\sin\left(30+\frac x2\right)$ $ \Longleftrightarrow$ $ x\in\emptyset$ .[/color]

Interesting problem

[hide="Solution"]

 [geogebra]3cd2588b3a8ec28b3adecf64c88d939dc79c0a23[/geogebra]

Define B' on ray AB such that BP = BB'.
Define C' on ray AC such that QB = QC'.
Let $ \alpha \equal{} \angle ABQ \equal{} \angle QBC$.

Then AB + BP = AQ + QB implies AB + BB' = AQ + QC', so that AB' = AC' with angle B'AC' = 60 degrees. Hence, $ \boxed{\triangle B'AC'\text{ is equilateral}}$.

Suppose that B, P, C' were not collinear.

By [hide="simple angle chasing,"]Since $ \angle ABC \equal{} 2 \alpha$, $ \boxed{\angle BB'P \equal{} \angle BPB' \equal{} \alpha}$. So this means that $ BQ \parallel{} B'P$. Since AB'C' is equilateral, $ \boxed{\angle PB'C' \equal{} 60^{\circ} \minus{} \alpha}$. Since $ PB'C'$ is isosceles (P is on the angle/perpendicular bisector of B'AC'), we have $ \boxed{\angle B'PC' \equal{} 60^{\circ} \plus{} 2\alpha}$.  Now

$ \angle BPC' \equal{} \angle BPB' \plus{} \angle B'PC' \equal{} \alpha \plus{} 60^{\circ} \plus{} 2\alpha \equal{} 60^{\circ} \plus{} 3\alpha$

$ \boxed{\boxed{\angle BPC' \equal{} 60^{\circ} \plus{} 3\alpha}}$

By the exterior angle theorem, $ \angle BQC' \equal{} \angle ABQ \plus{} \angle BAQ \equal{} \alpha \plus{} 60^{\circ}$, and since $ \triangle BQC'$ is isosceles, $ \boxed{\angle AC'B' \equal{} 60^{\circ} \minus{} \alpha/2}$. Also, since $ \angle C'B'B \equal{} 60^{\circ}$,

$ \angle PBC' \equal{}$
$ \angle PBB' \minus{} \angle C'BB' \equal{}$
$ \angle PBB' \minus{} (180^{\circ} \minus{} \angle CB'B \minus{} \angle BC'B') \equal{}$
$ \angle PBB' \minus{} (180^{\circ} \minus{} \angle CB'B \minus{} (\angle AC'B' \minus{} \angle AC'B')) \equal{}$
$ 180^{\circ} \minus{} 2\alpha \minus{} (180^{\circ} \minus{} 60^{\circ} \minus{} (60^{\circ} \minus{} (60^{\circ} \minus{} \alpha/2))) \equal{}$
$ 60^{\circ} \minus{} 3\alpha/2$

$ \boxed{\boxed{\angle PBC' \equal{} 60^{\circ} \minus{} 3\alpha/2}}$

The two double-boxed statements clearly imply that

$ \boxed{\boxed{\triangle BPC'\text{ is isosceles with }BP \equal{} PC'}}$[/hide] we can show that

$ \boxed{\triangle BPC'\text{ is isosceles with }BP \equal{} PC'}$.

Since QBC' is isosceles, the above boxed statement translates to
$ \boxed{QP\text{ bisects }\angle BQC}$

Now it is an [hide="easy exercise"]By the angle bisector theorem, $ BQ/QC \equal{} BP/PC$ and $ BA/AC \equal{} BP/PC$. So $ BQ/QC \equal{} BA/AC$, and combined with the law of sines,

$ \frac{\sin\angle QBC}{\sin\angle QCB} \equal{} \frac{\sin\angle ABC}{\sin\angle QCB} \implies \sin \angle QBC \equal{} \sin \angle ABC \implies \sin 2\alpha \equal{} \sin \alpha$

Therefore, $ 2\alpha \plus{} \alpha \equal{} 180^{\circ}$, so $ \angle ABC \equal{} 120^{\circ}$ and $ \angle ABC$ is obtuse. [In this special case where $ \angle BAC \equal{} 60^{\circ}$, the three vertices are collinear and $ ABC$ is degenerate].[/hide] to show that the above boxed statement, combined with the given, forces $ ABC$ to be obtuse (actually, degenerate). This is a contradiction.

So our initial assumption was false, and so B, P, C' are collinear. Now we can find the angles of ABC by noting that C = C' (both points are on lines BP and AQ - two lines intersect at one point (in Euclidean geometry :P )). This means that QB = QC, so that $ \alpha \equal{} \angle QCB$. Therefore, the three angles of ABC are $ 2\alpha$, $ \alpha$, and $ 60^{\circ}$. Solving for $ \alpha$ results in an $ \boxed{80\minus{}60\minus{}40\text{ triangle}}$.[/hide]

[quote="vishalarul"]Now

$ \angle BPC' = \angle BPB' + \angle B'PC' = \alpha + 60^{\circ} + 2\alpha = 60^{\circ} + 3\alpha$[/quote]The problem here is that $C'$ could be in between $Q$ and $C$, in which case we would have $360^\circ-\angle{BPC'}=\angle{BPB'}+\angle{B'PC'}$ instead.

Of course, this can be fixed by directed angles (counterclockwise being positive). Basic angle chasing using the facts that $\triangle{AB'C'}$ is equilateral, $BP=BB'$, $QB=QC'$, and $PB'=PC'$ tells us that $\angle{PBC'}=\angle{BC'P}=60^\circ-(3\angle{ABQ}/2)$, so $PB=PC'$. But $PB'=PC'$, so $PB'=PC'=PB=BB'$. This implies that $\triangle{PBB'}$ is equilateral, so $\angle{ABQ}=\angle{BB'P}=60^\circ$. But this forces the sum of the angles in $\triangle{ABC}$ to be more than $180^\circ$, which is impossible. So we must have $\angle{APQ}=40^\circ$.

[b]Edit:[/b] I also have a trigonometric solution that is somewhat shorter than [b]Virgil Nicula[/b]'s. Letting $2x=\angle{ABQ}$, we find by the Law of Sines that (for convenience, take all angles to be in degrees)
\[AB+\frac{AB\sin30}{\sin(150-4x)}=AB+BP=AQ+QB=\frac{AB\sin2x}{\sin(120-2x)}+\frac{AB\sin60}{\sin(120-2x)}.\]Dividing by $AB$ and simplifying, this is equivalent to
\[1+\frac{1}{2\sin(4x+30)}=\frac{\sin2x+\sin60}{\sin(2x+60)}=\frac{2\sin(x+30)\cos(x-30)}{2\sin(x+30)\cos(x+30)}=\frac{\cos(x-30)}{\cos(x+30)}.\]Clearing denominators, this is equivalent to
\[2\sin(4x+30)\cos(x+30)+\cos(x+30)=2\sin(4x+30)\cos(x-30),\]or
\begin{align*}
\cos(x+30) &= 2\sin(4x+30)[\cos(x-30)-\cos(x+30)]\\
&= 2\sin(4x+30)(2\sin{x}\sin30)\\
&= 2\sin(4x+30)\sin{x}\\
&= \cos(3x+30)-\cos(5x+30).
\end{align*}Thus
\[\cos(3x+30)=\cos(5x+30)+\cos(x+30)=2\cos(3x+30)\cos(2x).\]Note that $0<x<\angle{ABC}/4<45$. If $\cos(2x)=1/2$, then we must have $\angle{APQ}=2x=60$, making the sum of angles in $\triangle{ABC}$ greater than $180^\circ$. Thus $\cos(3x+30)=1\implies x=20$, so $\angle{ABC}=80$ and $\angle{ACB}=40$.

[quote="sayantanchakraborty"]Interestingly,both the geometry problems of IMO 2001 can be done by just a trigimetrical bash(though tiring and requires intelligence).Credit goes to those who will give a pure geometric approach.[/quote]
Actually I found it easier by pure geometry
Solved it in less than half an hour :wink:
The correct diagram is the key :)

The trigonometric solutions would have taken me ages :P

[hide="Solution"]
Let $\angle ABQ=\theta$. Note that $0< \theta < 60$. We will show that $\theta=40^\circ$ (Much of the calculation below is motivated by this answer).

Then, letting $AB=1$ gives that we must prove $1+\dfrac{\sin 30}{\sin(30+2\theta)}=\dfrac{\sin \theta}{\sin(60+\theta)}+\dfrac{\sin 60}{\sin(60+\theta)}$, or $\sin(60+\theta)(\sin 30 + \sin(30+2\theta))=\sin(30+2\theta)(\sin \theta +\sin 60)$. Now, we use the product to sum formula $\sin A \sin B = \dfrac{\cos(A-B)-\cos(A+B)}{2}$ to get, after some sign changes and grouping:

\[ [\cos(90-\theta)-\cos(30-2\theta)]+[\cos(-30)-\cos(150-3\theta)]+[\cos(30-\theta)-\cos(90-2\theta)]+[\cos(90-3\theta)-\cos 30]=0. \]

Now we use the sum to product formula $\cos A-\cos B=2\sin \left(\dfrac{A+B}{2}\right)\sin \left(\dfrac{A-B}{2}\right)$ to then get, after factoring:

\[ \sin\left(60-\dfrac{3\theta}{2}\right)\left[\sin\left(30+\dfrac{\theta}{2}\right)+\sin\left(-90+\dfrac{3\theta}{2}\right)+\sin\left(-30+\dfrac{\theta}{2}\right)+\sin\left(30-\dfrac{3\theta}{2}\right)\right]=0.\]

Clearly $\theta=40^\circ$ satisfies this, and if $\theta \neq 40^\circ$, then we must have $\left[\sin\left(30+\dfrac{\theta}{2}\right)+\sin\left(30-\dfrac{3\theta}{2}\right)\right]+\left[\sin\left(-90+\dfrac{3\theta}{2}\right)+\sin\left(-30+\dfrac{\theta}{2}\right)\right]=0$. By using the sum to product formula $\sin A+\sin B=2\sin \left(\dfrac{A+B}{2}\right)\cos \left(\dfrac{A-B}{2}\right)$, we then get

\[\sin\left(30-\dfrac{\theta}{2}\right)\cos \theta= \sin(60-\theta)\cos \left(30-\dfrac{\theta}{2}\right).\]

Using the $\sin$ and $\cos$ double angle-formulas allows us to simplify (Noting that $\theta<60^\circ$) to get $\cos \theta=1+\cos(60-\theta)$, but this never holds for $0<\theta<60$. Hence, $\theta=40^\circ$.
[/hide]

It  seems  to me that it can be solved by construction.

similar to #4

[hide=Solution]
Let $X,Y$ be points on $AB,AC$ respectively such that $BP=PX$ and $BQ=QY.$ Then we have
$$AX=AB+BX=AB+BP=AQ+QB=AQ+QY=AY,$$
so $AXY$ is equilateral.

[b]Claim.[/b] $B,P,Y$ are collinear.

[i]Proof.[/i] Suppose that $B,P,Y$ are not collinear. Note that $BPX,BQY$ are isoceles. Since $AP$ is an angle bisector of equilateral $AXY,$ we must have $PXY$ isoceles as well. Angle chasing, we have
\begin{align*}
\angle PYQ&=60^{\circ}-\angle XYP\\
&=60^{\circ}-\angle PXY\\
&=\angle PXB\\
&=\angle QBP,\\
\end{align*}
then $XB=PB=PY=PX,$ which means $BXP$ is equilateral, or $\angle B=120^{\circ},$ a contradiction.

Then we have
\begin{align*}
180^{\circ}&=\angle BQY+\angle QYB+\angle YBQ\\
&=60^{\circ}+\frac32\angle B,\\
\end{align*}
and $\boxed{\angle B=80^{\circ},\angle C=40^{\circ}},$ so we are done.
[/hide]

Wow did I just do a G8 by a pure trig bash?

Honestly, after applying the sine rule on both triangles and getting $\frac{\cos({m}+30)}{\cos({m}-30)}=\frac{\sin({30+4 m}) +\sin{30}}{\sin({30+4m})}$, it's almost mindless bashing... here is basically a sketch of my solution-

First, letting $\angle{ABC}= \alpha$ and $m= \frac{\alpha}{4}$, just mark all angles and apply sine rule on triangles $APB$ and $AQB$ to get the aforementioned formula $\frac{\cos({m}+30)}{\cos({m}-30)}=\frac{\sin({30+4 m}) +\sin{30}}{\sin({30+4m})}$.

Cross- multiply and simplify a bit to get a rather important step- $2 (\sin({60+3m})+ \sin{m}) \cos({2m}) = \sin{5m}+\sin({60+3m})$- this step is important because, if you weren't able to get a hint of what the angles may be through your diagram, the $60+3m$ is very telling of what the solutions could be...

Juggling now with the algebra a bit, the next important step we get to is $\sin({60+3m}) \cdot (2 \cos({2m})-1) = \sin({5m})+\sin({m})-\sin({3m})=2 \sin({3m}) \cos({2m})-\sin({3m})=\sin({3m}) \cdot (2 \cos({2m})-1)$

Aha! Noticing that as $m< 30=>cos({2m}) \neq \frac{1}{2}$, we get $\sin({3m})=\sin({60+3m})=>6m=120=>m=20 => \alpha=80$ ! Hence our angles are $60,80,40$, and we're done.

[hide=Note] I know it was a little ugly bash, but isn't this a little too easy for a G8, as there wasn't any real "Geometry" involved, at least in my solution... [/hide]

[b]Made with Blood, Sweat, and Tears[/b]
Let $\angle ABQ = \angle BQC = x$. If $BP = \sin 30$, then by the Law of Sines, we know that $$\frac{BP}{\sin (30)} = \frac{AB}{\sin(150-2x)} \Longleftrightarrow AB = \sin (150 -2x)$$
Furthermore, by the same reasoning as before,
$$\frac{AB}{\sin (120-x)} = \frac{BQ}{\sin (60)} \Longleftrightarrow BQ = \frac{\sin (150 -2x) \sin (60)}{\sin (120-x)}$$
$$\frac{AB}{\sin (120-x)} = \frac{AQ}{\sin (x)} \Longleftrightarrow AB= \frac{\sin (150 -2x) \sin (x)}{\sin (120-x)}$$
It follows that since we want $BP + AB = AQ + QB$, we need:
$$\frac{\sin (150-2x)(\sin(x) + \sin (60))}{\sin (120-x)} = \sin (30) + \sin(150 -2x)$$
We know that $\sin (x) + \sin (60) = 2\sin \left(\frac{x + 60}{2} \right)\cos \left(\frac{x - 60}{2} \right)$ and $$\sin (120-x) = \sin (60 + x) = 2\sin \left(\frac{60 + x}{2} \right) \cos \left( \frac{60 + x}{2} \right)$$
Therefore, we can reduce the problem to:
$$\frac{\sin (150-2x)\left( 2\sin \left(\frac{x + 60}{2} \right)\cos \left(\frac{x - 60}{2} \right)\right)}{2\sin \left(\frac{60 + x}{2} \right) \cos \left( \frac{60 + x}{2} \right)} = \sin (30) + \sin(150 -2x) \Longleftrightarrow$$
$$\frac{\sin (150 - 2x) \cos \left( \frac{x-60}{2} \right)}{\cos \left( \frac{x+60}{2} \right)} = \sin (30) + \sin(150 -2x) \Longleftrightarrow$$
$$\frac{\cos \left( \frac{x-60}{2} \right)}{\cos \left( \frac{x+60}{2} \right)} = \frac{\sin (30)}{\sin(150-2x)} + 1$$
This motivates us to consider:
$$\frac{\cos \left( \frac{x-60}{2} \right)}{\cos \left( \frac{x+60}{2} \right)} - 1 = \frac{\cos \left( \frac{x-60}{2}\right) - \cos \left(\frac{x+60}{2} \right)}{\cos \left(\frac{x+60}{2} \right)} = \frac{-2 \sin \left(\frac{x}{2} \right) \sin (-30)}{\cos \left(\frac{x+60}{2} \right)}$$
Putting this into our equation, we see that:
$$\frac{-2 \sin \left(\frac{x}{2} \right) \sin (-30)}{\cos \left(\frac{x+60}{2} \right)} = \frac{\sin (30)}{\sin(150-2x)} \Longleftrightarrow$$
$$\frac{2 \sin \left(\frac{x}{2} \right)}{\cos \left(\frac{x+60}{2} \right)} = \frac{1}{\sin (150-2x)} \Longleftrightarrow$$
$$2\sin \left(\frac{x}{2} \right) \sin (150 -2x)  = \cos \left(\frac{x}{2} + 30 \right)$$
Multiplying both sides by $2\cos \left(\frac{x}{2} \right)$, we get:
$$2\sin (x) \sin (150 -2x) = 2\cos \left(\frac{x}{2} + 30 \right) \cos \left(\frac{x}{2} \right) \Longleftrightarrow $$
$$- \cos \left(150 - x \right) - \cos (150 -3x) = \cos(x+30) + \cos(30) \Longleftrightarrow$$
$$\cos(150 - 3x) = \cos (30) \Longleftrightarrow x = 40$$
This means that the only such triangle has angles $60,40,80$.

Let $\angle ABQ=x$, so $\angle ABC = 2x$. By the angle bisector theorem and the Law of Sines,
\[AB+\frac{AB}{AB+AC}\cdot BC = AB+BP=AQ+BQ = AB \cdot \frac{\sin \angle ABQ+\sin \angle BAQ}{\sin \angle AQB}.\]
Divide through by $AB$ and apply the Law of Sines again to get
\[\frac{\sin 30^\circ+\sin (2x+30^\circ)}{\sin (2x+30^\circ)} = 1+\frac{\sin 60^\circ}{2\sin (2x+30^\circ)\cos (30^\circ)} = \]
\[1+\frac{\sin 60^\circ}{\sin (2x+60^\circ)+\sin 2x} = \frac{\sin (60^\circ+2x)+\sin 2x+\sin 60^\circ}{\sin (60^\circ+2x)+\sin 2x}=\frac{\sin \angle ACB+\sin \angle ABC+\sin \angle BAC}{\sin \angle ACB+\sin \angle ABC}=\]
\[\frac{AB+AC+BC}{AB+AC}=\frac{\sin \angle ABQ+\sin \angle BAQ}{\sin \angle AQB} = \frac{\sin x+\sin 60^\circ}{\sin (60^\circ+x)} = \frac{2\sin (x/2+30^\circ)\cos (x/2-30^\circ)}{2\sin (x/2+30^\circ)\cos (x/2+30^\circ)}=\]
\[\frac{\cos (x/2-30^\circ)}{\cos (x/2+30^\circ)}.\]
Cross-multiply to obtain
\[\sin (30^\circ)\cos (x/2+30^\circ)+\sin (2x+30^\circ)\cos (x/2+30^\circ) = \sin (2x+30^\circ)\cos (x/2-30^\circ).\]
Rewrite this as
\[\sin (30^\circ)\cos (x/2+30^\circ)=\sin (2x+30^\circ)(\cos (x/2-30^\circ) + \cos (150^\circ - x/2)) = \]
\[2\sin (2x+30^\circ)\cos (60^\circ)\cos (x/2-90^\circ) = 2\sin (2x+30^\circ)\cos (60^\circ)\sin (x/2).\]
Divide through by $\sin (30^\circ)=\cos (60^\circ)$ to get
\[\cos (x/2+30^\circ) = 2\sin (2x+30^\circ)\sin (x/2) = \cos (3x/2+30^\circ) - \cos (5x/2+30^\circ).\]
Rearrange as
\[\cos (3x/2+30^\circ) = \cos (x/2+30^\circ)+\cos (5x/2+30^\circ) = 2\cos (3x/2+30^\circ)\cos (x).\]
This implies that either $\cos x = 1/2$ or $\cos (3x/2+30^\circ) = 0$. The first implies $x=60^\circ \implies \angle ABC = 120^\circ \implies \angle ACB = 0^\circ$, absurd. Thus, we have that $3x/2+30^\circ \equiv 90^\circ\pmod{180^\circ}$, so $x/2\equiv 20^\circ\pmod{60^\circ} \implies \angle ABC \equiv 80^\circ \pmod{240^\circ}$. Thus, we get $\angle ABC = 80^\circ$ and $\angle ACB = 40^\circ$.

[hide=Solution?]Let $\angle ABQ=x$, so $\angle ABC = 2x$.Also let $QB=\sin 60.$ ( Scaling won't affect the angles)

Here we have, $AB+BP=AQ+QB.$ We will Note that by angle bisector theorem we have $AQ=\frac{AB}{AB+AC}\cdot BC.$

Now we find all the lenghts in terms of $\sin.$
Using $\sin$ law in $QBA,$ we get,
$$\frac{QB}{\sin 60}=\frac{AB}{\sin (120-x)}=\frac{AQ}{\sin x}.$$
Since we had $QB=\sin 60\implies AB=sin(120-x),~~AQ=sin(x).$.

Now, we find $AC,BC.$ Again using $\sin$ law on $\Delta ABC,$
we get $$\frac{BC}{\sin 60}=\frac{AB}{\sin(120-x)}=\frac{\sin(120-x)}{\sin(120-x)}\implies BC=\frac{\sin(120-x)\cdot \sin 60 }{\sin(120-2x)}$$
Similarly, we get $$ \frac{AC}{sin (2x)}=\frac{AB}{\sin(120-2x)}\implies AC=\frac{\sin(120-x)\cdot sin(2x)}{\sin(120-2x)}.$$

Now putting all the values we get
$$sin\left(60\right)+sin\left(x\right)=sin\left(120-x\right)\left(1+\frac{\left(\frac{sin\left(120-x\right)\left(sin\:60\right)}{sin\left(120-2x\right)}\right)}{\left(sin\left(120-x\right)+\left(\frac{sin\left(120-x\right)sin\left(2x\right)}{sin\left(120-2x\right)}\right)\right)}\right) $$
Simplifying the LHS, we get
$$sin\left(120-x\right)\left(1+\frac{\left(\frac{sin\left(120-x\right)\left(sin\:60\right)}{sin\left(120-2x\right)}\right)}{\left(sin\left(120-x\right)+\left(\frac{sin\left(120-x\right)sin\left(2x\right)}{sin\left(120-2x\right)}\right)\right)}\right)$$
$$=\sin(120-x)+\frac{\left(sin\left(120-x\right)\left(sin\:60\right)\right)}{\left(sin\left(120-2x\right)+sin\left(2x\right)\right)}$$
$$=sin\left(60+x\right)\left(1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}\right)$$
Note that $\sin(60+x)=2\sin(\frac{60+x}{2})\cos(\frac{60+x}{2}).$
 Also simplifying the RHS, we get $sin\left(60\right)+sin\left(x\right)=2\sin(\frac{x+60}{2})\cos(\frac{x-60}{2}).$
Since LHS=RHS we cancel out $\sin(\frac{x+60}{2})\ne 0.$

We get $$cos\left(\frac{x}{2}-30\right)=cos\:\left(\frac{x}{2\:}+30\right)\left(1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}\right) $$
So we have
$$ \frac{cos\left(\frac{x}{2}-30\right)}{cos\:\left(\frac{x}{2\:}+30\right)}=1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}$$
Note that here RHS is
$$ 1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}$$
$$=1+\frac{sin\left(60\right)}{\left(2sin\left(2x+30\right)+cos\left(30\right)\right)}$$
$$1+\frac{1}{2sin\left(2x+30\right)} $$
So we have $$ \frac{cos\left(\frac{x}{2}-30\right)}{cos\:\left(\frac{x}{2\:}+30\right)}=\frac{1+sin\left(2x+30\right)}{2sin\left(2x+30\right)}$$
Cross multiplying, we get
$$cos\left(\frac{x}{2}-30\right)2sin\left(2x+30\right)=cos\:\left(\frac{x}{2\:}+30\right)+2sin\left(2x+30\right)cos\:\left(\frac{x}{2\:}+30\right) $$
$$ =sin\left(\frac{5x}{2}\right)+sin\left(60+\frac{3x}{2}\right)=cos\:\left(\frac{x}{2\:}+30\right)+sin\left(\frac{5x}{2}+60\right)+sin\left(\frac{3x}{2}\right)$$
Now taking $x/2=y,$ we get
$\sin(5y+60)+\sin(60-y)+\sin(3y)=\sin(60+3y)+\sin(5y)\implies \sin(5y+60)-\sin(5y)+\sin(60-y)=\sin(60+3y)-\sin(3y).$

So, we have,
$$\cos (5y+30)+\sin(60-y)=\cos(3y+30)\implies \sin(60-5y)+\sin(60-y)=\cos(3y+30)$$ $$\implies 2\cdot \cos(3y+30)\cos(60-2y)=\cos (3y+30). $$

Note that here $0<y\le 30$
So $\cos(6-2y)=1/2$ is eliminated. Hence $\cos(30+3y)=0\implies y=20=x/2\implies \angle ABC=2x=80\implies \angle ACB=40.$[/hide]

[quote=orl]Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect  $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?[/quote]
Nice problem. Here's another solution.
Take Point $D$ such that
 $\angle ABD=60^\circ\implies \triangle ABD$ is equilateral, Let $AP\cap BD=H$ So $PD=BP$ Take Point $E\in BC$ such that $BE>BD$ and $DE=PD=BP$ Also take Point $F$ such that $BF>BA$ and $AF=BP$ and observe we have another equilateral $\triangle BFE$ Now take point $G\in BQ$ Such that $BG>BQ$ and $QG=AQ$

Claim 1-: $B$ is circumcenter of $\triangle FGE$
Prove-: it is Trivial observation As $BG=BE=BF$

Take Point $J$ Such that $BG=BJ$
Claim 2-: $GJ=FG=JE$
Prove-: First observe $\triangle BFG\cong \triangle BJE \implies FG=JE, \angle GFE=\angle JEF$ and also $FE||GJ$ So $GJEF$ is cyclic isosceles Trapezium. So $GJ=FG=JE$

Claim 3-: $\angle FBG=20^\circ$
Prove-: By Claim $2$ We get $\angle FEG=\frac{\angle FEJ}{2}$ and By Claim $1$

we get $\angle GEF=\frac{\angle FBG}{2}$ and

$\angle EFG=\angle FEJ=\frac{60-\angle FBG}{2}$
 So we get $\frac{\angle FBG}{2}=\frac{60-\angle FBG}{4}\implies \angle FBG=20^\circ$

Now By claim $3$ we get angles of $\triangle ABC$ are $\{40^\circ, 60^\circ, 80^\circ\}$ $\blacksquare$

Edit-: 200th Post in HSO :-)

I think bashing on this one is an insult to the problem. Here is a synthetic solution:

Let $B_{1}\in AC$ and $C_{1}\in AB$, so that $QB_{1}=QB$ and $BP=BC_{1}$ and $Q$ is between $A$ and $B_{1}$ and $B$ is between $A$ and $C_{1}$, and we have that:
$$AB_{1}=AQ+QB_{1}=AQ+QB=AB+BP=AB+BC_{1}=AC_{1}\Longrightarrow \angle C_{1}AB_{1}=60^{\circ}, AB_{1}=AC_{1}\Longrightarrow \triangle AB_{1}C_{1}\text{ is equilateral}$$
Notice that $\angle BC_{1}P=\angle BPC_{1}=\frac{\beta}{2}$ and $P$ lies on the angle bisector in an equilateral triangle, so $\triangle PAC_{1}\cong \triangle PAB_{1}$ $\Longrightarrow \angle PB_{1}A=\angle PC_{1}A=\frac{\beta}{2}$. Now assume that $P\not\in BB_{1}$. Then $QB=QB_{1}$ by construction and $\angle PBQ=\angle PB_{1}Q$, so since $P$ lies in the interior of $\triangle QBB_{1}$, then that'd mean that $P$ lies on the perpendicular bisector of $BB_{1}$, so $PQ\perp BB_{1}\Longrightarrow \angle BQP=\angle B_{1}QP$, so $P$ is the $A$-excenter for $\triangle ABQ$, which  implies that $\angle PBQ=\angle PBC_{1}\Longrightarrow \frac{3}{2}\beta=180^{\circ}\Longrightarrow \beta=120^{\circ}$, which implies that $\gamma=0$, which is obviously impossible. Therefore $P\in BB_{1}\Longrightarrow C\equiv B_{1}\Longrightarrow \gamma=\frac{\beta}{2}$, so $\beta=80^{\circ}$ and $\gamma=40^{\circ}$ and we're done!
[hide=Motivation]Basically whenever we're faced with equality of (sum of) segments like in this problem we try to translate the segments onto one or two lines and then angle chase, as shown above. Quite a beautiful problem in my opinion.[/hide]

lol.

Let $\cos \theta=x$ where $\theta=\angle ABQ=\angle CBQ$. Then we must solve $$3(4x^3-2x^2-2x+1)^2=(1-x^2)(4x^2-6x+2)^2$$ and we can factor out $(2x-1)^3$ since $x=\frac{1}{2}$ is extraneous which gives $$8x^3-6x+1=0\iff \cos 3\theta=-\frac{1}{2}$$ and since $\theta<60^{\circ}$ this means that $3\theta = 120^{\circ}$ and $\theta = 40^{\circ}$. Answer is $\angle A, \angle B, \angle C = 60^{\circ}, 80^{\circ}, 40^{\circ}$.

the sextic comes from bashing out $$1+\frac{\sin 30^{\circ}}{\sin(2\theta+30^{\circ})}=\frac{\sin \theta + \sin 60^{\circ}}{\sin(\theta + 60^{\circ})}$$

Yolo.

[img width=50]https://media.discordapp.net/attachments/925784397469331477/939643508531286127/Screen_Shot_2022-02-05_at_3.07.37_PM.png?width=1282&height=1169[/img]

Let $D$ be on $AB$ extended such that $ACD$ is equilateral. Let $C'$ be on $AC$ such that $QB=QC'$ and $A,C'$ are on opposite sides of $Q.$ Let $D$ be on $AB$ such that $BP=BD$ and $A,D$ are on opposite sides of $B.$ Note that $AB+BP=AQ+QB$ implies $AC'=AB+BD=AQ+QC'=AD$ so $AC'D$ is equilateral.

Since $AP$ is an angle bisector, $\angle AC'P=\angle ADP=\angle BPD=\frac{1}{2}\angle PBA=\angle PBQ.$ However, we have $QC'=QB,QP=QP,$ and $\angle QC'P=\angle QBP,$ so by quick sine law on $\triangle QPC',\triangle QPB$, we get that $\sin QPC'=\sin QPB.$

If they are supplementary then $C=C'$ and so $\angle BCD=\angle PDC=\frac{1}{2}\angle BPD=\frac{1}{2}\angle ACB$ and $\angle BCD+\angle ACB=60^\circ$ so $\angle ACB=40^\circ,\angle ABC=80^\circ.$

If they are the same then $PB=PC=PD,$ so $P$ is the circumcenter of $BCD$ which is impossible because $\angle CDB\neq 90^\circ.$ Thus, our only solution is $\{40^\circ,60^\circ,80^\circ\}$

Let $X$ and $Y$ be the points on ray $\overrightarrow{AB}$ and segment $\overline{AC}$ such that $BX=BP$ and $BQ=QY,$ respectively. Then, $\triangle APX\cong\triangle APY$ so $$\alpha=\angle AYP=\angle PXA=90-\tfrac{1}{2}\angle XBP=\angle PBQ=\angle QBP.$$ Suppose FTSOC that $Y$ does not lie on $\overline{PB}.$ Then, $$\angle PBY=\angle PBQ-\angle YBQ=\angle QYP-\angle QYB=\angle QYP$$ so $PX=PY=PB$ and $\alpha=60.$ Then, $\angle A+\angle B=60+2\alpha=180,$ meaning $\angle C=0,$ a contradiction. Hence, $B,C,$ and $P$ are collinear meaning $Y=C.$ Therefore, the sum of the angles of $\triangle ABC$ is $60+3\alpha$ so $\alpha=40$ and $\angle A=60,\angle B=80,\angle C=40.$ $\square$

[quote=pbed fuzi L]
aime p14 lvl bash in imo p5???

WOLOG let $R = \frac 12$ and $\angle ABQ = \angle QBC = x$.  By Law of Sines on $\triangle ABP$, $BP = \frac{\sin C}{2\sin(150^{\circ}-2x)}$, and by LoS on $\triangle ABQ$, $AQ = \frac{\sin C\sin x}{\sin(120^{\circ}-x)}, QB = \frac{\frac{\sqrt{3}}{2}\sin C}{\sin(120^{\circ}-x}$.  Thus, we can encode the given as\begin{align*}AB + BP &= AQ + QB \\ c + \frac{\sin C}{2\sin(150^{\circ}-2x)} &= \frac{\sin C\sin x}{\sin(120^{\circ}-x)} + \frac{\frac{\sqrt{3}}{2}\sin C}{\sin(120^{\circ}-x} \\ 1 + \frac{1}{2\sin(150^{\circ} - 2x)} &= \frac{\sin x + \sin 60^{\circ}}{\sin(120^{\circ}-x)}\\ 1 + \frac{1}{2\sin(150^{\circ} - 2x)} &= \frac{\cos\left(\frac x2 - 30^{\circ}\right)}{\cos\left(\frac x2 + 30^{\circ}\right)} \\ 2\cos\left(\frac x2 - 30^{\circ}\right)\sin(150^{\circ}-2x)&= 2\cos\left(\frac x2 + 30^{\circ}\right)\sin(150^{\circ}-2x) + \cos\left(\frac x2 +30^{\circ}\right) \\ \sin\left(120^{\circ}-\frac{3x}{2}\right) + \sin \left(\frac{5x}{2}\right) &= \sin\left(\frac{3x}{2}\right) + \sin\left(120^{\circ} - \frac{5x}{2}\right) + \cos\left(\frac x2 + 30^{\circ}\right) \\ \sin \left(\frac{5x}{2}\right) -\sin\left(\frac{3x}{2}\right) &= \sin\left(120^{\circ} - \frac{5x}{2}\right) - \sin\left(120^{\circ}-\frac{3x}{2}\right) + \cos\left(\frac x2 + 30^{\circ}\right) \\ 2\cos(2x)\sin\left(\frac x2\right) &= -2\cos(120^{\circ}-2x)\sin\left(\frac x2\right) + \cos\left(\frac x2 + 30^{\circ}\right) \\ 2\sin\left(\frac x2\right)\left[\cos(2x) + \cos(120^{\circ} - 2x)\right] &= \cos\left(\frac x2 + 30^{\circ}\right) \\ 2\sin\left(\frac x2\right)\cos(2x-60^{\circ}) &= \cos\left(\frac x2 + 30^{\circ}\right) \\ \sin x \cos(2x-60^{\circ}) &= \cos\left(\frac x2 + 30^{\circ}\right) \cos\left(\frac x2\right) \\ \sin(3x-60^{\circ}) &= \cos(30^{\circ}) \\ x &= 40^{\circ}, 60^{\circ}, 80^{\circ}.\end{align*}And we're FINALLY done.

godermit this problem sucks

[hide=Storage]
Let $x = \angle ABQ$ Note that by law of sines, $$\frac{AQ}{\sin x} = \frac{BQ}{\sin 60} = \frac{c}{\sin\angle AQB}$$ and $$\frac{BP}{\sin 30} = \frac{c}{\sin\angle APB}$$

As such, \begin{align*}AQ + QB = AB + BP &\iff \\ \frac{c}{\sin\angle AQB}(\sin x + \sin 60) = \frac{c}{\sin\angle APB}(\sin\angle 30 + \sin 2x + 30) &\iff \\ \frac{\sin x + \sin 60}{\sin(x + 60)} = 1 + \frac{1}{2\sin 2x + 30} \end{align*}

Then, since $\sin x + \sin 60 = \sin (\frac{x}{2} + 30) \cos(\frac{x}{2} - 30)$, $$\frac{\sin x + \sin 60}{\sin(x + 60)} = \frac{\sin (\frac{x}{2} + 30) \cos(\frac{x}{2} - 30)}{2\sin(\frac{x}{2} + 30)\cos(\frac{x}{2} + 30)} = \frac{\cos(\frac{x}{2} - 30)}{\cos(\frac{x}{2} + 30)} $$

Then, the equation becomes \begin{align*}
    \frac{\cos(\frac{x}{2} - 30) - \cos(\frac{x}{2} + 30)}{\cos(\frac{x}{2} + 30)} = \frac{1}{2\sin(2x + 30)} &\iff \\ \frac{\sin \frac{x}{2}}{\cos(\frac{x}{2} + 30)} = \frac{1}{2\sin(2x + 30)} &\iff \\
    2\sin(2x + 30)\sin \frac{x}{2} = \cos(\frac{x}{2} + 30) &\iff \\
    \cos\left(\frac{3x}{2} + 30\right) = \cos\left(\frac{5x}{2} + 30\right) + \cos\left(\frac{x}{2} + 30\right)
\end{align*}

Substitute $a = \frac{3x}{2} + 30$ to get $\cos a = 2 \cos a \cos x$. Then either $x = 60$ or $a = 90$. The first case implies $\angle C = 0$, impossible, so $\frac{3x}{2} + 30 = 90$ and $x = 40$. Thus, $\angle A = 60$, $\angle B = 80$, $\angle C = 40$.

[/hide]

[quote=orl]Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect  $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?[/quote]

Wow!! A G8!!
Can someone check if this is right (especially the second solution)?

WLOG BP=1.
By a ton of Law of Sines, we have <APB=150-B, <AQB=120-B. $\boxed{2\sin(30+B)}=\sin(30+B)BP/\sin30=AB=\boxed{\frac{AQ\sin(60+B)}{\sin(1/2B)}}=BQ\sin(120-1/2B)/\sin60=\boxed{\frac{2(AB+1-AQ)\sin(60+1/2B)}{\sqrt{3}}}$. [hide=my work, totally not used wolframalpha when first solving, don't click, it takes up the length of 10 solutions]Exactly, for this solution I am showing no work. I did indeed use wolfram alpha[/hide]
Simplifying this, we get angles of 40, 80, and 60 degrees.

Second solution: (Latexing this was a NIGHTMARE.) In these types of problems with little points and information, we must surely make a construction (for once I will make one!). The easiest way to do so is to make two segments' sum into two segments' sum which are collinear, since it is much simpler that way. Then we let B' on AB s.t. $BP=BB'$, C' on AC s.t. $QB=QC'$ (Remark: if this wasn't geogebra it would be hard to guess what type of diagram meaning it would be hard to tell C coincides with C'). We have $AB+BB'=AB+BP=AQ+QB=AQ+QC'$, or $AB'=AC'$ so $\triangle AB'C'$ is equilateral. Now, suppose B, P, and C' are not collinear. Then, we have $\angle ABC=2b$ gives $\angle BB'P=\angle BPB'=b$. By corresponding angles, BQ//B'P, and $\angle PB'C'=60-b$, and because B'PC' is isosceles, we have $\boxed{\angle{B'PC'}=60+2b}$. $\boxed{\angle BPC'}=\angle BPB'+\angle B'PC'=b+60+2b=\boxed{60+3b}$, $\angle BQC'=60+b$, and since BQC' is isosceles, $\angle BC'A=\angle BC'Q=\boxed{60-b/2}$; Finally, $\boxed{\angle{PBC'}}$=<PBB'-<C'BB'=<PBB'-(180-<BC'B'-<BB'C')=<PBB'-(180-<BB'C'-(<AC'B'-<BC'A))=$180-2b-(180-60-(60-(60-b/2)))=\boxed{60-3b/2}$. Then $\boxed{\angle BC'P=180-(60-3b/2)-(60+3b)=60-3b/2}$, so $\boxed{BP=PC'}$. We knew that QBC' was isosceles, so QP bisects <BQC, which gives BA/AC=BP/PC=BQ/QC. By Law of Sines, we have sin<QBC/sin<QCB=QC/QB=AC/AB=sin<ABC/sin<ACB, which gives sinb=sin<QBC=sin<ABC=sin2b, or 2b+b=180 b=60, and <ABC=120. However, this makes a degenerate triangle ABC, which is a contradiction. Then B, P, C' are collinear, so C=C' and therefore BQ=QC'=QC which means <QCB=<QBC=b, and 2b+b=180-60=120, so b=40 and then our angles are 40, 60, and 80 degrees. In fact, while the synthetic way is harder to see where to go, whereas trig you always know what to do, the synthetic would probably be quicker even though the solutions length is longer, because it is just a bunch of simple angle chasing, with one obvious construction. $\blacksquare$

Solved with [b]popop614[/b]

Let $P'$ be the reflection of $P$ over the exterior angle bisector of $B$ in triangle $ABC$, and let $B'$ be the reflection of $B$ over the angle bisector of $\angle BQC$. Then, the length condition means that $AB'P'$ is equilateral. Assume that $BPB'$ are not collinear. Now, note that $P$ is symmetric with respect to $AP'$ and $AB'$, so $\triangle PP'B'$ is isosceles. It is also symmetric with respect to $QB$ and $QB'$, so $BPB'$ is isosceles, implying $BP'=BP=PB'=PP'$. In other words, $BPP'$ is equilateral, so $\angle ABC = 120$, implying $ABC$ is degenerate. Therefore, $BPB'$ collinear, so $B'=C$. Therefore, $2\angle BCA = \angle ABC$. This implies that $ABC$ has angles $40$, $60$, and $80$.


Suppose $G$ is an infinite graph tht each finite dubgraph of $G$ is planar. Prove that $G$ is planar.

Define what you mean by "infinite graph" here, please. You have to have some bound on the cardinal of the set of vertices. If, for example, $G$ is the empty graph on more than $c$ (continuum cardinality) vertices, then every finite subgraph is the empty graph on finitely many vertices, and is thus planar, but there's no way you can place all the vertices of $G$ in different points of the plane (because there are more vertices of $G$ than there are points of the plane).

Maybe our graph is supposed to be countable? Or maybe it's supposed to have $\le c$ vertices?


geometry
geometric transformation
reflection
rotation

In how many ways can 8 people sit around a round table if Pierre and Thomas want to sit together, but Rosa doesn't want to sit next to either of them?

I asked about this question because:
the answer given is 480

I think the question is not clear enough

I am not sure if the seats are distinguishable/indistingushable, and if reflections count of the same arrangement count twice or once...

Also, I did not get 480 through any of the possibilities, but I think I messed up

We first seat Thomas, who has 8 spots to choose from. Then Pierre has 2 spots to choose after Thomas has been seated. Rosa cannot sit where Thomas or Pierre sit, nor in the 2 seats right next to them, giving Rosa 4 available spots. There are 5! ways to seat the remaining 5 people. Then we have to divide by 8 because we can rotate the table in a total of 8 possible ways. Thus, there are $ \frac{8\times 2\times 4\times 120}{8}\equal{}8\times 120\equal{}960$.

What did you get, hurdler? Can someone comfirm this?

Hmm...

I was thinking that since we want Pierre and Thomas to sit next to each other, we could view them sort of like one person. So for this "person" we have 7 choices. Rosa doesn't want to sit next to them, so there are 4 choices for her. The other people have $ 4!$ or $ 24$ ways. So:

$ \frac{(7 \times 2) \times 4 \times 24}{8} \equal{} \boxed{168}$

This seems way off...

This question is somewhat ambiguous, but I'm going to assume that it means that seating arrangements in which everyone is sitting next to the same person are not considered different arrangements.

Therefore, dragon96's solution is on the right track, however, you must divide by two, since the people can be seated next to the same people in either clockwise or counterclockwise directions.  This gives $ 960/2\equal{}\boxed{480}$.

Isabella's solution is on the right track, as well.  Treating them as one person is how I, personally, would have gone about finding the answer.  However, there were 5 more people to be seated, not 4.  Also, since you decided to treat two people as one person, there are only 7 rotations that give the same arrangement, not 8. And just like dragon96, you forgot to divide by 2.  So it would be:

$ \frac{(7*2)*4*120}{7*2}\equal{}\boxed{480}$

Ah, I see.

Thank you.

I am not sure what I was doing earlier but 480 makes sense now.

Please, in the future, thank someone in private, to reduce spam. Thanks!  :)


function
calculus
derivative
trigonometry
limit
absolute value
calculus computations

Could you provide of few examples of functions that are continuous on a certain interval but whose first order derivatives are discontinuous over that same interval.

What about $ f(x)\equal{}|x|$ on $ ]\minus{}\infty,\infty[$ ?

[quote="swede"]What about $ f(x) \equal{} |x|$ on $ ] \minus{} \infty,\infty[$ ?[/quote]

That's wrong ,  $ f(x) \equal{} |x|$ in not derivable in $ 0$.   :D  Take for exemple $ f(x)\equal{}x^2\sin \frac 1{x^2}, f(0)\equal{}0$ .  :cool:

For a function to be continuous at c the following must be true:

The limit for the left of f(x) is equivalent to the limit from the right of f(x).
The value f(c) exists.
The limit of f(x) as x approaches c is equal to the actual value of f(c).

So yes, $ f(x)\equal{}|x|$ works fine. The function itself is continuous while the value $ f^{\prime}(0)$ is nonexistent and the limits from the right and left of the derivative approach different values, making the derivative discontinuous.
Another justification for this is that differentiability implies continuity, and since the function isn't differentiable at x=0, nor is it continuous at x=0.

Observing the graphs of $ f(x)\equal{}x^2\sin\frac{1}{x^2}$ and its derivative we note that $ f^{\prime}(x)$ has either a relative maximum or minimum at the zeros of $ f(x)$. However, because $ y\equal{}\frac{1}{x^2}$ approaches the same value from the left and right of zero, we know that $ \cos{\frac{1}{x^2}}$ tends to some value $ C$ albeit by oscillation.

$ \displaystyle \lim_{x\to 0} f^{\prime}(x)\equal{}\lim_{x\to 0}\left[ \frac{2x^2\sin{\frac{1}{x^2}}\minus{}2\cos{\frac{1}{x^2}}}{x}\right]\equal{}\lim_{x\to 0} \left[ \minus{}\frac{2\cos{\frac{1}{x^2}}}{x}\right]\equal{}\lim_{x\to 0} \left[\minus{}\frac{C}{x}\right]$

So it looks like $ \lim_{x\nearrow 0} f(x)\equal{}\infty$ and $ \lim_{x\searrow 0} f(x)\equal{}\minus{}\infty$. This makes some sense taking the min/max observation from earlier into account. However, I can't be sure because the graph of the derivative seems to be crossing the axis in between every peak and is verifiably continuous for all values aside from $ x\equal{}0$. Whether or not this pattern holds or is negated by the fact that $ y\equal{}\frac{1}{x^2}$ is undefined at $ x\equal{}0$ is beyond me, though I suspect this is the case.

I think his example is valid. Here's a graph:

The fact that differentiably implies continuity is useless for showing $ |x|$ isn't differentiable at 0- of course the absolute value [i]is[/i] continuous at 0.

At points where things such as the chain rule and inverse rule fail, you're going to have to check using the limit definition. Some examples where it fails are:

- $ \sqrt[3]{x}$ at 0 (the derivative of $ x^3$ is 0 there)

- $ |x|\equal{}\sqrt{x^2}$ at 0 (the square root has issues at 0 similar to the one above)

- the function $ f$ Diogene gave ($ \frac{1}{x^2}$ clearly isn't differentiable at 0)

To see whether such functions are actually differentiable at 0, we must check the limit $ \lim_{h\rightarrow 0} \frac{f(h)\minus{}f(0)}{h}$. In the first case, for nonzero $ h$, $ \frac{f(h)\minus{}f(0)}{h}\equal{}h^{\minus{}2/3}$ and this limit doesn't exist (in the strict sense: i.e. $ \infty$ doesn't count)

In the second case, $ \frac{f(h)\minus{}f(0)}{h}\equal{}\frac{|h|}{h}$ which is 1 for positive $ h$ but -1 for negative $ h$. So that limit doesn't exist either.

But in the third case, $ \frac{f(h)\minus{}f(0)}{h}\equal{}h\sin \frac{1}{h^2}$. That's bounded in absolute value by $ |h|$, so it tends to 0. As a result, $ f^\prime (0)\equal{}0$ - even though $ f^\prime (x)$ is very badly behaved around 0. Note also that $ \frac{f(x)\minus{}f(0)}{x}$ and $ f^\prime (x)$ can be rather different functions.


Inequality
AMC
USAMO
three variable inequality

After the hint it was easy
LHS$\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$

And why is (I presume that's what you ment) $\frac 1{1+b+c} \leq \frac{abc }{bc(a+b+c)}$ ?  :huh:

Put $a=\ell A$, $b=\ell B$ and so on with $\ell\geq 1$ and $ABC=1$.

It's clear that

$\sum \frac{1}{1+b+c}\leq \sum \frac{1}{1+B+C}$

Put $A=x^3$, $B=y^3$ and so on.  It's clear that $xyz=1$

So we get

$\sum \frac{1}{1+B+C}=\sum \frac{1}{xyz+y^3+z^3}$

and our inequality is just USAMO 1997.

i can't say this is the most beautiful proof that can be done, but it definitely works!!!  :D

after 5 minutes of clearing the denominators ( :rotfl: ) we get to prove that
$2+2a+2b+2c\leq 2abc+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2$, we'll prove that
$2a+2b+2c\leq a^2b+a^2c+ab^2+b^2c+ac^2+bc^2$

by am-gm we have that $a^2b+a^2c\geq2\sqrt{a^3}\sqrt{abc}\geq2\sqrt{a^3}$, doing the same with the other terms yields
$a^2b+a^2c+ab^2+b^2c+ac^2+bc^2\geq2(\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3})$.
now by chebyschev we have that
$\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}\geq\frac{1}{3}(a+b+c)(\sqrt a+\sqrt b+\sqrt c)\geq(a+b+c)$,
and we're done!!! :P

This problem is equivalently with

$\"\ \triangle ABC,\ sr^2\ge 1\Longrightarrow 1+s\le 2Rsr\ \"$
$(\"\ a+b+c\le 2S^2\Longrightarrow 2+a+b+c\le abc\ \")$

Here is a algebrical solution of the initial inequality. The inequality is equivalently with the relation

$\sum a^2(b+c) +2abc\ge 2(a+b+c)+2$ [b][u](1).[/u][/b]

But $2abc\ge 2$ [b][u](2)[/u][/b] and $3+\sum a^2(b+c)=\sum (a^2b+a^2c+1)\ge$

$\ge 3\sum \sqrt [3] {a^2b\cdot a^2c\cdot 1}\ge$ $3\sum a\cdot \sqrt [3] {abc}\ge 3\sum a=$

$=2\sum a +\sum a\ge 2\sum a +3\sqrt [3] {abc}\ge 2\sum a +3\Longrightarrow \sum a^2(b+c)\ge 2\sum a$ [b][u](3).[/u][/b]

From the sum of the relations (2) and (3) we obtain the relation (1).

[quote="blahblahblah"]
It's clear that

$\sum \frac{1}{1+b+c}\leq \sum \frac{1}{1+B+C}$

[/quote]

Why is it?

$\frac{1}{1+b+c}=\frac{1}{1+\ell B+\ell C}\leq\frac{1}{1+b+c}\because \ell\geq 1$

Let $a,b,c$ be positive real numbers such that $abc= 1$. Prove that \[ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq 1.  \]([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=409496&hilit=Moscow+1997]Moscow Olympiads 1997[/url])

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=589699&p=3496195#p3496195]here[/url]

[quote="sqing"]The following inequality is also true.
Let $a,b,c$ be positive real numbers such that $abc\geq 1$. Prove that \[ \frac{b^2+c^2}{1+b+c}+\frac{c^2+a^2}{1+c+a}+\frac{a^2+b^2}{1+a+b}\geq 2.  \][/quote]
After using C-S we need to prove that $(a+b+c)^2\geq2(a+b+c)+3$, which is
${(a+b+c-3)(a+b+c+1})\geq0$, which is AM-GM.

Thanks.
The following inequality is also true.
Let $a,b,c$ be positive real numbers such that $abc\geq 1$. Prove that \[ \frac{b^2+c}{1+b+c}+\frac{c^2+a}{1+c+a}+\frac{a^2+b}{1+a+b}\geq 2.  \]

[hide="@sqing"]Edited long back... :P[/hide]

The following inequality holds for the case $abc \leq 1$!
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=602317]Inspired by "GMB-IMAR 2005, Juniors, Problem 1"[/url]

[quote="Valentin Vornicu"]Let $a,b,c$ be positive real numbers such that $abc\geq 1$. Prove that \[ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq 1.  \]
[hide="Remark"]This problem derives from the well known inequality given in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=185470#p185470]USAMO 1997, Problem 5[/url].
[/hide][/quote]

Remark: About Muirhead inequality
If $abc\geq 1$ than $[p_1,p_2,p_3]\geq [p_1-r,p_2-r,p_3-r]$, $\forall r\geq 0$
 So, [quote="campos"]after 5 minutes of clearing the denominators ( ) we get to prove that $ 2+2a+2b+2c\leq 2abc+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2 $
, we'll prove that $ 2a+2b+2c\leq a^2b+a^2c+ab^2+b^2c+ac^2+bc^2 $[/quote]
We can write $[1,0,0]\leq [2,1,0]$, which is true acording remark.

[quote=Valentin Vornicu]Let $a,b,c$ be positive real numbers such that $abc\geq 1$. Prove that \[ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq 1.  \]
[hide="Remark"]This problem derives from the well known inequality given in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=185470#p185470]USAMO 1997, Problem 5[/url].
[/hide][/quote]
[url=http://www.artofproblemsolving.com/community/c6h1309061p7008144]This is equivalent to[/url] \[ \frac{1}{1+a+b}+\frac{1}{1+b+c}\leq 1-\frac{1}{1+c+a}. \] Clear denominators to get \[ 2+2(a+b+c)\leq 2abc+a^2(b+c)+b^2(c+a)+c^2(a+b).\] This is obvious because $abc\geq 1$ and the rest by Muirhead.


geometry
geometric transformation
search

Can anyone in this forum please kindly translate the problems of Romanian District Olympiad 2007 from http://olimpiade.ro/informatii.php?etapa=2&materie=2&editie=6 ?

I requested the translation in http://www.mathlinks.ro/viewtopic.php?t=181940, but to no avail...

Please search on the site (for example :Romanian DMO works). The problems were posted last year by felixman.

Thanks. What about the national olympiad for Grade 7 and 8? Have they been posted?

If you look carefully, you will see that the 7th and 8th grades' problems were posted :D.

Well, I mean the national olympiad, not the district. I think I have looked carefully, but I really have not found it. Please enlighten me  :D

Oh, i'm sorry. I have misunderstood the question. The NMO 2007 for 7th and 8th grades was not posted.


Determine the number of perfect squares in the infinite sequence :

[i][b]1! , 1!+2! , 1!+2!+3! , ..... , 1!+2!+...+n!, ...[/b][/i]

any clue?

[quote="ZetaX"]Hint:
[hide]look $\mod 5$.[/hide][/quote]

Thanks... I get it now.

Er,emmmm, Sorry i dont. Plz tell something more if u have time....

Thanx and have a nice day

Is it some wide spread problem in India¿ ;)

Further hint:
[hide]Squares are $\equiv 0,1,4 \mod 5$, numbers of type $1!+2!+3!+...+n!$ for $n \geq 4$ are not.[/hide]


Find all triples $ (m,n,k)$ of possitive integer numbers such that:
\[ \left[\frac {m^2}{kn}\right] \plus{} \left[\frac {n^2}{km}\right] \equal{} \left[\frac {m}{kn} \plus{} \frac {n}{km}\right] \plus{} mn,
\]
where $ [x]$  denotes the largest integer not larger than $ x$.

[i]Edited by Myth[/i]

Please again, take care about where you are posting... ;)
According to what you are saying, this one should be in the proposed and own problems section...  :?

Pierre.

[quote="nhat"]is it very hard?it makes me  two days to have the solution.[/quote]

It may be. The special case $k=1$ was in the ISL 1996.

Pierre.


When Roslyn writes her name repeatedly in a 5 by 5 table as shown, the bottom right corner contains the letter R. If she repeats the process in a 20 by 20 table, what letter would occupy the bottom right corner?

[asy]for(int i=0; i<6; ++i)
{
	draw((0,i)--(5,i),linewidth(2));
}
for(int i=0; i<6; ++i)
{
	draw((i,0)--(i,5),linewidth(2));
}

filldraw((4,0)--(5,0)--(5,1)--(4,1)--(4,0)--cycle,gray);

for(int i=0; i<5; ++i)
{
	label("R",(i+.5,4.5-i));
}
for(int i=0; i<4; ++i)
{
	label("N",(i+.5,3.5-i));
}
for(int i=0; i<3; ++i)
{
	label("Y",(i+.5,2.5-i));
}
for(int i=0; i<2; ++i)
{
	label("L",(i+.5,1.5-i));
}
for(int i=0; i<1; ++i)
{
	label("S",(i+.5,.5-i));
}
for(int i=1; i<5; ++i)
{
	label("O",(i+.5,5.5-i));
}
for(int i=2; i<5; ++i)
{
	label("S",(i+.5,6.5-i));
}
for(int i=3; i<5; ++i)
{
	label("L",(i+.5,7.5-i));
}
for(int i=4; i<5; ++i)
{
	label("Y",(i+.5,8.5-i));
}[/asy]

The last column looks like:
$ O,L,N,O,L,N,...,O,L,N,O,L$ so it's L.

Or, there are $ 400$ letters, and $ 400 \equiv 4 \pmod{6}$, so it's the $ 4$th letter, which is $ \boxed{\text{L}}$.


search
real analysis
real analysis unsolved

In this post http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=916749100&t=147136 I do not understand how to prove that set which puuhikki defined is a $ \sigma$-algebra.    How can I prove that if I have a countable collection of sets in this set, then the union is in this set?  given $ \epsilon >0$ , what are $ G$ and $ F$ in this case?

Thanks

For instance, let $ \Phi\equal{} \{A: A$ is measurable and given $ \epsilon >0$ there exist $ G$ open and $ F$ closed such that $ F \subset A \subset G$ and $ m(G \backslash F) < \epsilon\}$.    Now, let $ A_n$ be a collection of sets in $ \Phi$.    Given $ \epsilon >0$ there exists $ G_n$ open and $ F_n$ closed such that $ F_n \subset A_n \subset G_n$ and $ m(G_n \backslash F_n) < \epsilon$, for all $ n$.    I was thinking to take $ G\equal{} \cup_n G_n$ and $ F\equal{}closure\{\cup_n F_n\}$, but the problem is that I am stuck trying to prove that $ m(G \backslash F) < \epsilon$.   Am I right taking this $ F$ and $ G$ ?

Thanks in advance.

No. You cannot afford taking the closure of the union, it may be too big. Instead, you have to take sufficiently large finite union. Also you need to choose $ F_n$ and $ G_n$ such that $ \mu(G_n\setminus F_n)\le\varepsilon_n$ where $ \sum_n\varepsilon_n < \varepsilon$.


inequalities unsolved
inequalities
algebra
highschoolmath

[quote="arqady"]Let $a\geq b\geq c\geq0.$
Thence, $(a^2-ac+c^2)(b^2-bc+c^2)\leq a^2b^2.$
Hence, $(a^2-ab+b^2)(a^2-ac+c^2)(b^2-bc+c^2)\leq(a^2-ab+b^2)a^2b^2=$
$=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2\leq12.$
Equality when $a=2,b=1,c=0$ and sym. :)[/quote]

LOL, GREAT solution.
This problem may look trivial but I think it's quite tough and not standart at all :lol:

Thanks Loser,you are great and the solution is too.Loser,do you have some problems like my problem?If you have ,you can post it.We can propose :D .I think the style of this problem is very nice and maybe hard :P

Hi, the solution is same as me in my book.!

To tamtrantrong: Dung co post bai cua thay len day ma. Em phai tu lam di chu. Ma co post thi cung nen ghi nguon goc cua no. OK?

[quote="tamtrantrong"]let a,b,c be nonnegative real numbers satisfy a+b+c=3.Prove that:
         (a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)<=12[/quote]
[i]Solution.[/i] WLOG $a\ge b\ge c\ge 0$. Let $f(a,b,c)=\prod_{cyc} (a^2-ab+b^2)$. Verify that $f(a,b+c,0)-f(a,b,c)\ge 0$. Thus, it suffice to maximize $f(a,b,0)$. We have $a+b=3$, and by AM-GM, \begin{align*}
f(a,b,0) & =a^2b^2(a^2-ab+b^2)\\ &=a^2b^2(9-3ab)\\ &=\frac{4}{9}\cdot \frac{3ab}{2}\cdot \frac{3ab}{2}(9-3ab)\\ &\le \frac{4}{9}\left(\frac{9}{3}\right)^3\\ &=\boxed{12}.\end{align*} $\blacksquare$

Arkan

[quote="tamtrantrong"]Let $x,y,z$ be nonnegative real numbers satisfy $x+y+z=3$. Prove that:
\[(y^{2}-yz+z^{2})(z^{2}-zx+x^{2})(x^{2}-xy+y^{2})\leq12. \]
[/quote]
[url=https://artofproblemsolving.com/community/c6h550011p3191659]h[/url] [url=https://artofproblemsolving.com/community/c6h498067p2798739]h[/url] [url=https://artofproblemsolving.com/community/c6h2038741p14409539]h[/url]
Let $a, b,c,d,e$ be nonnegative numbers such that $a+b+c+d+e =5.$ [url=https://artofproblemsolving.com/community/c6h1738323p14387776]Prove that[/url]$$(a^2-ab+b^2)(b^2-bc+c^2)(c^2-cd+d^2)(d^2-de+e^2)(e^2-ea+a^2)\leq 192.$$


Ok so:
+/- post number
If you double a number OR zero a number, you get one gold coin.
Everyone starts with 0 grenades; you have to buy them.
When you fire a grenade, it will go off 2+(binary digital sum) turns later.
You can bump a grenade if you have one; you may NOT bump it on the thrower.
You may NOT bump a grenade if you are set to lose that same turn.
You die if you reach -4(current post number). You can revive at the cost of 2 gold coins.
You may option to obtain a gold coin if you reach 4(current post number), your points will reset to 0.
You may join in any time of the game. You start with 3 gold coins.

Invincibility star: Lasts (binary digital sum) turns. Cost: 3 gold coins.
Grenade: Cost: 3 gold coins.
Grenade Shield: Lasts (binary digital sum) turns. Cost: 2 gold coins.
Mushroom: Can give/lose points to self. Cost: 1 gold coin.

Starting now:
NAME; Coins; Points; Status
james4l; 3; 0;

NAME; Coins; Points; Status
james4l; 3; 0;
evilhamster; 3; 3;

[To kill: -12]

The game has already started!

Thank you for making a reasonable game.

james4l; 3; -1;
evilhamster; 3; 3;
chenhsi; 3; 0

james4l: 3 coins, 5 points, alive
evilhamster: 3 coins, 3 points, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive

james4l: 3 coins, -2 points, alive
evilhamster: 3 coins, 3 points, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive

james4l: 3 coins, -10 points, alive
evilhamster: 3 coins, 3 points, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive

james4l: 3 coins, -10 points, alive
evilhamster: 3 coins, 12 points, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive

james4l: 3 coins, 0 points, alive
evilhamster: 4 coins, 12 points, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive

james4l: 3 coins, 0 points, alive
evilhamster: 4 coins, 1 point, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive
cf249: 3 coins, 0 points, alive

james4l: 3 coins, 0 points, alive
evilhamster: 4 coins, -11 point, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive
cf249: 3 coins, 0 points, alive[/quote]

james4l: 3 coins, -13 points, alive
evilhamster: 4 coins, -11 points, alive
chenhsi: 3 coins, 0 points, alive
ernie: 3 coins, 0 points, alive
cf249: 3 coins, 0 points, alive

james4l: 3 coins, 0 points, alive
evilhamster: 4 coins, -11 points, alive
chenhsi: 4 coins, 0 points, alive
ernie: 3 coins, 0 points, alive
cf249: 3 coins, 0 points, alive[/quote]

... james4l has 1 point and chenhsi has 3 coins...

james4l: 3 coins, 1 points, alive
evilhamster: 4 coins, -11 points, alive
chenhsi: 3 coins, -15 points, alive
ernie: 3 coins, 0 points, alive
cf249: 3 coins, 0 points, alive

How much is a grenade?

That wasn't nice...

james4l: 3 coins, 119 points, alive
evilhamster: 0 coins, -61 points, alive
chenhsi: 2 coins, -93 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 267 points, alive

Death [-376]

That wasn't nice...

james4l: 3 coins, 119 points, alive
evilhamster: 0 coins, -61 points, alive
chenhsi: 2 coins, -188 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 267 points, alive

Death [-380]

Hey, please stop attacking me.

james4l: 3 coins, 213 points, alive
evilhamster: 0 coins, -61 points, alive
chenhsi: 2 coins, -188 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 267 points, alive

Death [-384]

Fine, but only if you heal me.

james4l: 3 coins, 116 points, alive
evilhamster: 0 coins, -61 points, alive
chenhsi: 2 coins, -188 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 267 points, alive

Death [-384]

james4l: 3 coins, 116 points, alive
evilhamster: 0 coins, -61 points, alive
chenhsi: 2 coins, -188 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 150 points, alive

Death [-388]

James, actually, before we kill each other, lets kill evilhamster since he is inactive.
james4l: 3 coins, 116 points, alive
evilhamster: 0 coins, -159 points, alive
chenhsi: 2 coins, -188 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 150 points, alive

Death [-392]

Why? He's inactive.

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -159 points, alive
chenhsi: 2 coins, -188 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 150 points, alive

Death [-396]

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -159 points, alive
chenhsi: 2 coins, -288 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 150 points, alive

Death [-400]

I feel evil

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -159 points, alive
chenhsi: 2 coins, -288 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive

Death [-404]

SMB: remember that it is the post you are posting not the previous.

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -159 points, alive
chenhsi: 2 coins, -391 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive

Death [-412]

I could say the same to you sup3rcrash3r

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -263 points, alive
chenhsi: 2 coins, -391 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive

Death [-416]

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -263 points, alive
chenhsi: 0 coins, 0 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive
sup3rcrash3r: 3 coins, 0 points, alive

death:  -420

lol forgot to add myself.

Does killing someone count as zeroing?

Yeah, its when you don't have enough coins to revive. sup3rcrash3r, alliance?

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -263 points, alive
chenhsi: 0 coins, -106 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive
sup3rcrash3r: 3 coins, 0 points, alive

death: -424

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -263 points, alive
chenhsi: 0 coins, -213 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive
sup3rcrash3r: 3 coins, 0 points, alive

death: -428

sure...

But I just killed chenhsi once. Do I get a coin?

james4l: 3 coins, 215 points, alive
evilhamster: 0 coins, -263 points, alive
chenhsi: 0 coins, -321 points, alive
ernie: -1 coin, OUT, dead
cf249: -1 coin, OUT, dead
lotsofmath: 3 coins, -173 points, alive
SuperMathBoy: 3 coins, 252 points, alive
sup3rcrash3r: 3 coins, 0 points, alive

death: -432

No, read the first post to see when you get a coin.


A biologist has found that the number of branches on a certain rare tree in its first few years of life can be modeled by the polynomial $ b(y) \equal{} 4y^2\plus{} y$. The number of leaves on each branch can be modeled by the polynomial $ l(y) \equal{} 2y^3 \plus{} 3y^2 \plus{} y$, where $ y$ is the number of years after the tree reaches  height of $ 6$ feet. Write a polynomial describing the total number of leaves on the tree.

[quote="JRav"]In your first equation, do you mean $ 4y^{3}$?[/quote]
Sorry seems I made a major error I mean $ b(y)\equal{}4y^2\plus{}y$


[hide]Easy...but if not grasped properly may cause great confusion in future,

 5-2(3n-2)
=5-(6n-4)
=5-6n+4
=-6n+1

Did it in more steps to simplify it.[/hide]

I think this is an excellent problem to solve for this forum....
[hide]
5-6n+4 = 9-6n[/hide]

[quote="ying'n'yang"]

=5-6n+4
=-6n+1

[/quote]

Check your step. How come +5+4 got into 1?

And since this is discussion of distributive property, solve this: (from 2002 Cayley)

5x+2(4+x) can be simplifed to what?

[quote]5x+2(4+x) [/quote]

which is 5x + 8 + 2x

[quote]a(b - c(a - b)) = ab - ac - abc[/quote]

this might be alittle advanced....

a(b-c(a-b)) = a(b-ca+cb) = ab-a^2c+abc

ab-a^2c+abc  :!=: (doesn't equal sign)ab - a^2c - abc

[quote="Treething"]The main thing to remember that the negative sign distributes through the terms inside the parentheses.

Extension:

Prove or disprove this:
a(b - c(a - b)) = ab - a:^2:c - abc[/quote]

$a(b-c(a-b))=ab-ac-abc$
$a(b-ca+cb)=ab-ac-abc$
$ab-ca^2+abc=ab-ac-abc$
$-ca^2+2abc$ is not $-ac$

[quote="byrdgyrl"]Rewrite 5-2(3n-2) using the distributive property (get rid of the parenthesis.)  I know this is easy but I'm bad at thinking up hard questions... I also don't have an imagination like Treethings.[/quote]

eh... I just found this question, and I had the urge to do it because I always do these kinda things wrong on a tournament and screw up my score:
[hide]
5-6n-(-4)
5-6n+4
-6n+9[/hide]

[quote="Silverfalcon"]
5x+2(4+x) can be simplifed to what?[/quote]

doing random problems... I don't really know how to explain problems very well, so i"m just gonna do them and show my step and cross my fingers that someone'll understand me....

5x + 8 + 2x = 7x+8

k this is how i did it
let me give a full explanation
i mean TRY TO!
5-2(3n-2)
5-6n+4             +4 because when u time -2*-2 it will give u +4
-6n+9

[quote="jli"][hide]5-2(3n-2)
=5-6n+4
[b]=-6n+1 [/b][/hide][/quote]
[hide]To get your answer, the parentheses would have to be different:
5-(6n+4)
That is how it would have to be.  The answer a is 9-6n.[/hide]


Hey guys, i just wanted to clarify: what's the minimum value of $ 5x^2 \plus{} \frac{1}{x^4}$? do i break down $ 5x^2\equal{}4x^2\plus{}x^2$ and use AM-GM? or do i break it down in another way? thanks!

We have $ 5x^2\plus{}\frac{1}{x^4}\equal{}\frac{5}{2}x^2\plus{}\frac{5}{2}x^2\plus{}\frac{1}{x^4}\geq 3\sqrt[3]{\frac{25}{4}}$
If you break down $ 5x^2\equal{}x^2\plus{}4x^2$ and use AM-GM, you can't find the minimum value because $ x^2\neq 4x^2$

why does it need to be equal?

OH yeah, it's a condition before using AM-GM. thanks!:)

AM-GM says that AM >= GM, with equality only if the terms are equal. If the terms are never equal, you're not getting equality.


limit
function
integration
real analysis
real analysis unsolved

Need a hint for this well known problem:
\[ \lim_{p\rightarrow \infty}\|f\|_p=\|f\|_{\infty}  \]

especially for proving this direction; $\|f\|_\infty\leq \lim_{p\rightarrow \infty}\|f\|_p$

[quote="Kent Merryfield"]Find a 1-term simple function that is smaller than $|f|$ everywhere but almost as big as $\|f\|_{\infty}$ on a set of positive measure - the set doesn't have to be big.

Oh, and while it is quite clear what you meant, you'd be better off thinking of what you're trying to prove as $\|f\|_\infty\leq \liminf_{p\rightarrow \infty}\|f\|_p$[/quote]

If we can find such 1-term simple function, say $\alpha 1_A$, then $\alpha\leq |f|$. Meanwhile $f$ is not always bounded below by $\alpha>0$. Maybe I don't understand what you mean.

How if I do like this?
Let $\alpha\in(0,1)$, then clearly $\alpha\|f\|_\infty<\|f\|_\infty$. It follows for $A: =\{x\mid |f|>\alpha\|f\|_\infty\}$ we have $m(A)>0$. Then
\[ \|f\|_p^p\geq \int_A|f|^p\geq \alpha^p\|f\|_\infty^pm(A)  \]
or
\[ \|f\|_p\geq \alpha\|f\|_\infty m(A)^{1/p}  \]
Taking $\liminf_{p\rightarrow \infty}$ and $\alpha\rightarrow 1$ we obtain $\liminf_{p\rightarrow \infty}\|f\|_p\geq \|f\|_\infty$

That is precisely what I meant for you to do. I assume that you have already proved the other side, namely $\limsup_{p\to\infty}\|f\|_p\le\|f\|_{\infty}.$


How would you answer this question:

In the attachment there should be a photo of a figure. In that figure there would be a square were all the sides are equal to 1.

If the triangle CMN is a equilateral triangle what is the area of that triangle?

[hide]
Since $ ABCD$ is a square, $ BC \equal{} CD$, $ \angle B \equal{} \angle D \equal{} 90$, and $ CMN$ is equilateral, $ CM \equal{} CN$. Therefore $ CMD$ is congruent to $ CNB$, and so $ \angle BCN \equal{} \angle DCM$.

Connect $ AC$, $ \angle DCA \equal{} \angle BCA$, so $ \angle MCA \equal{} \angle NCA \equal{} \frac {1}{2}*60 \equal{} 30$. Thus, $ CA$ perpendicularly bisect $ MN$ at, say $ H$.

Let the side length of $ CMN$ equal $ x$, then $ AH \equal{} MH \equal{} x\sin30$, and $ CH \equal{} x\cos30$. Because $ AH \plus{} CH \equal{} AC \equal{} \sqrt2, x \equal{} \frac {\sqrt6 \minus{} \sqrt2}{2}$

And then the area is $ \frac {\sqrt3}{4}x^2 \equal{} \frac {2\sqrt3 \minus{} 3}{4}$

[/hide]


floor function
logarithms
algebra
polynomial
combinatorics open
combinatorics

Does anyone know of a relatively fast way to determine the largest root and exponent in a very large integer -- on the order of 1,000+ digits -- that does not exceed the value of the integer. For example: for the number 1234, 4^^5 is the largest root/exponent (1024).

So you want a way to find the largest $a^{b}$ such that $a^{b}\le n$, right?

Also, $4^{5}$ is the same as $2^{10}$; which one is considered "bigger"?

Is it given the root, find the largest exponent or given the exponent, find the largest root?

I am actually trying to determine BOTH the root and exponent that most closely approaches a given number, without exceeding that number.

Just to rephrase: we are searching for the largest perfect power less than a given integer.

The "obvious" method is to do the following: given the integer $n$, compute $k = \lfloor\log_{2}n\rfloor$.  Then compute $a_{i}= \lfloor n^{1/i}\rfloor^{i}$ for $2 \leq i \leq k$.  Then take the largest of the $a_{i}$.  Assuming you can efficiently implement logs and roots, this is efficient (polynomial in the number of digits of $n$).  It's also the case that you only really need to check the primes less than $k$, since every $pq$-th root is also a $p$-th root and a $q$-th root.  For numbers with 1000 binary digits, this requires you to know a list of the primes up to 1000.  For 1000 decimal digits, this requires you to know a list of the primes up to $\log_{2}10^{1000}\approx 3321$, of which there are 467.  So, if you have to do this many times, you can save yourself a factor of 7 in the running time if you're willing to put in the time to list the first few hundred primes.  (If you're going to have to do this for numbers of arbitrary large length, this is probably not worthwhile.  I don't actually know a running time for it, though.  It looks polynomial in $k$?  That actually good enough, I guess, but it doesn't make it clear whether the time savings are worth it.  Of course, if we really want the asymptotics (i.e., $n$ is going to $\infty$, not just restricted to numbers of 1000 or so digits) we should be able to decide if the tradeoff is worth it using the Prime Number Theorem.)

Thank you.  This all seems to make sense. Unfortunately, I'm not a mathematician, so some of finer points are lost on me. Can you give me an example of the process using a random integer of 100 decimal digits?  Thanks in advance.

Bill

No, but only because I don't want to write out a bunch of 100-digit numbers.  Here's an example with a 10-digit, not particularly random number:

Take $n = 1234567890$.  Then $\log_{2}n \approx 30.2$.  Thus $k = 30$.  The primes less than or equal to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.  Taking the appropriate roots of $n$ and rounding down gives, respectively, 35136, 1072, 65, 19, 6, 5, 3, 3, 2, 2.  Raising these to their appropriate powers gives 1234538496, 1231925248, 1160290625, 893871739, 362797056, 1220703125, 129140163, 1162261467, 8388608 and 536870912.  The largest of these is $1234538496 = 35136^{2}$, the largest perfect power at most 1234567890.

Excellent! I think I finally get it.  Thanks.

Bill


okay this is what it says...

white+water=picnic

each distinct letter represents a different digit.
find the number represented by the answer PICNIC

OKAY couple things plz dont call it hard
i know where that problem is from
im not COMPLETELY sure it is MATHCOUNTS related
bsides, your gettin the answer 2morrow...

hmmm... i got two solutions using http://bach.istc.kobe-u.ac.jp/cgi-bin/crypt/crypt.cgi?crypt=white+water+picnic =D
85642 + 83427 = 169069
83642 + 85427 = 169069

thats from purple comet, not mathcounts.
post it somewhere else.


Slove this equation: $ 2\sqrt{5x\minus{}1}\equal{}9x^2\minus{}5x$

[quote="thanhnam2902"]Slove this equation: $ 2\sqrt {5x \minus{} 1} \equal{} 9x^2 \minus{} 5x$[/quote]
[hide="AHH"]Squaring results in $ 20x\minus{}4\equal{}81x^{4}\minus{}90x^{3}\plus{}25x^{2}$
$ 81x^{4}\minus{}90x^{3}\plus{}25x^{2}\minus{}20x\plus{}4\equal{}0$
Using synthetic division...
$ (x\minus{}1)(81x^{3}\minus{}9x^{2}\plus{}16x\minus{}4)$.
Having tried RRT and synthetic a painful amount of times I have not been able to factor out the bigger polynomial D:
So I think 1 is the solution.[/hide]

Yes. But how we find $ x$ from $ 81x^3\minus{}9x^2\plus{}16x\minus{}4$ ?

$ 81x^3\minus{}9x^2\plus{}16x\minus{}4\equal{}(9x\minus{}2)(9x^2\plus{}x\plus{}2)\rightarrow x \equal{} \frac{2}{9}$

[quote="Brut3Forc3"]$ 81x^3 \minus{} 9x^2 \plus{} 16x \minus{} 4 \equal{} (9x \minus{} 2)(9x^2 \plus{} x \plus{} 2)\rightarrow x \equal{} \frac {2}{9}$[/quote]
O_O, nice I didn't think about fractional stuff =D

Oh! This problem is easy and nice, because Equation $ \Longleftrightarrow (1\plus{}\sqrt{5x\minus{}1})^2\equal{}(3x)^2$
$ \Longleftrightarrow 1\plus{}\sqrt{5x\minus{}1}\equal{}3x$  [size=150][color=darkblue]OR[/color][/size]  $ 1\plus{}\sqrt{5x\minus{}1}\equal{}\minus{}3x$

OK?


How will this Theorem effect math in any way?  I mean, it's nice to know there are no solutions for $x^{n}+y^{n}=z^{n}$, where $n \ge 3$. But it seams kind of pointless for Wiles to virtually kill himself over this problem, it didn't really change things that much. Or am I missing something :?:

Did you know that Fermat wrote he had found a "wonderous proof" to his theorem, but he didn't have enough room to write it down in the margin of a book were he wrote the famous(or infamous) equation?

Shouldn't this be in 'other topics' rather than MathCounts?

He connected elliptic curves and modular forms by proving the Taniyama-Shimura conjecture, which has serious implications for the study of both.  FLT was just a consequence of the theorem (essentially because a solution "could be" one, but not the other).

Strictly speaking, he didn't prove Taniyama-Shimura.

Perhaps there are not any great applications to FLT, but Wiles discovered several things of significant mathematical interest en route to his proof, which may or may not have applications (most likely in cryptography)

[quote="blahblahblah"]Strictly speaking, he didn't prove Taniyama-Shimura.[/quote]

In what sense didn't he?

Uhh, I think he only proved the semi-stable case, whatever that means.

Wiles did not actually prove Taniyama-Shimura. He only proved a special cases that is equivalent to Fermat's Last Theorem. The full Taniyama-Shimura was finally proven in 1999.

all of what I will say is from: [i]Fermat's Enigma[/i] (great book)
as said above, Wiles didn't prove Taniyama-Shimura... he proved special case... and to the pt. about Wile's pratically killing himself for the proof: it's like what they say about people climbing mt. everest; they do it because it is there.
Also, there about500 - 1000 mathematicians in the world who actually understand every part of Wiles' proof and the mathematics surrounding it.


Hello, is there any way to draw a sphere or an ellipsoid in Asymptote? Sorry, I'm new and just learning 3D.
Thank you!

With Asymptote 1.45:
[code]import three;
size(5cm,0);
draw(unitsphere, paleblue);
draw(shift(-Z)*scale(3,2,1)*unitsphere, lightgrey);[/code]

I mean [url=https://sourceforge.net/forum/forum.php?forum_id=878482]Asymptote version 1.46[/url].
You can use a [url=http://en.wikipedia.org/wiki/Aggregator]RSS agreggator[/url] to keep informed.
The addresses of Asymptote rss feeds are [url]https://sourceforge.net/export/rss2_project.php?group_id=120000[/url].

Wait...a new version comes out like every week?!?!?!

Every time I try to draw a 3D image, this message comes up and an animation window comes instead of an image with this message:
freeglut (asy): Unable to create direct context rendering for window ''
This may hurt performance.
freeglut (asy): Unable to create direct context rendering for window 'Asymptote [Double click right button for menu]'
This may hurt performance.
freeglut (asy): Unable to create direct context rendering for window 'freeglut menu'
This may hurt performance.
This happened when I installed 1.46. Can anyone help me with this?
Thank you!


[quote="mmrproblem"]Find the centroid of the perimeter of any convex quadrilateral.Is it intersection of diagonals[/quote]

Obviously not (check what happens when two vertices coincide).

In fact, I have never seen a reasonable characterization of the perimeter centroid for [i]generic[/i] quadrilaterals.

  Darij

[quote="mmrproblem"]Is it possible the centroid of the four vertices A,B,C,D is at the midpoint of either segment joining the midpoints of AB and CD  or AC and  BD?[/quote]

Yes, this is a trivial property of vectors. But you are speaking about the vertex centroid, not about the perimeter centroid here.

  Darij


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a}{2a^2\plus{}3b\plus{}2}\plus{}\frac{b}{2b^2\plus{}3c\plus{}2}\plus{}\frac{c}{2c^2\plus{}3a\plus{}2} \le \frac{3}{7}.\]
Equality holds if and only if $ a\equal{}b\equal{}c\equal{}1$.

P/s: I see this inequality is very nice and I would like to see a difference proof with mine. :)

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac {a}{2a^2 \plus{} 3b \plus{} 2} \plus{} \frac {b}{2b^2 \plus{} 3c \plus{} 2} \plus{} \frac {c}{2c^2 \plus{} 3a \plus{} 2} \le \frac {3}{7}.
\]
Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$.

P/s: I see this inequality is very nice and I would like to see a difference proof with mine. :)[/quote]
$ \frac {a}{2a^2 \plus{} 3b \plus{} 2} \plus{} \frac {b}{2b^2 \plus{} 3c \plus{} 2} \plus{} \frac {c}{2c^2 \plus{} 3a \plus{} 2}\le\frac {a}{4a \plus{} 3b} \plus{} \frac {b}{4b \plus{} 3c} \plus{} \frac {c}{4c \plus{} 3a}$
let $ x \equal{} \frac {b}{a},y \equal{} \frac {c}{b},z \equal{} \frac {a}{c}$,so it  comes to $ \frac {1}{4 \plus{} 3x} \plus{} \frac {1}{4 \plus{} 3y} \plus{} \frac {1}{4 \plus{} 3z} \le \frac {3}{7}$
 $ xyz \equal{} 1$

[quote="NextPeace"][quote="ye109"]
$ \frac {1}{4 \plus{} 3x} \plus{} \frac {1}{4 \plus{} 3y} \plus{} \frac {1}{4 \plus{} 3z} \le \frac {3}{7}$
 $ xyz \equal{} 1$[/quote]

But it's not true  :wink:  try x=y=10, z=1/100.[/quote]
i try it ,but it is right
$ \frac{1}{34}\plus{}\frac{1}{34}\plus{}\frac{1}{4\plus{}\frac{3}{100}}\minus{}\frac{3}{7}\equal{}\minus{}\frac{5832}{47957}$

$ \frac {1}{4 \plus{} 3x} \plus{} \frac {1}{4 \plus{} 3y} \plus{} \frac {1}{4 \plus{} 3z} \le \frac {3}{7}$
 $ xyz \equal{} 1$
suppose  $ x\leq\ y\le z$ so$ x\le1 z\ge1$$ g(x,y) \equal{} \frac {1}{4 \plus{} 3x} \plus{} \frac {1}{4 \plus{} 3y} \minus{} \frac {2}{4 \plus{} 3\sqrt {xy}} \equal{} \frac {3(\sqrt {x} \minus{} \sqrt {y})}{4 \plus{} 3\sqrt {xy}}(\frac {\sqrt {y}}{4 \plus{} 3y} \minus{} \frac {\sqrt {x}}{4 \plus{} 3x})$
$ if y\le\frac {4}{3},g(x,y)\le0$ so $ f(x,y,z)\leq\ f(\sqrt {xy},\sqrt {xy},z) \equal{} f(\sqrt {\frac {1}{z}},\sqrt {\frac {1}{z}},z)$
i$ f y\ge\frac {4}{3}$,fixd  y let x biger  and z smaller  ,$ f(x,y,z) \minus{} > f(\frac {xz}{y},y,y) \equal{} f(\frac {1}{y^2},y,y)$
and use unitary derivativeness

[quote="ye109"][quote="NextPeace"][quote="ye109"]
$ \frac {1}{4 \plus{} 3x} \plus{} \frac {1}{4 \plus{} 3y} \plus{} \frac {1}{4 \plus{} 3z} \le \frac {3}{7}$
 $ xyz \equal{} 1$[/quote]

But it's not true  :wink:  try x=y=10, z=1/100.[/quote]
i try it ,but it is right
$ \frac {1}{34} \plus{} \frac {1}{34} \plus{} \frac {1}{4 \plus{} \frac {3}{100}} \minus{} \frac {3}{7} \equal{} \minus{} \frac {5832}{47957}$[/quote]

Sorry,  :blush: I have a typo. try x=y=1/10, z=100, we have
$ \frac {1}{4\plus{}\frac{3}{10}} \plus{} \frac {1}{4\plus{}\frac{3}{10}} \plus{} \frac {1}{304} \minus{} \frac {3}{7} >0$.

yes, you are right .i got it wrong  :wink:

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac {a}{2a^2 \plus{} 3b \plus{} 2} \plus{} \frac {b}{2b^2 \plus{} 3c \plus{} 2} \plus{} \frac {c}{2c^2 \plus{} 3a \plus{} 2} \le \frac {3}{7}.
\]
Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$.

P/s: I see this inequality is very nice and I would like to see a difference proof with mine. :)[/quote]
To prove this nice inequality, we need the following one variable inequality (with no equality point)
\[ \frac {1}{2\sqrt {2(3x \plus{} 2)}} \plus{} \frac {2x^2 \plus{} 7x \plus{} 2}{26x^2 \minus{} 9x \plus{} 26} < \frac {3}{7}\quad \forall x > 0.
\]
;) :)

Aqardy, VasC, where are you? :( I think you both will have nice solution for this. :)


Find $x;y;z;t \in R$ such that:
${\frac{xy+yz+xz}{xyz}=\frac{11}{2}}$
${\frac{yz+zt+ty}{yzt}=\frac{8}{3}}$
${\frac{xt+zt+tz}{xzt}=\frac{5}{4}}$
${\frac{xy+yt+zy}{xyt}=\frac{6}{5}}$

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!


calculus
integration
calculus computations

Just make the substitution $y=\frac{1}{x}$.

[quote="afcweswarrior"]what do you mean, substitute it for what[/quote]if $y=\frac{1}{x}$, then $dy=-\frac{1}{x^{2}}dx$

$I = \int^{2}_{1}\frac{e^{1/x}}{x^{2}}$.

A substitution is not needed if we note that $\left(\frac{1}{x}\right)' =-\frac{1}{x^{2}}$. So,

[hide="Answer"]$I = \left[-e^{1/x}\right]^{2}_{1}= e-\sqrt{e}$.[/hide]


[quote="quark1005"]What strategy should be used when solving tiling problems, such as

[i]In how many ways can an area 3*4 be covered in tiles of dimensions (1*2)?[/i]

How should such problems be tackled? There are of course much harder ones...[/i][/quote]

http://en.wikipedia.org/wiki/Domino_tiling

http://portal.acm.org/citation.cfm?id=1030333.1030336


Prove that for all graphs of $ n$ vertices such that every pair of distinct vertices has exactly 2 distinct path connecting them, the graph has $ n$ edges.
e.g.

the graph
[code]
o---o
| /
|/
o
[/code]
has 2 distinct paths connecting every pair of distinct vertices.

Note: unknown difficulty. May be very trivial or very difficult. probably the former.

Well, no vertex may exist in multiple cycles.

BTW

[code]
A../D
|\C
|/.\E
B

[/code]

From D to E, do DCBACE and DCABCE count?

@james41: Paths can't go through the same vertex twice.

@Problem: The graph obviously can't contain v-1 edges, or else it would have no cycles. Assume that it has at least v+1 edges. Then it is a tree of v vertices plus two edges. Note that a graph has one cycle iff it has v edges (remove one edge from the cycle and it is a tree, add an edge to a tree and it is a cycle), and a graph has no cycles iff it has v-1 edges. Therefore our graph with v+1 edges has at least two cycles. Let these cycles be C_1, C_2, and let them be connected by path P (P is unique or else there would be more cycles). Consider point A on C_1 and B on C_2, where neither A nor B are on P. Then A can reach P in 2 ways, and B can reach P in 2 ways, so A can reach B in 4 ways, a contradiction. Therefore the graph cannot have more than v edges, and it cannot have v-1 edges, so it has exactly v edges.


geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
projective geometry
geometry solved

Given a circle $(O)$ (with center $O$) and an arbitrary line $(l)$. Let $H$ be the foot of the perpendicular from the point $O$ to the line $(l)$. Let $M$ be a fixed point on the circle $(O)$. Let $A$ and $B$ be arbitrary points on the line $(l)$ such that the point $H$ is the midpoint of the segment $AB$. Let the lines $MA$ and $MB$ intersect the circle $(O)$ at the points $P$ and $Q$, respectively (apart from $M$). Prove that the line $PQ$ passes through a fixed point when the points $A$ and $B$ wander along the line $(l)$.

Let MH meet (O) again at N. Let NB meet (O) again at R. Since $H \equiv AB \cap MN,\ AH = BH$ and $AB \perp OH,$ by the butterfly theorem, the diagonal intersection $MN \cap PR$ of the cyclic quadrilateral MPNR is H. Let the circle tangent NN at N meet PQ at X. Consider the 6 concyclic points $M, P, N, N, Q, R$ (N is a tangency point, i.e., a double point). By Pascal's theorem, the intersections $X \equiv NN \cap PQ,$ $B \equiv NR \cap MQ,$ and $H \equiv MN \cap PR$ are collinear, i.e., $X \in BH \equiv (l).$ But the tangent NN and the line (l) are fixed, hence their intersection $X \equiv NN \cap PQ \cap (l)$ is also fixed.

Comment: If the line (l) intersects the circle (O), the cyclic quadrilateral MPNR is convex, otherwise it is reflex.

See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108346]Interesting problem[/url] for the special case $H \equiv O.$

I think it's also doable by a simple pespective transformation. The (projective) transformation, mapping $A$ into $P$ (from $l$ to $(O)$) with the center in $M$ also maps $B$ into $Q$. Similiar, the transformation with the center in $H$, which maps $A$ into $B$, is a projective transformation. From that we conclude that $P \mapsto Q$ ($(O) \mapsto (O)$) is a perspectivity, as $MH \cap (O)$ maps into itself. So $PQ$ goes through a fixed point (not hard to determine).

[quote="bui_haiha2000"]Given a circle $ (O)$ (with center $ O$) and an arbitrary line $ (l)$. Let $ H$ be the foot of the perpendicular from the point $ O$ to the line $ (l)$. Let $ M$ be a fixed point on the circle $ (O)$. Let $ A$ and $ B$ be arbitrary points on the line $ (l)$ such that the point $ H$ is the midpoint of the segment $ AB$. Let the lines $ MA$ and $ MB$ intersect the circle $ (O)$ at the points $ P$ and $ Q$, respectively (apart from $ M$). Prove that the line $ PQ$ passes through a fixed point when the points $ A$ and $ B$ wander along the line $ (l)$.[/quote]

Similar to Yetti's solution, but with a more direct characterization of the fixed point:

Let the lines PQ and (l) intersect at a point X. We want to show that this point X is fixed.

Let the tangent to the circle (O) at M intersect (l) at a point Y. Consider our configuration from the following viewpoint:

We have a triangle MQP with the circumcenter O. The line (l) lies in the plane, and the point H is the orthogonal projection of the point O on this line (l). The line (l) intersects the lines QP, PM, MQ at the points X, A, B. The point Y is the point of intersection of the line (l) with the tangent to the circumcircle of triangle MQP at the point M.

Hence, after Theorem 10 in the note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i], the point H is the midpoint of the segment AB if and only if the point H is the midpoint of the segment XY.

But we know that H is the midpoint of AB. Thus, it follows that the point H is the midpoint of XY. Hence, the point X is the reflection of Y in the point H. Since both points Y and H are fixed (by their construction - they don't depend on A and B), we thus conclude that X is fixed, qed.

  Darij