geometry
IMO Shortlist
reflection
trigonometry
circumcircle
Isogonal conjugate

I am going to give two solutions, but first I rewrite the problem.

In fact, the equation < NAC = < MAB shows that the line AN is the isogonal of the line AM wrt the angle CAB. Thus, the point N lies on the isogonal of the line AM wrt the angle CAB. Similarly, < NBC = < MBA shows that the point N lies on the isogonal of the line BM wrt the angle ABC. Hence, the point N is the point of intersection of the isogonals of the lines AM and BM wrt the angles CAB and ABC. But it is known that the isogonals of the lines AM and BM wrt the angles CAB and ABC intersect at the isogonal conjugate of the point M wrt triangle ABC. Hence, the point N is the isogonal conjugate of the point M wrt triangle ABC. Therefore, solving the problem is equivalent to proving the following theorem:

[color=blue][b]Theorem 1.[/b] Let M be a point in the plane of a triangle ABC, and let N be the isogonal conjugate of the point M wrt triangle ABC. Then,

$\frac{AM \cdot AN}{AB \cdot AC}+\frac{BM \cdot BN}{BC \cdot BA}+\frac{CM \cdot CN}{CA \cdot CB}= 1$.[/color]

[i]First proof of Theorem 1.[/i] Since N is the isogonal conjugate of the point M wrt triangle ABC, the line AN is the isogonal of the line AM wrt the angle CAB. Hence, < NAB = < CAM (in fact, with directed angles modulo 180°, we would have < (AB; AN) = - < (CA; AM), but we don't use directed angles here).

We will use the notation $\left[P_{1}P_{2}...P_{n}\right]$ for the (non-directed) area of an arbitrary polygon $P_{1}P_{2}...P_{n}$.

Let X', Y', Z' be the reflections of the point M in the lines BC, CA, AB. Since Z' is the reflection of M in the line AB, we have AZ' = AM, < BAZ' = < BAM and [AZ'B] = [AMB]. Thus, < NAZ' = < NAB + < BAZ' = < CAM + < BAM = < CAB.

Since the area of a triangle equals $\frac12$ times the product of two of its sides times the sine of the angle between them, we have $\left[ABC\right]=\frac12\cdot AB\cdot AC\cdot\sin\measuredangle CAB$ and $\left[NAZ^{\prime}\right]=\frac12\cdot AZ^{\prime}\cdot AN\cdot\sin\measuredangle NAZ^{\prime}$. Hence,

$\frac{\left[NAZ^{\prime}\right]}{\left[ABC\right]}=\frac{\frac12\cdot AZ^{\prime}\cdot AN\cdot\sin\measuredangle NAZ^{\prime}}{\frac12\cdot AB\cdot AC\cdot\sin\measuredangle CAB}=\frac{\frac12\cdot AM\cdot AN\cdot\sin\measuredangle CAB}{\frac12\cdot AB\cdot AC\cdot\sin\measuredangle CAB}=\frac{AM\cdot AN}{AB\cdot AC}$.

Similarly, $\frac{\left[NBZ^{\prime}\right]}{\left[ABC\right]}=\frac{BM\cdot BN}{BC\cdot BA}$. Thus,

$\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}=\frac{\left[NAZ^{\prime}\right]}{\left[ABC\right]}+\frac{\left[NBZ^{\prime}\right]}{\left[ABC\right]}=\frac{\left[NAZ^{\prime}\right]+\left[NBZ^{\prime}\right]}{\left[ABC\right]}$
$=\frac{\left[NAZ^{\prime}B\right]}{\left[ABC\right]}=\frac{\left[AZ^{\prime}B\right]+\left[ANB\right]}{\left[ABC\right]}$,

what, using [AZ'B] = [AMB], becomes

$\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}=\frac{\left[AMB\right]+\left[ANB\right]}{\left[ABC\right]}$.

Similarly,

$\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$;
$\frac{CM\cdot CN}{CA\cdot CB}+\frac{AM\cdot AN}{AB\cdot AC}=\frac{\left[CMA\right]+\left[CNA\right]}{\left[ABC\right]}$.

Summing up these three equations, we obtain

$2\cdot\left(\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}\right)$
$=\frac{\left[BMC\right]+\left[BNC\right]+\left[CMA\right]+\left[CNA\right]+\left[AMB\right]+\left[ANB\right]}{\left[ABC\right]}$.

Thus,

$\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}$
$=\frac{\left[BMC\right]+\left[BNC\right]+\left[CMA\right]+\left[CNA\right]+\left[AMB\right]+\left[ANB\right]}{2\cdot\left[ABC\right]}$
$=\frac{\left(\left[BMC\right]+\left[CMA\right]+\left[AMB\right]\right)+\left(\left[BNC\right]+\left[CNA\right]+\left[ANB\right]\right)}{2\cdot\left[ABC\right]}$
$=\frac{\left[ABC\right]+\left[ABC\right]}{2\cdot\left[ABC\right]}=1$,

and Theorem 1 is proven.

[i]Second proof of Theorem 1.[/i] First remember the easy fact that if a quadrilateral has perpendicular diagonals, then its area equals $\frac12$ of the product of the two diagonals.

In the same way as in the First proof of Theorem 1, we show that the point N is the isogonal conjugate of the point M wrt triangle ABC. Hence, by a well-known property of isogonal conjugates (Theorem 6 in [b][1][/b]), it follows that if X, Y, Z are the orthogonal projections of the point M on the lines BC, CA, AB, then $AN\perp YZ$, $BN\perp ZX$ and $CN\perp XY$.

Hence, the quadrilaterals AYNZ, BZNX and CXNY have perpendicular diagonals, and thus, their areas are

$\left[ AYNZ\right] =\frac12 \cdot AN\cdot YZ$;
$\left[ BZNX\right] =\frac12 \cdot BN\cdot ZX$;
$\left[ CXNY\right] =\frac12 \cdot CN\cdot XY$.

Hereby, [F] denotes the area of a figure F.

Since < AYM = 90° and < AZM = 90°, the points Y and Z lie on the circle with diameter AM; thus, the segment AM is a diameter of the circumcircle of the triangle AYZ. Hence, by the Extended Law of Sines, $YZ = AM \cdot \sin \measuredangle YAZ = AM \cdot \sin A$. Hence,

$\left[ AYNZ\right] =\frac12 \cdot AN\cdot YZ = \frac12 \cdot AN \cdot AM \cdot \sin A$.

But if S is the area of triangle ABC, then $S = \frac12 \cdot AB \cdot AC \cdot \sin A$. Thus,

$\frac{AM\cdot AN}{AB\cdot AC}= \frac{\frac12 \cdot AN\cdot AM \cdot \sin A}{\frac12 \cdot AB \cdot AC \cdot \sin A}= \frac{\left[ AYNZ\right] }{S}$.

Similarly,

$\frac{BM\cdot BN}{BC\cdot BA}= \frac{\left[ BZNX\right] }{S}$;
$\frac{CM\cdot CN}{CA\cdot CB}= \frac{\left[ CXNY\right] }{S}$.

Altogether,

$\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}= \frac{\left[ AYNZ\right] }{S}+\frac{\left[ BZNX\right] }{S}+\frac{\left[ CXNY\right] }{S}$
$= \frac{\left[ AYNZ\right]+\left[ BZNX\right]+\left[ CXNY\right] }{S}= \frac{\left[ABC\right]}{S}= \frac{S}{S}= 1$,

what proves Theorem 1 again.

[b]References[/b]

[b][1][/b] Darij Grinberg, [i]Isogonal conjugation with respect to a triangle[/i], avaliable at http://de.geocities.com/darij_grinberg/ and http://www.mathlinks.ro/Forum/viewtopic.php?t=18472.

  Darij

[quote="orl"]Let $ M$ and $ N$ be points inside triangle $ ABC$ such that
\[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC.
\]
Prove that
\[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1.
\]
[/quote]
Sorry to revive an old topic, but here is another solution:
[hide="Alternate Solution"]
It is quite well-known that $ \angle MCA \equal{} \angle NCB$, which can be proven by taking the sine version of Ceva's Theorem on $ M$ and $ N$ on dividing (I won't do that here.)  Consider points $ P$, $ Q$, and $ R$ on the extensions of $ AM$, $ BM$, and $ CM$ respectively such that $ \triangle ANB\sim \triangle ACP$, $ \triangle BAQ\sim \triangle BNC$, and $ \triangle CNA\sim \triangle CBR$.  Now, since $ \angle MPC \equal{} \angle APC \equal{} \angle ABN \equal{} \angle CBM$, $ CPBM$ is cyclic, meaning that $ P$ is where $ AM$ meets the circumcircle of $ \triangle CMB$.  Similarly, $ Q$ is where $ BM$ meets the circumcircle of $ \triangle ACM$ and $ R$ is where $ CM$ meets the circumcircle of $ \triangle ANB$.  Now, by http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=851551707&t=191082, we have that $ \frac {BM}{BQ} \plus{} \frac {AM}{AP} \plus{} \frac {CM}{CR} \equal{} 1$.  Since \[ \triangle ANB\sim \triangle ACP\implies \frac {AB}{AN} \equal{} \frac {AP}{AC}\implies \frac {AB\cdot AC}{AN} \equal{} AP\implies \frac {AM}{AP} \equal{} \frac {AM\cdot AN}{AB\cdot AC}\] Similarly, we have that $ \frac {BM}{BP} \equal{} \frac {BM\cdot BN}{AB\cdot BC}$ and $ \frac {CM}{CP} \equal{} \frac {CM\cdot CN}{AC\cdot BC}$.  Hence,\[ \frac {BM\cdot BN}{AB\cdot BC} \plus{} \frac {CM\cdot CN}{AC\cdot BC} \plus{} \frac {AM\cdot AN}{AB\cdot AC} \equal{} \frac {AM}{AP} \plus{} \frac {BM}{BP} \plus{} \frac {CN}{CP} \equal{} 1\]
as desired. [/hide]

[b]Lemma:[/b] In $\triangle ABC$, let $a = BC$, $b = CA$, and $c = AB$, and let $P$ be a point in the interior of $\triangle ABC$ with barycentric coordinates $(p,q,r)$ (with $p$ corresponding to $A$, $q$ corresponding to $B$, and $r$ corresponding to $C$.) Let $U$ and $V$ be points such that $\angle UAV = \pi - \angle A$, $AU = rb$ and $AV = qc$. Then $AP (p+q+r) = UV = \sqrt{(br)^2 + (cq)^2 + (rq)(b^2 + c^2 - a^2)}$.

[i]Proof:[/i] Let $AP$ hit $BC$ at $D$. By Stewart's theorem, $a \cdot AD^2 + a \cdot BD \cdot DC = (AB)^2(DC) + (AC)^2 (BD)$. $BD = \frac{ar}{q+r}$ and $DC = \frac{aq}{q+r}$, so $AD^2 a + \frac{a^3 qr}{(q+r)^2} = \frac{ac^2 q + ab^2 r}{q+r}$, so $AD^2 = \frac{c^2 q + b^2 r}{q+r} - \frac{a^2 qr}{(q+r)^2} = \frac{(br)^2 + (cq)^2 + rq(b^2 + c^2 - a^2)}{(q+r)^2}$.

By the law of cosines, $b^2 + c^2 - a^2 = 2 bc \cos A$. Hence,
$\begin{align*}
(br)^2 + (cq)^2 + (rq)(b^2 + c^2 - a^2)
&= (br)^2 + (cq)^2 + 2(rq)(bc) \cos A \\
&= (br)^2 + (cq)^2 - 2(br)(cq) \cos(\pi - A),$
which by the law of cosines is equal to $UV^2$. Therefore, $AD(q+r) = UV$.

From properties of barycentric coordinates, $\frac{AP}{PD} = \frac{q+r}{p}$, so $\frac{PD}{AP} = \frac{p}{q+r}$, so $\frac{AD}{AP} = 1 + \frac{PD}{AP} = \frac{p+q+r}{q+r}$, so $AP = \frac{AD(q+r)}{p+q+r} = \frac{UV}{p+q+r}$, as desired.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

In $\triangle ABC$, let $a = BC$, $b = CA$, and $c = AB$, and let the barycentric coordinates of $M$ be $(p,q,r)$. The trilinear coordinates of $M$ are thus $(\frac{p}{a}, \frac{q}{b}, \frac{r}{c})$. The conditions of the problem imply that $N$ is the isogonal conjugate of $M$,  so the trilinear coordinates of $N$ must be $(\frac{a}{p}, \frac{b}{p}, \frac{c}{r})$. It follows that the barycentric coordinates of $N$ are $(\frac{a^2}{p}, \frac{b^2}{q}, \frac{r^2}{q})$.

Let $X$ and $Y$ be points in space such that $\angle XAY = \pi - \angle A$.  Pick $B_1$ and $C_1$ on rays $\overrightarrow{AX}$ and $\overrightarrow{AY}$, respectively, so that $AB_1 = rb$ and $AC_1 = qc$. Pick $B_2$ and $C_2$ on rays $\overrightarrow{AX}$ and $\overrightarrow{AY}$, respectively, so that $AB_2 = \frac{c^2}{r} b$ and $AC_2 = \frac{b^2}{q} c$. Note that $\frac{AC_2}{AB_2} = \frac{\frac{b^2}{q} c}{\frac{c^2}{r} b} = \frac{br}{cr} = \frac{AB_1}{AC_1}$, so $\triangle AB_1 C_1 \sim \triangle A C_2 B_2$. Hence, $\frac{B_1 C_1}{AB_1} = \frac{B_2 C_2}{AC_2}$, so ${(B_1 C_1)(B_2 C_2) = (B_1 C_1)^2 \frac{AC_2}{AB_1} = (B_1 C_1)^2 \frac{\frac{b^2}{q}c}{rb}} = (B_1 C_1)^2 \frac{bc}{qr}$.

By our lemma, we find that
\begin{align*}
\frac{AM \cdot AN}{AB \cdot AC}
&= \frac{(B_1 C_1)(B_2 C_2)}{(p + q + r)\left(\frac{a^2}{p} + \frac{b^2}{q} + \frac{c^2}{r}\right) \cdot bc} \\
&= \frac{(B_1 C_1)^2 \frac{bc}{qr}}{(p + q + r)\left(\frac{a^2}{p} + \frac{b^2}{q} + \frac{c^2}{r}\right) \cdot bc} \\
&= \frac{p(br)^2 + p(cq)^2 + pqr(b^2 + c^2 - a^2)}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)}.
\end{align*}

Hence,
\begin{align*}
\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB}
&= \sum_{cyc} \frac{p(br)^2 + p(cq)^2 + pqr(b^2 + c^2 - a^2)}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \\
&= \frac{1}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \left( \sum_{cyc} b^2 p^2 r + \sum_{cyc} c^2 p q^2 + pqr \sum_{cyc} (b^2 + c^2 - a^2) \right) \\
&= \frac{1}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \left( \sum_{cyc} b^2 p^2 r + \sum_{cyc} b^2 p^2 q + pqr(a^2 + b^2 + c^2) \right) \\
&= \frac{(a^2 qr + b^2 pr + c^2 pq)(p + q + r)}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \\
&= 1,
\end{align*}
as desired.

[hide="Solution"]Define $X$ to be a point such that $\triangle BAM$ is directly similar to $\triangle BXC$. Since $\angle XBC=\angle ABM=\angle NBC$, we have that $X, N, B$ are collinear.

According to a property of spiral similarity, we also have that $\triangle BMC\sim \triangle BAX$.

From these similarities, we have that $\frac{MA}{BM}=\frac{CX}{BC}\implies CX=\frac{MA\cdot BC}{BM}$ and that $\frac{MC}{BM}=\frac{AX}{AB}\implies AX=\frac{AB\cdot MC}{BM}$.

Furthermore, since $\angle NAC=\angle BAM=\angle NXC$, quadrilateral $AXCN$ is cyclic.

Now, applying Ptolemy's Theorem on quadrilateral $AXCN$ gives $AN\cdot CX+AX\cdot CN=AC\cdot NX$.

Substituting in values for $AX$ and $CX$ above and noting that $NX=BX-BN$, we have that $\frac{AM\cdot AN\cdot BC}{BM}+\frac{CM\cdot CN\cdot AB}{BM}=AC(BX-BN)$.

Now, multiplying both sides of the above expression by $\frac{BM}{AB\cdot BC\cdot CA}$ and noting that $BM\cdot BX=BA\cdot BC$ gives the desired result. We are done. $\blacksquare$ [/hide]

Just a small note: since $M,N$ are clearly isogonal conjugates,
\[\frac{(m-a)/(b-a)}{(c-a)/(n-a)}\in\mathbb{R},\]which is thus equal to its magnitude. The result quickly follows. The identity involved here also shows by the triangle inequality that
\[\sum\frac{AM\cdot AN}{AB\cdot AC}\ge1\]for arbitrary points $M,N$ in the plane, with equality (I think?) iff $M,N$ are isogonal conjugates.

[quote=sayantanchakraborty]This problem can be killed of without the use of isogonal conjugates(expect the fact that if three cevians are concurrent,their isogonal lines are also concurrent).Just apply sine rule in $\triangle{AMB},\triangle{BNC},\triangle{CNA},\triangle{BMC}$to get the ratios $\frac{AM}{AB},\frac{AN}{AC},\frac{BM}{AB},\frac{BN}{BC},\frac{CN}{AC},\frac{CM}{BC}$ in terms of the angles and noting that $\angle{BCM}=\angle{ACN}$ and the result will follow.

[hide]One might say it is bashing but sometimes bashing helps when the calculations are not too long and boring,as is the case with this problem.Anyway I think that Darij's method is the best one.[/hide][/quote]

Could anyone show the complete steps for this solution? I don't see how the result follows once we get everything in terms of the sines those angles.

[quote=1998 G4]Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that
\[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC.
\]
Prove that
\[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1.
\][/quote]

Funny solution :)

Note that $M$ and $N$ are isogonal conjugates in $\triangle ABC$.

[b]Lemma.[/b] Let line $AM$ meet $\odot(BMC)$ again at $M_A$. Then $\triangle AM_AB \sim \triangle ACN$.

[i]Proof.[/i] Clearly, $\angle BAM_A=\angle NAC$ and $\angle AM_AB=\angle MM_AB=\angle MCB=\angle ACN$ proving the similarity. $\blacksquare$

Thus, $\sum_{\text{cyc}} \frac{AM \cdot AN}{AB \cdot AC}=\sum_{\text{cyc}} \frac{AM}{AM_A}$. Now we apply inversion at $M$, to convert this into the equivalent problem.

[quote=inverted problem]
Point $M$ lies in the interior of triangle $ABC$ and lines $AM, BM, CM$, meet sides $BC, CA, AB$ at $M_A, M_B, M_C$, respectively. Then show $\sum_{\text{cyc}} \frac{MM_A}{AM_A}=1$.
[/quote]

However, this is obvious: if $M$ has barycentric coordinates $(x:y:z)$ then we are just summing $\frac{x}{x+y+z}$ and the conclusion follows. $\blacksquare$

Let $[\cdot]$ denote area. Also, let $\triangle DEF$ be the pedal triangle of $M$ with respect to $\triangle ABC.$

[b][color=red]Claim:[/color][/b] $\frac{AM\cdot AN}{AB\cdot AC}=\frac{[AENF]}{[ABC]}$
[i]Proof.[/i] Notice $\overline{EF}\perp\overline{AN}$ as $$\angle NAE+\angle AEF=\angle FAM+\angle AMF=90.$$ By LoS, $EF=AM\cdot\sin\angle BAC,$ and we also know $[ABC]=\tfrac{1}{2}\cdot AB\cdot AC\cdot\sin\angle BAC.$ Thus, $$2[AENF]=AN\cdot EF=AM\cdot AN\cdot\sin\angle BAC=\frac{2[ABC]\cdot AM\cdot AN}{AB\cdot AC}.$$ $\blacksquare$

$$\frac{AM\cdot AN}{AB\cdot AC} + \frac{BM\cdot BN}{BA\cdot BC} + \frac{CM\cdot CN}{CA\cdot CB}=\frac{[AENF]+[BDNF]+[CDNE]}{[ABC]}=1.$$ $\square$

Let ray $AN$ meet $(BNC)$ again at $A_1$, ray $BN$ meet $(CNA)$ again at $B_1$, and ray $CN$ meet $(ANB)$ again at $C_1$. Clearly, $M$ and $N$ are isogonal conjugates wrt $ABC$.

Observe that $$\angle BAM = \angle NAC = \angle A_1AC$$ and $$\angle ABM = \angle NBC = \angle NA_1C = \angle AA_1C$$ yielding $ABM \sim AA_1C$. Thus, $$\frac{AM \cdot AN}{AB \cdot AC} = \frac{AC}{AA_1} \cdot \frac{AN}{AC} = \frac{AN}{AA_1}.$$ Analogous processes give $$\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = \frac{AN}{AA_1} + \frac{BN}{BB_1} + \frac{CN}{CC_1}$$ so we're done by [url=https://artofproblemsolving.com/community/c618937h1648194p17497791]BAMO 2008/6[/url], which is also [i]EGMO 8.26[/i]. $\blacksquare$

[b]Remarks:[/b] The sole purpose for constructing $A_1$, $B_1$, and $C_1$ is to get rid of $M$ entirely.

Also, [url=https://artofproblemsolving.com/community/c618937h1622635_problem_318]EGMO 3.18[/url] is the inverted version of [i]EGMO 8.26[/i].

Let $BN$ intersect $(ANC)$ again at $P$. Note that $\angle BPC=\angle NAC=\angle BAM$ and $\angle ABM=\angle NBC$ implying that $\triangle BAM\sim \triangle BPC$. Similarly, $\triangle BCM\sim \triangle BPA$.
\\$~$\\
Thus, we have
\begin{align*}
&~~~~\frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB}\\
&= \frac{AM}{AB}\cdot \frac{AN}{AC}+\frac{BM}{BA}\cdot \frac{BN}{BC}\cdot \frac{CM}{CB}\cdot \frac{CN}{CA}\\
&= \frac{PC}{PB}\cdot \frac{AN}{AC}+\frac{BC}{BP}\cdot \frac{BN}{BC}+\frac{AP}{BP}\cdot \frac{CN}{CA}\\
&= \frac{PC\cdot AN + BN\cdot AC+AP\cdot CN}{BP\cdot AC} \\
&= \frac{AC\cdot PN + AC\cdot BN}{BP\cdot AC} = 1
\end{align*}
as desired.

Employ barycentric coordinates on $\triangle ABC$. Note that $M$ and $N$ are isogonal conjugates, so letting $M=(p:q:r)$ gives $N=\left(\frac{a^2}{p}:\frac{b^2}{q}:\frac{c^2}{r}\right)$.

Applying distance formula and simplifying, we have $$\frac{AM\cdot AN}{AB\cdot AC} = \frac{-a^2p q r + b^2 p q r + b^2p r^2 + c^2 p q^2 + c^2 p q r}{a^2 p q r + a^2 q^2 r + a^2 q r^2 + b^2 p^2 r + b^2 p q r + b^2 p r^2 + c^2 p^2 q + c^2 p q^2 + c^2 p q r}.$$ From here we can see that we will add back the $a^2pqr$ twice, and similar with the other variables, so the numerator will contain $a^2pqr+b^2pqr+c^2pqr$. Then, by symmetry, all the terms similar to $b^2pr^2$ will be counted once in our sum, so the numerator is exactly the denominator, which means that the LHS is equal to $1$, as desired.


If the volume of a sphere inscribed in a cube is $ \frac{\pi}{6}$ cubic inches what is the number of cubic inches in the volume of the cube?

The diameter of the sphere is equal to the length of the cube. If the volume of the sphere is $ \frac{\pi}{6}$, then its radius is:

$ \frac{\pi}{6} \div \frac{4}{3} \equal{} \frac{\pi}{3}$

$ \sqrt[3]{\frac{\pi}{3}} \times 2 \equal{} \text{length}$

Cube that, and you get $ \boxed{\frac{8\pi}{3}}$.

The formula for a volume of a sphere is $ \frac {4}{3}\pi{r^3} \equal{} V$

$ \frac {4}{3}\pi{r^3} \equal{} \frac {\pi}{6}$
$ 8\pi{r^3} \equal{} {\pi}$
$ r^3 \equal{} \frac {1}{8}$
$ r \equal{} \frac {1}{2}$

If we picture the sphere, we can see that the diameter is the same length as a side of the cube.

$ d \equal{} 2r$
$ d \equal{} 1$

If $ 1$ is the side of the cube, then the volume of the cube is $ 1^3 \equal{} 1$

[quote="isabella2296"]The diameter of the sphere is equal to the length of the cube. If the volume of the sphere is $ \frac {\pi}{6}$, then its radius is:

$ \frac {\pi}{6} \div \frac {4}{3} \equal{} \frac {\pi}{3}$

$ \sqrt [3]{\frac {\pi}{3}} \times 2 \equal{} \text{length}$

Cube that, and you get $ \boxed{\frac {8\pi}{3}}$.[/quote]

Here: $ \frac {\pi}{6} \div \frac {4}{3} \equal{} \frac {\pi}{3}$
You should divide by $ \frac{4\pi}{3}$
And you should get $ \frac{1}{8}$


probability
MATHCOUNTS
number theory
relatively prime
AMC

A coin that comes up heads with probability $ p > 0$ and tails with probability $ 1\minus{}p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $ \frac{1}{25}$ of the probability of five heads and three tails. Let $ p \equal{} \frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.

[hide="solution"] From the given information, we have $ \binom{8}{3}p^3(1\minus{}p)^5\equal{}\frac{1}{25}\binom{8}{5}p^5(1\minus{}p)^3$ or $ 1\minus{}p\equal{}\frac{1\cdot{p}}{5}$. So, we find $ p\equal{}\frac{5}{6}$ and our answer is $ \boxed{11}$.[/hide]

[hide="Answer"]
So from the problem the given is that $ p^3(1 \minus{} p)^5 \equal{} p^5(1 \minus{} p)^3/25$ so $ (1 \minus{} p)^2 \equal{} p^2/25$; $ (1 \minus{} p)/p \equal{} 1/5$ after square rooting so clearly $ p \equal{} 5/6\implies \boxed{11}$.[/hide]

Wow this was a mathcounts countdown level

bleargh, i missed it too. :P i had 8C3/p^3(1-p)^5 and stuff so i got 1/6...

Quick!
Just form an equation.
$$\binom{8}{3}p^3(1 - p)^5 = \frac{1}{25}\binom{8}{5}p^5(1 - p)^3$$
Canceling out some exponents yields:
$$\binom{8}{3}(1 - p)^2 = \binom{8}{5}p^2$$
Using the identity $\binom{n}{k} = \binom{n}{n - k}$, $\binom{8}{3} = \binom{8}{5}$. Remove them.
$$(1 - p)^2 = \frac{1}{25}p^2$$
Take the square root of both sides.
$$1 - p = \frac{p}{5}$$
$$\implies p = \frac{5}{6}$$
Hence our answer is:
$$5 + 6 = \boxed{11}$$

[quote=Math4Life7][quote=OlympusHero]p3: We have $\binom{8}{3}p^3(1-p)^5 = \frac{1}{25} \binom{8}{3}p^5(1-p)^3$. Then $(1-p)^2 = \frac{1}{25}p^2$, so $1-p=\frac{1}{5}p \implies p =\frac{5}{6} \implies \boxed{11}$.[/quote] How in the world did you get this deep in the aops archives...[/quote]

Probably just searched old threads like these.


Three schools $A, B$ and $C$ , each with ﬁve players denoted $a_{i}, b_{i}, c_{i}$ respectively, take part in a chess tournament. The tournament is held following the rules:
(i) Players from each school have matches in order with respect to indices, and defeated players are eliminated; the ﬁrst match is between $a_{1}$ and $b_{1}$.
(ii) If $y_{j}\in Y$ defeats $x_{i}\in X$ , his next opponent should be from the remaining school if not all of its players are eliminated; otherwise his next oponent is $x_{i+1}$ . The tournament is over when two schools are completely eliminated.
(iii) When $x_{i}$ wins a match, its school wins $10^{i-1}$ points.
At the end of the tournament, schools $A, B, C$ scored $P_{A}, P_{B}, P_{C}$ respectively. Find the remainder of the number of possible triples $(P_{A}, P_{B}, P_{C})$ upon division by $8.$


geometry
pigeonhole principle
perimeter
combinatorial geometry
combinatorics proposed
combinatorics

Consider a square of sidelength $ n$ and $ (n\plus{}1)^2$ interior points. Prove that we can choose $ 3$ of these points so that they determine a triangle (eventually degenerated) of area at most $ \frac12$.

What`s your meaning?Can someone give a solusion more clear

Let the convex hull of our $ (n\plus{}1)^2$ points be a $ k$-gon $ F$ for some integer $ k$.

First case. $ k\ge 4n$. Then the perimetr of a polygon $ F$ does not exceed $ 4n$ (since it lies in a square with perimetr $ 4n$), hence the sum $ a\plus{}b$ of some two consecutive sides $ a$, $ b$, of $ F$ does not exceed 2, hence $ ab\le 1$ and the area of triangles with sides $ a$, $ b$ does not exceed $ 1/2$.

Second case. $ k<4n$. Triangulate $ F$ onto $ k\minus{}2$ triangles (say, draw all diagonals from some specific vertex). Then mark one by one all other $ (n\plus{}1)^2\minus{}k$ points inside $ F$ and for each of them join it with the vertices of triangle, into which it is situated on the step. So, we get $ k\minus{}2\plus{}2((n\plus{}1)^2\minus{}k)\equal{}2n^2\plus{}4n\minus{}k>2n^2$ triangles in our triangulation, hence the area of one of them does not exceed $ 1/2$.

See  [url=http://www.rmm.lbi.ro/_dwl/solutions.pdf]here[/url]. There are some remarks about that problem.

Note that in a unit square, any 3 points determine a triangle with area at most $\frac{1}{2}$. This is because if any of the 3 points is not at a corner of the square, we can always move that point to one of the corners such that the area is not decreased. Thus we can assume that each of the 3 points is at a corner of the square and calculate that the area is at most $\frac{1}{2}$.

We conclude that the problem statement is trivial for $n = 1$.

We will now prove the problem statement for $n \ge 2$. Assume for the sake of contradiction that every 3 points determine a triangle with area greater than $\frac{1}{2}$. Clearly, no 3 points are collinear. Let $\mathcal{P}$ be the convex hull of the interior points and let $k$ be the number of vertices of $\mathcal{P}$.  We will prove an upper bound and a lower bound for $k$ that have no intersection, thus providing our desired contradiction.

Let $V_1, V_2, \cdots, V_k$ be the vertices of $\mathcal{P}$, in that order, and let $\theta_i$ denote $\angle V_{i-1}V_iV_{i+1}$ for each $i$. Let $a_i$ denote the length of side $V_iV_{i+1}$, for each $i$. Note that for each $i$, triangle $V_{i-1}V_iV_{i+1}$ has area strictly greater than $\frac{1}{2}$. But this is equivalent to \[a_{i-1}a_i \sin \theta_i>1.\] Since $\sin x < x$ for all $x > 0$, we have $a_{i-1}a_i\theta_i > 1$, implying \[\theta_i > \frac{1}{a_{i-1}a_i}.\] Noting that the sum of the $\theta_i$'s is $2\pi$, we sum over $i$ to get \[ 2\pi > \sum_i \frac{1}{a_{i-1}a_i}. \]

Now we rewrite this as \[\frac{k}{2\pi} < \frac{k}{\sum_i \frac{1}{a_{i-1}a_i}}\] and use AM-GM-HM to conclude that \[\frac{k}{2\pi} < (a_1a_2\cdots a_k)^{2/k} \le (\frac{a_1+a_2 +\cdots + a_k}{k})^2. \] Thus \[a_1 + a_2 + \cdots + a_k > \frac{k\sqrt{k}}{\sqrt{2\pi}}.\]

Note that the perimeter of $\mathcal{P}$, which is $a_1 + a_2 + \cdots + a_k$, is at most the perimeter of the square of side length $n$. Thus \[a_1 + a_2 + \cdots + a_k \le 4n.\] Putting this together, we conclude that \[\frac{k\sqrt{k}}{\sqrt{2\pi}} < 4n,\] or equivalently \[k < \sqrt[3]{32\pi n^2}.\] This is our upper bound on $k$.

Now let $r$ be the number of points inside $\mathcal{P}$; i.e. $r = (n+1)^2 - k$. It's easy to prove by induction that we can partition $\mathcal{P}$ into $k+2r-2$ triangles such that the vertices of the triangles are points and such that no point is inside any triangle. Doing this, we conclude that the area of $\mathcal{P}$ is greater than \[\frac{k+2r-2}{2} = \frac{2(n+1)^2-k-2}{2} = (n+1)^2 - 1 - \frac{k}{2} = n^2 + 2n - \frac{k}{2}.\] It's also clear that the area of $\mathcal{P}$ is at most the area of the square; i.e. \[n^2 + 2n - \frac{k}{2}  < n^2.\] Thus \[k > 4n.\] This is our lower bound on $k$.

Putting the two bounds together, we conclude that \[4n < \sqrt[3]{32 \pi n^2}\] and thus \[64n^3 < 32\pi n^2,\] implying \[2n < \pi,\] contradicting $n \ge 2$. This is our desired contradiction and we are done. $\blacksquare$

[b]Note[/b] This proof shows that the $(n+1)^2$ figure can be improved to $n^2 + \sqrt[3]{4\pi n^2} + 1$.

Clearly, if any 3 points are collinear, they form a degenerate triangle, implying the result. Thus we assume that no three points are collinear. We use the following lemma:

Consider a polygon (convex or not) with $x$ vertices and $y$ points inside it, and a triangulation with the vertices of the triangles the $x+y$ points. Then there are $2y+x-2$ triangles in the triangulation.

We double count the total sum of angles of triangles. Denote by $T$ the number of triangles in the triangulation. The sum of angles of all of them is clearly $T\pi$. On the other hand, the angles around each of the interior points add up to $2\pi$, while the angles around the $n$ points of the polygon add up to the sum of interior angles of the polygon, that is, $(n-2)\pi$. It follows that: $$T\pi=2m\pi+(n-2)\pi\implies T = 2m+n-2.$$

Consider the convex hull of the $(n+1)^{2}$ points in the square. This will be a convex polygon with, say, $k$ vertices, $3 \leq k \leq(n+1)^{2}$. Thus the area of the convex hull is at most $n^{2}$. This along with our lemma our claim for $x=k, y=(n+1)^{2}-k$ gives that there exists a triangle with area at most $$\frac{n^{2}}{-k-2+2(n+1)^{2}}=\frac{n^{2}}{2 n^{2}+4 n-k}.$$ Consider the perimeter of the convex hull. This is at most $4n$. Now consider a triangle with sides on the hull, say $l_1$ and $l_2$. These must be consecutive. Then the area is at least $$\textrm{Area}=\frac{1}{2} l_{1} l_{2} \sin \left(\angle l_{1} l_{2}\right) \leqslant \frac{1}{2} l_{1} l_{2} \overset{AM-GM}{\leqslant} \frac{1}{2}\left(\frac{l_{1}+l_{2}}{2}\right)^{2} \leq \frac{(4 n)^{2}}{k^{2}} \cdot \frac{1}{2}=\frac{8 n^{2}}{k^{2}}.$$ Combining the two we get $$\frac{8 n^{2}}{k^{2}} \leqslant \frac{n^{2}}{2 n^{2}+4 n-k} \Leftrightarrow \frac{k-4 n}{n-2 n(n+2)} \leqslant 0.$$ As $k<2 n(n+2) \implies k \geqslant 4 n$. Setting $k=4n$, both bounds give $\frac{1}{2}$ which is the required bound.

Construct the convex hull of the $(n+1)^2$ points. Suppose $k$ of the $(n+1)^2$ points are on the convex hull. It's well-known that the convex hull must have perimeter less than $4n$ because it's inside a square with perimeter $4n$. If $k\ge 4n$, we can find two adjacent sides of the convex hull which sum to less than $2$, because if not we would get a perimeter of $\ge 4n$. Let these two sides be $AB$ and $BC$. Then $[\triangle ABC]=\frac{1}{2}(AB)(BC)\sin \angle ABC\le \frac{1}{2}\left(\frac{AB+BC}{2}\right)^2\sin\angle ABC\le \frac{1}{2}$ by the sine area formula and AM-GM. Otherwise, if $k<4n$, we triangulate the convex hull, then form disjoint (besides sharing a side) triangles with the points not on the convex hull. All in all, $k-2+2(n+1)^2-2k=2(n+1)^2-k-2>2(n+1)^2-4n-2=2n^2$ triangles have been formed. Since the triangles all together have area $\le n^2$, we get there must be at least one triangle with area at most $\frac12$, and we are done.

Solution from [i]Twitch Solves ISL[/i]:

Assume no $3$ points are collinear (else they are area $0$). Suppose the convex hull of our polygon has $k \ge 3$ points. We consider two cases.
-------------------
[color=blue][b]First case: $k \le 4n$[/b][/color] If $k \le 4n$, we use the following algorithm: draw the entire convex hull, and then repeatedly draw non-intersecting segments between pairs of points. The end result is a division of the $k$-gon into triangles.
By double-counting interior angles, the number of triangles is given exactly by \[ \frac{360^{\circ} \cdot \left( (n+1)^2-k \right) + 180^{\circ} \cdot (k-2)}{180^{\circ}} = 2n^2 + 4n - k \ge 2n^2. \] The sum of the areas of the triangles is at most $n^2$ (the total area of the square), so we're done.
-------------------
[color=blue][b]Second case: $k \ge 4n$[/b][/color] We will show that we can take some three consecutive vertices of the convex hull.
Let $a_1$, \dots, $a_k$ be the side lengths of the convex hull in order. A convex polygon contained inside a square of side length $n$ has perimeter at most $4n$, so we have $a_1 + \dots + a_k \le 4n$. Using the lemma with $k \ge 4n$, there should be an index $i$ with $a_i + a_{i+1} \le 2$. It follows that $a_i a_{i+1} \le 1$ for some index $i$ by AM-GM. The two associated sides then carve out a triangle of area at most $\frac{1}{2}$.

Suppose no three points are collinear, else the problem is obvious. Let the convex hull of the points contain $k$ vertices.

It is well-known that if a convex hull completely contains another, the perimeter of the smaller hull is at most that of the larger. Thus if $k \geq 4n$ the convex hull has perimeter at most $4n$, so there should be some two consecutive edges whose lengths sum to at most $2$, hence have a product of at most $1$. The triangle with these edges as sides thus has area at most $\tfrac{1}{2}$.

If $k \leq 4n$, we divide the convex hull into triangles as follows: first ignore the interior points, and triangulate the convex hull into $k-2$ triangles. Then, reintroduce one point at a time. Each point should lie strictly inside exactly one pre-existing triangle; draw the segments connecting the point to its vertices, increasing the number of triangles by $2$. Thus we have a total of $(k-2)+2((n+1)^2-k)=2n^2+4n-2 \geq 2n^2$ triangles, none of which overlap, hence some has area at most $\tfrac{1}{2}$. This finishes the problem. $\blacksquare$

[quote=Fedor Petrov]It is classical subject. If $ N$ points are indside unit square, how large may be the least are of triangles they form? Heilbronn conjectured that the answer is $ \mathcal{O}\left(\frac{1}{n^2}\right)$, but it turns out that it's possible that all areas are logarithmically larger. So now refined Heilbronn's conjecture is that the least area is $ \mathcal{O}(n^{\epsilon\minus{}2})$ for any $ \epsilon>0$. The best known estimate is (as I know) $ n^{\epsilon\minus{}5/4}$.

As for the problem proposed, it may be solved by triangulation the convex hull of these points. There are two cases, when convex hull contains "many" points on its boundary (in this case we find a triangle with vertices on the convex hull boundary) or "not many" points, in this case triangulation has many triangles and the Pigeonhole Principle works.[/quote]

Update but the best known estimate is now $n^{-\frac{8}{7} - \frac{1}{2000}}$ btw.

Suppose the convex hull of the $(n+1)^2$ points has $k$ points.

If $k > 4n$, then let the vertices of the convex hull $\mathcal P$ be $A_1, A_2, \dots, A_k$ and consider triangles of the form $A_iA_{i+1}A_{i+2}$. The product of the areas of these triangles is equal to $$\prod_{i=1}^k [A_iA_{i+1}A_{i+2}] \leq \frac 1{2^k} \prod_{i=1}^k (P_iP_{i+1})^2 \leq \frac {P(\mathcal P)^{2k}}{(2k^2)^k} < \left(\frac{16n^2}{2k^2}\right)^k < \left(\frac 12\right)^k$$ because the perimeter of $\mathcal P$ is at most $4n$ by convexity. Hence one of these triangles has area at most $\frac 12$.

If $k \leq 4n$, then we perform a triangulation of the $k$-gon into $k-2$ triangles. Every one of the remaining $n^2+2n+1-k$ points must belong in one of the triangles; thus, when we incorporate it into the triangulation, we create exactly $2$ more triangles. Upon complete triangulation, there are $$k-2+2(n^2+2n-k+1) = 2n^2+4n-k \geq 2n^2$$ triangles that cover an area of at most $n^2$. Hence one of these triangles has area at most $\frac 12$ again.

[color=#00f][b]Remark:[/b][/color] We omitted the $\sin(\angle A_iA_{i+1}A_{i+2})$'s in our computation for the first part. If we include them, we actually get a better bound of $\frac{16\pi \cdot n^2}{k^3}$, but this is not necessary (or entirely that helpful either) for the problem in question.


calculus
integration
function
geometry
absolute value
calculus computations

How to show that: $\int_{0}^\infty cos(t^{2})dt=\int_{0}^\infty sin(t^{2}) dt$

I'm not an expert, so could someone explain to me how can
$\int_{0}^{+\infty}\cos(t^{2}) dt$ exist?

cause when $t \rightarrow+\infty$, $-1 \le \cos(t^{2}) \le 1$, but to integrate a function at $+\infty$ we need that it $\rightarrow 0$...

[quote="hydro"]to integrate a function at $+\infty$ we need that it $\rightarrow 0$...[/quote]

That is not in fact true.  For instance, the characteristic function of the integers $f(x) = 1$ if $x \in \bf Z$ and $f(x) = 0$ otherwise has an integral defined and equal to 0 over any interval (including those that go to $\pm \infty$).  In this particular function, it oscilates so quickly for large t that we are adding only smaller and smaller pieces.  This means that it could at least potentially converge.

Just to add on to JBL's point, here's a proof that the integrals converge:

First, $\int_{0}^{1}\cos(x^{2})\,dx$ and $\int_{0}^{1}\sin(x^{2})\,dx$ are proper integrals and obviously exist.
We consider the convergence of $\int_{1}^\infty\cos(x^{2})\,dx$ and $\int_{1}^\infty\sin(x^{2})\,dx$.

Integrating by parts, $\int_{1}^{M}\sin(x^{2})\,dx=\int_{1}^{M}\frac1{2x}\cdot 2x\sin(x^{2})\,dx$
$=\left[-\frac1{2x}\cdot\cos(x^{2})\right]_{1}^{M}-\int_{1}^{M}\frac1{2x^{2}}\cdot\cos(x^{2})\,dx$.
Taking the limt as $M\to\infty$, the first part converges because $\frac1{2M}\cos(x^{2})\to0$ and the remaining integral converges absolutely by comparison to $\frac1{2x^{2}}$, which converges for the same reasons.

Integration by parts is a very powerful technique for this sort of argument, especially when dealing with oscillating functions.
Similarly, $\int_{1}^{M}\cos(x^{2})\,dx= \left[\frac1{2x}\cdot\sin(x^{2})\right]_{1}^{M}+\int_{1}^{M}\frac1{2x^{2}}\cdot\sin(x^{2})\,dx$.

I've shown the equality while evaluating Fresnel integral with $e^{i z^{2}}$.

We have

$\int_{0}^{\infty}e^{i z^{2}}=\int_{0}^{\infty}coz(z^{2})+i\int_{0}^{\infty}sin(z^{2})$

and we have the value of $\int_{0}^{\infty}e^{i z^{2}}$ (as it was shown

at http://en.wikipedia.org/wiki/Fresnel_integral (there where Error

function was)) 'after integrating the function  $e^{i z^{2}}$ (that's my

function) around a pizza-slice shaped area beginning in the point $(0, 0)$

(on the complex plane), then going out to $(R, 0)$, up along the arch of the

circle centered in $(0, 0)$ and with radius R to the point $R e^{i\pi/4}$

and back to $(0, 0)$

in a straight line'

$e^{i z^{2}}$ is analytic on $C$ so integral araound all that pizza-slice

shaped area is 0.

Integral from $(0,0)$ to $(R,0)$ gives us (after taking limits) our integral

(that from the beginning),

and that up along the arch of the circle is zero (we show that absolute value tends to zero)

Integral  from $(R e^{i\pi/4},0)$ to $(0,0)$ is $-e^{\frac{i \pi}{4}}\frac{\sqrt{\pi}}{2}$

And after all we have

$\int_{0}^{\infty}coz(z^{2})+i\int_{0}^{\infty}sin(z^{2})= \int_{0}^{\infty}e^{i z^{2}}=e^{\frac{i \pi}{4}}\frac{\sqrt{\pi}}{2}=\frac{\sqrt{2}}{2}\frac{\sqrt{\pi}}{2}+i\frac{\sqrt{2}}{2}\frac{\sqrt{\pi}}{2}$

We compare real and imaginary parts and that gives equality.

I think I shouldn't have placed this problem in 'Computations and

Tutorials', but this equality seemed easy :)


geometry
circumcircle
trapezoid
rectangle
geometric transformation
reflection
geometry solved

Let P be the other end of the diameter through C and let PA, PB meet the circle (O) again at Q, R. Let the circle tangent at C meet QR at S.  Let a parallel to PQ through C meet the circle again at D. PC is a circle diameter, $\angle PQC = 90^\circ$ is rigtht, hence the cyclic trapezoid PQCD is a rectangle and its diagonals PC, QD meet at the circle center O. Let CD meet PR at a point B'. Consider the cyclic pentagon PQCRD as a degenerate cyclic hexagon PQCC'RD with the double point $C' \equiv C,$ so that $CC' \equiv CS$ is the circle tangent.  By Pascal theorem, the intersections $O \equiv PC \cap DQ,\ B' \equiv C'D \cap PR,\ S \equiv QR \cap CC'$ are collinear, i.e., SO meets PR at B'. But the intersection of PR with SO is B, so that $B' \equiv B$ are identical. Thus the line $OS \equiv AB$ with $A \in PQ,\ B \in CD$ cuts the rectangle PQCD through its diagonal intersection, which means AO = BO.

My solution isn't nice.
Let a tangent to a circle be $k$ and cirle $o$. Denote:
$CD$-diameter of $o$
$DA\cap o=E$
$DB\cap o=F$
$DA\cap k=G$
$DB\cap k=H$
$AB\cap k=P$
Points$E, F, H, G$ lie on a circle, thus
$PH\cdot PG=PE\cdot PF=PC^{2}$
$AO=BO\Leftrightarrow BC \parallel DG\Leftrightarrow\frac{HB}{BD}=\frac{HC}{CG}$.
By Menelaos theorem
$\frac{HB}{BD}=\frac{PH}{PC}$.
$\frac{PH}{PC}=\frac{PH\cdot(PG-PC)}{PC\cdot CG}=\frac{PC^{2}-PC\cdot PH}{PC\cdot CG}=\frac{HC}{CG}$. :D

PS  EDSON_  from where have you this problem?

I have just told some other people not to use Imageshack for attaching graphics here...

The problem is:

Let $XYZ$ be a triangle with circumcenter $O$. Let $C$ be the point diametrically opposite to $Z$ on the circumcircle of triangle $XYZ$. The tangent to this circumcircle at the point $C$ intersects the line $XY$ at a point $W$; the line $WO$ meets the lines $YZ$ and $ZX$ at two points $B$ and $A$. Prove that $AO=OB$; in other words, prove that the point $O$ is the midpoint of the segment $AB$.

In fact it seems that I have solved this problem somewhere on ML but I don't remember where. If my memory doesn't betray me, it's from St. Petersburg MO 2002.

[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=110996 for this problem.

  darij

DARIJ:
REMEMBER WHERE YOU SAW THIS QUESTION ? :idea:

J555:

  YOUR SOLUTION IS THE MOST BEAUTIFUL    :)

Another proof:

Let $C'$ be the reflection of $C$ in the line $AB$. Also, let's give names to some of the points in the picture: let $D$ be the antipode of $C$, and let $X$ be the intersection of the tangent through $C$ to the circle with $AB$. Finally, let $M,N$ be the intersections of $DA,DB$ respectively with the circle.

The tangent to the circle in $C'$ is the reflection in $AB$ of that through $C$, so they intersect in $X$. This means that $CMNC'$ is a harmonic quadrilateral, which, in turn, implies that the lines $DM, DN, DC, DC'$ form a harmonic quadruple. The same must then be true of the four points of intersection of these lines with $AB$, and these points are $A,B,O$ and the infinity point of $AB$ respectively. The conclusion follows.

With Darij's notations:
Take $D,E$ the antipodes of $X,Y$ so $XYDE$ is a rightangle. If $YD, XE$ cut $AB$ at $P,Q$ respectively then: $\angle PYB=\angle OXZ=\angle OZX=\angle CYX=\alpha$, $\angle AXQ=\angle OYZ=\angle OZY=\angle CXY=\beta$.
Further:
$\frac{WX}{WY}=\frac{WQ}{WP}=\frac{XQ}{YP}=\frac{CX^{2}}{CY^{2}}=\frac{sin^{2}\alpha}{sin^{2}\beta}$ .. (1)
$\triangle XQO: \frac{OA}{AQ}=\frac{OX.sin\alpha}{XQ.sin\beta}$ ..(2)
$\triangle YPO: \frac{OB}{PB}=\frac{OY.sin\beta}{YP.sin\alpha}$ ..(3)
Combining (1),(2),(3):
$OA.PB=AQ.OB$
$\frac{OA}{AQ}=\frac{OB}{PB}$
$\frac{OA}{OA+AQ}=\frac{OB}{OB+PB}=\frac{OA}{OQ}=\frac{OB}{OP}$
$OA=OB$


trigonometry
vector
geometry
AMC
AMC 10
AIME
complex numbers
\/closed

Is the Intermediate Trigonometry text coming out this year?

for any wondering, it comes out the end of this summer!

[quote="rrusczyk"]Ah, sorry I missed this post earlier (I was out of the country!).

I'm working on the book now and hope to finish it this summer.[/quote]

i need this book!

Would you rather have a good book that comes out at the end of summer, or a rushed not-so-good book that came out now?

[quote="benzi455"]Or you could steal the manuscript.[/quote]

When you find it, let me know, so I don't have to write it.

any news about Trigonometry Textbook yet?

Is their a intermediate trigonometry class?

Our Precalc book will be out around year-end if all goes as planned.  It will contain trig, complex numbers, and vectors and matrices in 2 and 3 dimensions.  (We will go much deeper into all of these areas than a typical precalc book does.)

Wait, I'm confused...so there will be a precalc book and the course is called trigonometry right?

No, the course is now called Precalculus.

[quote="Blazer123"]Is there a trigonometry textbook text but no class?[/quote]

There is currently no trigonometry class but Precalc covers lots of trigonometry that you need (although you of course should master basic trigonometry for right triangles)

No trig book, either.  Our Precalc book will cover that.

hmmm... Is the trig covered in AoPS volume 2 sufficient enough for AMC10-AIME?

For the most part, yes.  We go to greater depth in the precalc book.


Two boxes are each filled with colored socks. Box  has 2 red socks and 4 blue socks. Box has 4 red socks and 4 blue socks. A box is chosen at random, and a sock is drawn from the box. Given that the sock chosen is blue, what is the probability that it came from box ?

are you trying to ask "what is the probability it came from box" A or B.

You didn't give the complete question...


I've seen some amazing videos over the internet of children no more than 12 years of age that are able to compute massive calculations in their mind! I'm talking huge things like $ 7.045697 \times 2364.0145$

I'm nearly 16 and I was wondering if anyone out there has used the Soroban (Japanese abacus)? I find it really fascinating how fast these computations can be done. I was just wondering if its worth it for me to start learning how to use the abacus, or is it too late for me?

I'd like to enhance my arithmetic, my problem solving and memorization abilities.

Also could you recommend any good books on it or any websites?

[url]http://www.youtube.com/watch?v=wIiDomlEjJw[/url]

this is very old post but still i am trying to find out answers for same

Try joining Aloha Abacus courses (if they are available at your city). At some point, I too could do 3x3 digit multiplications mentally, very fast. You only have to imagine an abacus in your mind.


I have a phpBB3 install and all but when I went thorugh the mod that was at http://www.mayer.dial.pipex.com/tex.htm, it only supports tex tags

When I modified the hook file as

[code]    preg_match_all("#\[tex\](.*?)\[/tex\]#si",$message,$tex_matches);[/code]

->

[code]    preg_match_all("#\\$(.*?)\\$#si",$message,$tex_matches);[/code]

still dollar signs don't work, but they do work when you preview.

how do you fix this up so it works when viewing the message?

You have to escape the escape for the dollar signs which makes it look odd. So try
[code]preg_match_all("#\\\$:$bbcode_uid(.*?)\\\$:$bbcode_uid#si",$message,$tex_matches);[/code]
If you want to allow \ [ ... \ ] it gets even more insane:
[code]preg_match_all("#\\\\\[:$bbcode_uid(.*?)\\\\\]:$bbcode_uid#si",$message,$tex_matches);[/code]
I have only tried this with phpBB2 but have made the changes above to get the phpBB3 syntax. I am attaching  phpBB2 copy of phpbb_hook_2.php. It allows for tex tags, dollar tags and \ [...\ ]
To make it phpBB3 compatible you must replace all occurences of:
[code]$uid by $bbcode_uid
$text by $message[/code]

Hmm it still doesn't work.... o.O

I think its because phpBB is weird and requires it to be in the custom BBcode in the admin panel for it to work

I knew someone else had done it in phpBB3 and have finally found his instructions [url=http://www.phpbb.com/community/viewtopic.php?f=72&t=653165&p=4250675#p4250675]here[/url].

Yay. I got mine working but didn't really use his directions. I just saw that you had to edit message_parser.php and it worked

Great! Do feel free to publish your method on that forum for the benefit of others. That's after all how the GPL code gets better.

Oh no when I click edit on my message it automatically becomes blank :(


[quote="Franz Joseph"]Try Sylvester's Elimination Method.[/quote]

what exactly is that ?

It's a useful method for solving problems involving two functions $f(x)$ and $g(x)$ such that they have a common root. You can probably find it on MathWorld, it involves determinants.

There IS another value of $a$ on $(1,2)$ because for $f(a)=a^{3}+5a^{2}-a-13$ we have $f(1)<0,f(2)>0$

[quote="bugmens"][quote="shyong"]Determine the value of $a$ such that the equation

$x^{3}+ax-1=0$ and $x^{3}-3x+a=0$

have a common root .[/quote]
[hide]t is the common root.
$t^{3}+at-1=0$ and $t^{3}-3t+a=0$
$t=\frac{1+a}{a+3}$
$(\frac{1+a}{a+3})^{3}+a \frac{1+a}{a+3}-1=0,(a+2)(a^{3}+5a^{2}-a-13)=0,a=-2$[/hide][/quote]

That's what I did, I used a graphics calculator for the last part though. If i remember correctly, there were two other (ugly) values of a that work.

Yup, two more values on $(-5,-4)$ and on $(-2,-1)$


In the figure $ \overline{AB} \equal{} \overline{AC}$, angle $ BAD \equal{} 30^{\circ}$, and $ \overline{AE} \equal{} \overline{AD}$.
[asy]unitsize(20);
defaultpen(linewidth(.8pt)+fontsize(8pt));
pair A=(3,3),B=(0,0),C=(6,0),D=(2,0),E=(5,1);
draw(A--B--C--cycle);
draw(A--D--E);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E);
label("$D$",D,S); label("$E$",E,NE);[/asy]Then angle $ CDE$ equals:

$ \textbf{(A)}\ 7\frac {1}{2}^{\circ} \qquad\textbf{(B)}\ 10^{\circ} \qquad\textbf{(C)}\ 12\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 15^{\circ} \qquad\textbf{(E)}\ 20^{\circ}$

Let $ x \equal{} \angle ABC$ and $ y \equal{} \angle ADE$. Because $ \triangle ABC$ is isosceles, $ x \equal{} \angle ABC \equal{} \angle ACB$. Because $ \triangle ADE$ is isosceles, $ y \equal{} \angle ADE \equal{} \angle AED$. It follows that the vertex angle $ \angle EAD \equal{} 180 \minus{} 2y$.

We know that $ \angle BAC \equal{} \angle BAD \plus{} \angle DAC \equal{} 30 \plus{} 180 \minus{} 2y$. We also know that it is the vertex angle of $ \triangle ABC$ so it's also $ 180 \minus{} 2x$. Equating, we have

$ 180 \minus{} 2x \equal{} 30 \plus{} 180 \minus{} 2y\\
2y \minus{} 2x \equal{} 30\\
y \minus{} x \equal{} 15$

$ \angle BDA \equal{} 180 \minus{} (x \plus{} 30)$ and so it follows that $ \angle CDE \equal{} 180 \minus{} (180 \minus{} (x \plus{} 30) \plus{} y) \equal{} x \plus{} 30 \minus{} y \equal{} \minus{}(y \minus{} x) \plus{} 30 \equal{} \boxed{(D)15^\circ}$.

[hide="Alternative last step to TachyonPulse solution"]Consider $ \triangle CDE$ and use Exterior Angle Theorem to yield $ \angle CDE\plus{}x\equal{}y$.[/hide]


Let $Z$ be the integers. The function $f : Z\rightarrow Z$ satisfies $f(n) = n-3$ for $n > 999$ and $f(n) = f( f(n+5) )$ for $n < 1000$. Find $f(84)$.

ch1n353ch3s54a1l, please use the RESOURCES section or use SEARCH. It is OK to revive that old thread if you have any questions about the solution(s) presented there.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=392618#p392618[/url]


analytic geometry
number theory proposed
number theory

Let $P$ be the set of all points with rational coordinates like $(x_1,...,x_n)$
In $P$ we can move from a point to another is the distance between them be equal to $1$
$\text{Distance betweeen two points like}$,$ A=(x_1,...,x_n),B=(y_1,...,y_n)$ $\text{is equal to}$,$ \sum (x_i-y_i)^2$.

Prove:

1.for $n \le 4$ there exist a point that we can not reach it by finitely moves.
2.For $n>4 $ ,we can reach any point by finitely moves.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=lagrange&t=2635

Pierre.


Suppose an unclothed man could survive in the air for 2 hours (120 minutes) in 45 degrees F. How long can he survive in the water at 45 F????? Refer to the Thermal conductivity chart at the bottom (i don't remember the exact stuff but air was .026 and water was .06).

I put 3 minutes after time was up because he can only hold his breath that long  :rotfl: ...

how would you do this problem?

[quote="t0rajir0u"]The assumption is that the man will die after losing a certain amount of thermal energy.  You calculate how much that energy is in the first case, and use that to calculate the amount of time needed to conduct the same amount in the second.[/quote]

I still believe that he can only live 3 minutes maximum (no matter how much thermal energy is lost) from drowning  :rotfl:

[quote="gregx007"]That is making the assumption that he is both (a) completely submerged and (b) has no scuba or snorkel apparatus, neither of which are stated in the problem.[/quote]

exactly...so that question shouldn't count as part of the test  :huh:

Well, I would have to disagree.  If you want to be technical, he question asks how long he CAN survive in water, and the given is how long he COULD survive in open air.  Thus, it follows that you are allowed to give the guy in the water any advantages that you can also give him in the open air--for example, you could imagine him with a scuba tank in both situations without changing the given info.  In contrast, you couldnt give him, say, a thermal suit, since that would change the problem.


Let $\{a_{n}\}_{n\geq 1}$ and $\{b_{n}\}_{n\geq 1}$ be two sequences of natural numbers such that $a_{n+1}=na_{n}+1$ and $b_{n+1}=nb_{n}-1$ for every $n\geq 1$ . Prove that these two sequences can have only a finite number of terms in common.

This is problem 3, 3rd Romanian Selection Test, 1992, proposed by L. Panaitopol.
Hint: prove that $a_{n+1}=n!\left(a_{1}+\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}\right)$ and $b_{n+1}=n!\left(b_{1}-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{n!}\right).$ Next, use that the sequence $e_{n}=\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}$ increases and converges to $e-1$.

If we have infty sequence $n_{1}<n_{2}<...$ of positive integers such that $a_{n_{k}+1}=b_{n_{k}+1}\;\forall k\geq 1$ then $b_{1}-a_{1}=2(\sum_{i=1}^{n_{k}}\frac{1}{i!})\;\forall k\geq 1$, here let $k\to\infty$ we have $b_{1}-a_{1}=2(e-1)$, in particular $e\in\mathbb{Q}$, contradiction!.

[quote="enescu"]No, actually, the problem asks to prove that there are finitely many pairs $(h,k)$ such that $a_{h}=b_{k}.$[/quote]
Sorry  :blush: But solution is similary  as i post above! Do you agree?  :P

Not quite!
Suppose $a_{n+1}=b_{m+1}$ for some positive integers $n,m.$ Then $n!(a_{1}+e_{n})=m!(b_{1}-e_{m})$, hence $\frac{n!}{m!}=\frac{b_{1}-e_{m}}{a_{1}+e_{n}}.$ Since $1\le e_{n}<2$, we obtain $\frac{b_{1}-1}{a_{1}+2}<\frac{n!}{m!}<\frac{b_{1}}{a_{1}}$. The conclusion follows from this obvious result: given the positive real numbers $A$ and $B$, with $A<B$, there are finitely many pairs $(n,m)$ of positive integers such that $A<\frac{n!}{m!}<B.$

[quote="enescu"]given the positive real numbers $A$ and $B$, with $A<B$, there are finitely many pairs $(n,m)$ of positive integers such that $A<\frac{n!}{m!}<B.$[/quote]
Can you post prove of this assertion?  :oops:

[quote="N.T.TUAN"][quote="enescu"]given the positive real numbers $A$ and $B$, with $A<B$, there are only finitely many pairs $(n,m)$ of positive integers such that $A<\frac{n!}{m!}<B.$[/quote]
Can you post prove of this assertion?  :oops:[/quote]
Take $k\in\mathbb{N}$ with ${1\over k}<A<B<k$. Then there are only finitely many $(m,n)\in\mathbb{N}^{2}$ with $\max\{m,n\}<k$.
Assume $\max\{m,n\}\geq k$.
If $m>n$, then $\frac{n!}{m!}=\frac{1}{(n+1)\cdots m}\leq{1\over m}\leq{1\over k}<A$.
If $n>m$, then $\frac{n!}{m!}=(m+1)\cdots n\geq n\geq k>B$.
So there are only finitely many $(m,n)\in\mathbb{N}^{2},m\not =n$ with $A<\frac{n!}{m!}\leq B$.
[quote="enescu"]Suppose $a_{n+1}=b_{m+1}$ for some positive integers $n,m$. Then $n!(a_{1}+e_{n})=m!(b_{1}-e_{m})$, hence $\frac{n!}{m!}=\frac{b_{1}-e_{m}}{a_{1}+e_{n}}.$
Since $1\le e_{n}<2$, we obtain $\frac{b_{1}-1}{a_{1}+2}<\frac{n!}{m!}<\frac{b_{1}}{a_{1}}$.[/quote]
$1\leq e_{m},e_{n}<e-1$, so $\frac{b_{1}+1-e}{a_{1}+e-1}<\frac{n!}{m!}\leq \frac{b_{1}-1}{a_{1}+1}$.
$b_{n+1}\in\mathbb{N}$, so $0<{b_{n+1}\over n!}=b_{1}-e_{n}\mathop{\longrightarrow}_{n\to\infty}b_{1}+1-e$. So $b_{1}\geq 2$ and $0<\frac{b_{1}+1-e}{a_{1}+e-1}<\frac{b_{1}-1}{a_{1}+1}$.
This gives only finitely many $(m,n)\in\mathbb{N}^{2},m\not =n$ with $a_{n+1}=b_{m+1}$.
But there are also only finitely many $n\in\mathbb{N}$ with $a_{n+1}=b_{n+1}$, as shown by [b]N.T.TUAN[/b].
This gives the result.

[quote="olorin"]Take $k\in\mathbb{N}$ with ${1\over k}<A<B<k$. Then there are only finitely many $(m,n)\in\mathbb{N}^{2}$ with $\max\{m,n\}<k$.
Assume $\max\{m,n\}\geq k$.
If $m>n$, then $\frac{n!}{m!}=\frac{1}{(n+1)\cdots m}\leq{1\over m}\leq{1\over k}<A$.
If $n>m$, then $\frac{n!}{m!}=(m+1)\cdots n\geq n\geq k>B$.
So there are only finitely many $(m,n)\in\mathbb{N}^{2},m\not =n$ with $A<\frac{n!}{m!}\leq B$.
[/quote]
Nice! You are true, thank you very much  :wink:


Whas is the minimum number of squares that one needs to draw on a white sheet in order to obtain a complete grid with n squares on a side?

If these squares must be contained within the $ n \times n$ grid, then the answer is:

[hide]
$ 2n-1$

Consider an $ n+1 \times n+1$ [i]lattice[/i].

Draw a box around the outside to enclose it (1 square).

Consider the points along one diagonal. There are $ n-1$ such points.
Each such point can be thought of as the intersection between two gridlines, and the point can be thought of as dividing each into two parts, for a total of four segments.

These can be grouped into two pairs that are necessarily equal. For each pair, construct a square such that the segments of the pair form two of the sides.

Hence, for each point on the diagonal, it will take 2 squares to cover the two gridlines that pass through it.

We then have $ 2 \times (n-1)+1 = 2n-1$[/hide]

[b]Edit:[/b]

[hide="continued"]
Note that except for the surrounding box, any square that covers an entire gridline cannot touch any other gridlines.
Hence, either each gridline must be taken individually (apart from the outside ones), or we must create squares that cover multiple gridlines at a time.

In the first case, we have $ n-1$ interior gridlines of each orientation, plus the surrounding box, for a total of $ 2(n-1)+1 = 2n-1$ squares [b]with spillover[/b].

In the second case, each gridline must be broken into at least two parts. For optimization, let's break each into only two parts. Because we are creating squares, we must cut the gridline in a place such that a square that occupies one part will not divide another gridline into three parts. This forces the cut points to be along a common diagonal, resulting in the situation in the first part, which also requires $ 2n-1$ squares.

[/hide]


I was wondering if anyone out there knew their way around PARI and could help me with a problem I've been having.  I've been trying to get PARI to read from a text file, but it won't let me.  The file is merely a sequence of integers separated by line breaks, and I am trying to use the command readvec().  When I type "readvec(test.txt)" it replies "unknown member function", and when  I rename test.txt to just test (without a file extension, but the same file), and try readvec(test), it gives me "error opening input file".  I am running PARI 2.3.1 (the GP calculator mode), on a windows XP machine.  And I cannot figure out what I am doing wrong.

Help!

Thanks in advance,
ian

problem solved.  PARI was looking in a strange directory for input files, and hence could not find them.

ian


In unit 1-9 using the casio calculator: solution 9 I'm having a hard
time getting my casio to give me the right answer even when following the
given instructions.

(64 shift x2 - 32)+(126 x -3) =

the answer supposible is 24. However, my casio is not giving me that
number. I'm getting -379

What am I doing wrong?


inequalities
number theory unsolved
number theory

Let $ p_1 \equal{} 2,p_2 \equal{} 3......,p_n$ is the firstn prime number,there $ n\geq 3$ Prove that
$ \frac {1}{p_1^2} \plus{} \frac {1}{p_2^2} \plus{} .....\frac {1}{p_n^2} \plus{} \frac {1}{p_1p_2p_3.....p_n}\leq \frac {1}{2}$
Help

[b] 1). Bertrand-Cebyshev Theorem:[/b] $ (\forall) n > 1 (\exists)$ p=prime and:  $ n < p < 2n$;
[view book [i]Niven, Zuckerman, Montgomery[/i]: [b]An Introduction to the Theory of Numbers[/b](1991, fifth edition); [b]p.367, theorem 8.7[/b]].
[b]2). [/b]$ (\forall)n > 1 (\exists)$ p=prime and: $ 2^n < p < 2^{n \plus{} 1}.$

Unfortunately, that gives an inequality in the wrong direction.

[quote="t0rajir0u"]$ 2^{n \minus{} 1} \le p_n \le 2^n$ is demonstrably false (try $ n \equal{} 4$); it also flatly contradicts the Prime Number Theorem.  Bertrand-Chebyshev says nothing about how many primes there are between $ n$ and $ 2n$, just that there's [i]at least[/i] one.[/quote]
I agree with you,we only know there's [i]at least[/i] one prime between $ (n,2n)$ but we dont know $ k$ (in $ p_k$ sorry because my EL is bad)
To mihai miculita: can you detail $ \sum_{i \equal{} 1}^n \frac {1}{p_i^2} < \sum_{k \equal{} 1}^n \frac {1}{4^k} < \frac {1}{3}$ Thanks al ot


function
probability
calculus
derivative
integration
probability and stats

A random variable X is uniformly distributed on (0,1]. Another Random Variable Y is defined as Y=(1/X)-1. Find the distribution function of Y, and hence its probability density function. Show that the mean of Y doesn't exist.

The cumulative distribution function transforms by simple substitution; $ P(Y>a)\equal{}P(\frac1X\minus{}1>a)\equal{}P(X<\frac1{a\plus{}1})$. The density is the derivative of the cumulative distribution, and transforms like an integrand.

The integral you get for the mean will diverge to $ \infty$.

[quote="jmerry"]The integral you get for the mean will diverge to $ \infty$.[/quote]
Just one comment: you can determine that without solving for the details of the distribution of $ Y.$

$ E(Y)\equal{}E(1/X\minus{}1)\equal{}\int_0^1\left(\frac1x\minus{}1\right)\,dx\equal{}\infty.$

So the two parts of the problem can be worked in either order. Of course, now that we know this, it can serve as a check once we do know the density for $ Y.$


Let $n$ be a positive even integer. In how many ways can we fill an $n \times n$ board with $1$’s and $-1$’s such that the sum of each row and column is zero?

I found this reference.  It does not appear to have a nice formula.

[url]http://www.research.att.com/~njas/sequences/A058527[/url]

Hmm, I see. I thought I had seen something like it on the forum before, but I may have been mistaken and it could be a programming problem.


does the singer of gipsey kings gruop has an arabia wife?


Jeff decides to play with a Magic 8 Ball. Each time he asks it a question, it has a 2/5 chance of giving him a positive answer. If he asks it 5 questions, what is the probability that it gives him exactly 2 positive answers?

This is $ \binom{5}{2} \times \left (\frac{2}{5} \right )^2 \times \left (\frac{3}{5} \right )^3\equal{}\frac{10 \times 4 \times 27}{3125}\equal{}\frac{2 \times 4 \times 27}{625}\equal{}\boxed{\frac{216}{625}}$.


I am faced with a problem.I registered for the SAT yesterday for the October 10, 2009 date on the collegeboard.com and after paying for the registration, I came to know that I need to bring along a valid passport for identification.Well I had thoroughly read the collegeboard instructions and I am even a member on the collegeconfidential.com but I never knew about this rule.As of today I do not have a passport and I have 36 days left.

Can you tell if it is absolutely necessary to have a passport or is there any other accepted photo identification.

Or do you have any idea how long it takes to get a passport?

And by the way , I will be writing the SAT from Kolkata .

Thanks in advance.I would really appreciate any sort of help.

Call College Board.  That's the best way to get an answer.

The passport is a MUST. I know people who get sent back from their centres coz they had a photo identification other than passport.


Prove that for any $ n\in N$,
$ \sqrt{1\plus{}\sqrt{2\plus{}....\plus{}\sqrt n}}<2$

[quote="KDS"]Prove that for any $ n\in N$,
$ \sqrt {1 \plus{} \sqrt {2 \plus{} .... \plus{} \sqrt n}} < 2$[/quote]

See here : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1594050#1594050


What happened to these subforums? Are they lost in cyberspace?

They seem to have be joined in with the Archives forum, something you get after you've taken classes.

Actually, now I don't see anything. When I click on "Online School" it says there are no subforums. Is it supposed to be like this?

So the Middle School, High School, and Teachers fora don't exist?


How many ounces of $ 20\%$ acid solution must be mixed with a $ 30\%$ acid solution to obtain 45 ounces of $ 24\%$ acid solution?

1) Set up an equation and solve!

If we set the number of ounces of the 20% solution to be x, we can safely express the number ounces of the 30% solution to be 45-x since the total number of ounces adds up to 45.

x(0.2)+(45-x)(0.3)=45(2.4)
20x+30(45-x)=45(24)
20x+1350-30x=1080
-10x=1080-1350
-10x=-270
x=27

x is what we are looking for since it represents the ounces in the 20% solution!
Answer is [u][i][b]27[/b][/i][/u] :D


The following two ineqs are simmilar:

Prove that $ \forall x,y,z>0$ we have:
$ \sum \frac{x}{x+2y+3z} \geq  \frac{1}{2}$ .
(***)

Prove that $ \forall x,y,z>0$ we have:
$ \sum \frac{x}{x+2y+z} \geq \frac{3}{4}$.
(made by Eckstein Gh.)

I have solved them using Jensen we define $ f: (0, \infty) -> R^{+}$ and $ f(x)=\frac{1}{x}$  (convex) and WLOG $x+y+z=1$.

we solve both problems this way.

do you have other solutions?

cheers! :D :D

I have another solution for you.By Cauchy,we have:
 \sum [x(x+2y+3z)][x/(x+2y+3z)] \geq (x+y+z) 2
But  \sum x(x+2y+3z)
=(x+y+z) 2 +3(xy+xz+yz) \leq 2(x++y+z) 2
So we have what we need. :)


induction
graph theory
combinatorics proposed
combinatorics

Suppose we color the edges of the complete graph $K_n$ on $n$ vertices such that each color is used at most $k$ times.

(a) Prove that if $k\le n-2$, then there must be a triangle with edges of different colors;

(b) Prove that the conclusion at (a) doesn't necessarily hold for $k=n-1$.

I think the idea is that if such triangles do not exist, then there is a color such that the subgraph determined by the edges of that color is connected, which is clearly enough to solve (a). This goes easily by induction on the number of vertices. Suppose it is true for $n$ and add one extra vertex $v$. Call "blue" the color with that property for the first $n$ vertices. If there is a blue edge emerging from $v$, then we are done. If not, it is not hard to prove that all edges from $v$ have the same color so again our claim is true. Indeed, if $w$ and $w'$ are 2 vertices from the first $n$, take $w=w_1,w_2,...w_k=w'$ an elementary chain between $w$ and $w'$ and then $vw_i$ and $vw_{i+1}$ have the same color for all $i$, so this is also true for $vw$ and $vw'$.
(b) is clear since you can take the color of $ij$ to be $min(i,j)$, which works.


Find all real values of $ x>1$ which satisfy:

$ \frac{x^2}{x\minus{}1} \plus{} \sqrt{x\minus{}1}  \plus{}\frac{\sqrt{x\minus{}1}}{x^2} \equal{} \frac{x\minus{}1}{x^2} \plus{} \frac{1}{\sqrt{x\minus{}1}} \plus{} \frac{x^2}{\sqrt{x\minus{}1}}$

[hide="Solution"]Let $ a\equal{}\frac{x^{2}}{x\minus{}1}$, $ b\equal{}\sqrt{x\minus{}1}$ and $ c\equal{}\frac{\sqrt{x\minus{}1}}{x^{2}}$. Then $ abc\equal{}1$ and $ a\plus{}b\plus{}c\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$.

So $ a\plus{}b\plus{}\frac{1}{ab}\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}ab\Rightarrow (a\minus{}1)(b\minus{}1)(ab\minus{}1)\equal{}0$. So now we just have to solve $ a\equal{}1$, $ b\equal{}1$ and $ c\equal{}1$. Solving each equation and noting $ x>1$ is a real, we easily get $ x\equal{}2$ as the only solution.[/hide]


The points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ are  variables at two constant parallel lines $e_{1},e_{2}$ respectively .
If $P(x_{1},x_{2})$ and $M(y_{1},y_{2})$ prove that the line $PM$ passes through a fixed point and find its coordinates.

Given $A(a,0),B(b,c)$; the values $a$ and $b$ are variables, the value $c$ is fixed.

Points $P(a,b)$ and $M(0,c)$

It's clear that $M$ is a fixed point of $PM$.


Let $ x \plus{} y \plus{} z \equal{} 1,{x,y,z}\in{R^ \plus{} },$prove
\[ \sum_{cyc}{\frac {x}{\sqrt {x \plus{}2 y}}}\leq{\sqrt {6\sqrt {3} \minus{} 9}}.
\]

As I have known, this one was created by hungkhtn and canhang_2007 at least 2 years ago  :wink:
The equality holds on
\[ (a,b,c)\equal{}\left(\frac{2(\sqrt{3}\minus{}1)}{\sqrt{3}},\frac{2\minus{}\sqrt{3}}{\sqrt{3}},0\right)\]

It was killed by Mixing Variable method  :)

As I have known,

\[ \sum_{cyc}{\frac {x}{\sqrt {x \plus{}2 y}}}\leq{\sqrt {\frac{3}{2}}}\]

 was  proved by  canhang_2007 ,see here:

http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=48&Id=14349

[quote="fjwxcsl"]As I have known,
\[ \sum_{cyc}{\frac {x}{\sqrt {x \plus{} 2 y}}}\leq{\sqrt {\frac {3}{2}}}
\]
was  proved by  canhang_2007 ,see here:

http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=48&Id=14349[/quote]
Yes, I posted that inequality last year, but I have also solved the sharper inequality (which you posted now) at the same time, and my solution, you can find it at: [i][b]Old And New Inequalities 2[/b][/i], Vo Quoc Ba Can - Cosmin Pohoata, GIL publishing house 2008.
Regards
:)


If f(x) is a fifth degree polynomial, and f(0)=3, f (1)=5, f (2)=6, f (3)=10, f (4)=11, and  f (5)=12, find f (6).

A. 32		B. 15		C. 14		D.  44		E.  NOTA

How do you do this problem quickly (i.e. without finding the polynomial)? I would prefer not to have to solve a huge system of equations....although I do know that the term cancel quite easily, but it still takes quite a bit of time, and this is a 25 question 3 tiebreaker one hour test.

This was number 5 from the Comprehensive level VHHS test of 2007, which can be found at this link:
[url]http://www.mistacademy.com/vhhstourney.htm[/url]

To check your answer:

Answer:
[hide]A. 32[/hide]

Thanks in advance

Use the method of finite differences.  A fifth degree polynomial has constant fifth difference, so all you have to do is compute some values in a table.  It's not that bad.

As a shortcut, the following identity holds for all polynomials of degree less than $ n$ and for any $ x$.

$ \sum_{i \equal{} 0}^{n} ( \minus{} 1)^i {n \choose i} P(x \plus{} n \minus{} i) \equal{} 0$

$ x \equal{} 6, n \equal{} 6$ gets an expression with the answer.

For proof and more details and t0r's suggestion, I will do a [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]shameless plug[/url].


Hello AoPS,

I have a MATHCOUNTS problem that is really bugging our team.  Our cases never turn out correct.

The problem is:

Two integers are randomly selected from the set of integers greater than or equal to -5 and less than or equal to 5.  The two numbers need not be different.  What is the probability that the sum of the two integers is less than their product?'

Thanks!

--Coach Lee

Hmm... I know of a few ways to tackle this problem. Since the numbers have a relatively small range (I use relatively loosely), you could make a chart and test them all out, old fashioned casework. If you are low on time, however, you could try to form a pattern between the numbers.

First of all, there are 121 ways to choose a number between -5 and 5, then another number (If they can be the same number). So, our answer will be $ \frac {\text {some number}} {121}$ .

We can see that one number can't be positive and the other be negative, because $ xy$ would be negative. While the sum could be negative as well, it would not be any smaller than the product. Thus, The numbers must have the same sign (or include the number 0). Through a little experimentation, we find any pair of negative numbers, or a negative number and 0, works. This adds (Number of numbers between -5 and 0) times (Number of numbers between -5 and 0) options, or [b]36 [/b]options.

Through a little more experimentation, we find 0 nor 1 can work with any other numbers. this leaves (Number of numbers between 2 and 5) times (Number of numbers between 2 and 5), 16 options. However, (2,2) does not work, since its sum is its product, thus adding [b]15 [/b]options.

Add these, and $ \frac {36\plus{}15} {121}$ $ \equal{}$ $ \boxed {\boxed {\boxed {\frac {51} {121}}}}$ is our answer.


LaTeX
analytic geometry
Pythagorean Theorem
geometry
power of a point

In circle $ O$, the midpoint of radius $ OX$ is $ Q$; at $ Q$, $ \overline{AB} \perp \overline{XY}$. The semi-circle with $ \overline{AB}$ as diameter intersects $ \overline{XY}$ in $ M$. Line $ \overline{AM}$ intersects circle $ O$ in $ C$, and line $ \overline{BM}$ intersects circle $ O$ in $ D$. Line $ \overline{AD}$ is drawn. Then, if the radius of circle $ O$ is $ r$, $ AD$ is:
[asy]defaultpen(linewidth(.8pt));
unitsize(2.5cm);

real m = 0;
real b = 0;

pair O = origin;
pair X = (-1,0);
pair Y = (1,0);
pair Q = midpoint(O--X);
pair A = (Q.x, -1*sqrt(3)/2);
pair B = (Q.x, -1*A.y);
pair M = (Q.x + sqrt(3)/2,0);

m = (B.y - M.y)/(B.x - M.x);
b = (B.y - m*B.x);

pair D = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b));

m = (A.y - M.y)/(A.x - M.x);
b = (A.y - m*A.x);

pair C = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b));

draw(Circle(O,1));
draw(Arc(Q,sqrt(3)/2,-90,90));
draw(A--B);
draw(X--Y);
draw(B--D);
draw(A--C);
draw(A--D);
dot(O);dot(M);

label("$B$",B,NW);
label("$C$",C,NE);
label("$Y$",Y,E);
label("$D$",D,SE);
label("$A$",A,SW);
label("$X$",X,W);
label("$Q$",Q,SW);
label("$O$",O,SW);
label("$M$",M,NE+2N);[/asy]$ \textbf{(A)}\ r\sqrt {2} \qquad \textbf{(B)}\ r\qquad \textbf{(C)}\ \text{not a side of an inscribed regular polygon}\qquad \textbf{(D)}\ \frac {r\sqrt {3}}{2}\qquad \textbf{(E)}\ r\sqrt {3}$

The solution was thought of by [b]stickfigure[/b], but as he doesn't know much LaTeX, I'm typing it for him.

[hide="Solution"]
We have $ OQ \equal{} \frac{r}{2}$. And drawing quadrilateral $ BXAO$, we find that it is a rhombus. It follows that $ \triangle BQO$ is a $ 30\minus{}60\minus{}90$ right triangle because $ OB \equal{} r \equal{} 2OQ$. Therefore, $ BQ \equal{} \frac{\sqrt{3}}{2}r \equal{} QM \equal{} AQ$. Therefore, $ \triangle BQM$ is a $ 45\minus{}45\minus{}90$ right triangle. It follows that $ BM \equal{} \frac{\sqrt{3}}{2}r\cdot \sqrt{2} \equal{} \frac{\sqrt{6}}{2}r \equal{} AM$ (by reflective symmetry).

We know that $ OM \equal{} QM \minus{} OQ \equal{} \frac{\sqrt{3}}{2}r \minus{} \frac{1}{2}r \equal{} \frac{\sqrt{3} \minus{} 1}{2}r$. By Power of a Point,

$ (MB)(MD) \equal{} (MX)(MY)\implies MD\cdot \frac{\sqrt{6}}{2}r \equal{} (OX \plus{} OM)(OX \minus{} OM) \equal{} OX^2 \minus{} OM^2\implies MD \equal{} \frac{\sqrt{2}}{2}r.$

We know that $ \triangle AMD$ is right because $ \angle BMA$ is an inscribed angle in a semicircle. Therefore, by the Pythagorean Theorem,

$ AD \equal{} \sqrt{\left(\frac{\sqrt{6}}{2}\right)^2 \plus{} \left(\frac{\sqrt{2}}{2}\right)^2} \equal{} \sqrt{\frac{6 \plus{} 2}{4}r^2} \equal{} \boxed{r\sqrt{2} \text{ (A)}}$.

[/hide]

This problem could have been solved using coordinates, but it involves a lot of ugly algebra.


The Sides of a regular icosagon inscribed in a circle have length 6. To the nearest integer,what is the diameter of the circle?

[hide]Taking two adjacent vertices of the 20-gon, we have:
\[ R \equal{} \frac {\frac {6}{2}}{\sin 9^\circ} \approx 19.177 \Rightarrow 19
\]The diameter is twice that, or $ \boxed{38}$.
[/hide]

how did u get $ \frac{\frac{6}{2}}{sin9}$?

Draw a triangle with two adjacent vertices of the icosagon and the circumcenter (or incenter). Now bisect that angle. You'll see two congruent right triangles that share a common leg.


AMC USA(J)MO USAMO algebra polynomial quadratics

Prove that the product of the two real roots of x^4 + x^3 - 1 = 0 is a root of x^6 + x^4 + x^3 - x^2 - 1 = 0.

No one has tried this?  But this is a fun problem!

A hint, for those of you who like fancy names:

[hide]Viete's formulae are rather useful for this one.[/hide]

grr patience i have to do a character analysis on a guy who loves his cousin (in french) and write a horrific short story and write a film review on mendel and some other stuff i think, solution maybe tomorrow

Sorry, didn't realize this had only been around for one day.  But it was already being pushed down to the bottom . . .

i'll try it.

[hide]

let a, b, c, d be the roots of the quartic.  We know that a+b+c+d=-1, ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd = 0, abcd=-1.  We'll construct a 6th degree polynomial with ab, ac, ad, bc, bd, cd as the roots and show that it is the 6th degree polynomial that we were given.  We have x^6+k_1*x^5+k_2*x^4+k_3*x^3+k_4*x^2+k_5*x+k_6.  -k_1 is the sum, which we know is 0.  Also, the product of the roots, (abcd)^3=-1, so k_6 is -1.  Then it gets a little messy... theres probablly a better way.  k_2 = the sum of the roots taken 2 at a time... so there are 15 terms, i won't write them out, but it simplifies to a( sum of the roots of the quartic taken 3 at a time)+b(sum of the roots of the quartic 3 at a time)+...d(sum of the roots of the quartic taken 3 at a time - abc).  So it's 0+0+0-abcd, which is 1.  So k_2=1.  I think that's right so far, but this probablly isnt the best way, since k_3 will have 20 terms that i'll have to wade through... oh well.   But i suspect if i were to go through this i'd have the right answer, but im not sure if i have the right idea, making a 6th degree with ab, ac, ad, bc, bd, etc. as the roots.  I just dont know how else to do it...  [/hide]

cats, that was basically what I meant.  You can wade throught the 20 terms -- they aren't that bad.  The other cases can actually be done without really having to write it all out -- you can just say, "look, if we multiply these two things together and add abcd, we get the same thing as if we just took them all pairwise."  The choose-4 case will be very closely related, since it's just picking 2 to leave out.  The 1 and 5 cases are very simple.  So it's just that middle coefficient that turns out ugly, and even it isn't that bad.

I also thought of another way of doing it, and I don't actually remember which way is nicer.  [hide]We could take the two equations m:^4: + m:^3: - 1 = 0 and n:^4: + n:^3: - 1 = 0, and then use that to show that the expression we get plugging mn into the sixth  degree polynomial is equal to zero.  That does look pretty ugly, though.  I'll go try it now and report back.[/hide]

Either I don't remember how to do this the best way, or it is in fact a very messy method to do.

Or, another thought -- [hide]that 6th degree equation is solvable.  That's gonna be really ugly, though.  Backwards-plug-in.  It will be very nasty to take the 6 solutions to the sextic and decompose them into their 4 product-parts and show they satisfy the original equation.[/hide]

sextic is name for 6th degree polynomial?

Im not sure i understand how you would solve the sextic... could you explain a little more?  thanks.

Oh, sorry, left that bit out.  Well, this is really [i]not[/i] the way you want to approach this problem.  But we note that x = 0 is not a solution, so we can divide through by x:^3: to get
(x^3 - x^(-3)) + (x - x^(-1)) + 1 = 0.  Then, you can make the substitution y = x - x^(-1) to help convert this to a cubic equation in y, and then you will see 1 obvious root of the cubic, which reduces it to a quadratic.  Then you can solve the quadratic, which will give you three quadratics in x, from which you can derive the roots.  Then it's just a matter of finding out which are the a, b, c and d.  (You're maybe aided by the fact that pairs of roots of the sextic will be negative reciprocals of each other, because abcd = -1)

Does that make sense?

Reasonably certain that "sextic" means "sixth degree" -- even if it doesn't, it should :).

Anyhow, I'm pretty sure the method I outlined first (and that you have done much of) is the way to go.  I'm not 100% sure of that, but it feels right.

yeah that makes sense, thanks.  i remember seeing that method before, but didnt think to apply it.


email logarithms AMC USA(J)MO USAMO trigonometry ARML

Do any of you know Sameer Mishra, the spelling bee champ this year??? (I think he was from indiana).... sorry for posting on your forum but mine is dead and I was just wondering.....

*cough*whataboutmustit?*cough*
...jerk.

I go to his school too.

[quote="cognos599"][quote="dynamo729"][quote="cognos599"]umm... he goes to my school and he's one of my best friends. i eat lunch with him. ive known him since like 4th grade. so yes i do know him. neway great job sameer, i don't see him until school starts so yeah. mathandknowledge, logic812, and mustit also go to my/sameer's school.[/quote]

Cool. Is he a total nerd, or is he cool? (I was on the bus with who I think was Justin Ahmann and he was talking about homo sapien superiors and Chuck Norris microwaving burritos.....)[/quote]

no way, i have talked to Justin Ahmann. he would never talk about homo sapien superiors and chuck norris. sameer is pretty cool, but just like me, he's a huge nerd.[/quote]

He [i]was[/i] really funny on the bee..... my mom predicted him to win from the top like 24 or something...

i used to be good at spelling. last year, i was 3rd in the school to sameer and someone else. i stopped doing spelling in 6th grade and i started doing math. i started math when i was 3. but i stopped doing it in 5th and 4th grade but continued in 6th. that's why i am so bad at math.

I had the honor of harassing him with my tie (WHIP) at a concert when we (my friends and I, as well as other orchestra/band members) were ushering as volunteers :D We sat at the back of the stands so no one really noticed until one of my friends snorted really loudly <_<

Is sameer suffering trauma from that concert?

er and why are you awake at this hour (unless you're in korea? - I've noticed koreans tend to migrate during the summer months too :|)

[quote="logic812"]He tends to follow me, cognos, and MathAnd around.[/quote]Good one.  Don't lie; nobody's gonna buy it.  You spend all lunch looking for someone, anyone in fact, to be with.  Also, I thought you used to be friends with Sameer.  Now you act like he's inferior to the rest of the school?  Is this jealousy?  Or a fact that you can't admit to NOT having any friends?  Ok, I'm sorry I'm being so harsh when I say this, but still...that quote is SO hypocritical.

EDIT: Also, gmail isn't a verb.  It's an online email service offered by Google.

[quote="logic812"]cognos, since when was sameer *nerdy*? (MathAnd is the real nerd)
more like a spell whiz that almost failed his part of the spanish oral exa
He tends to follow me, cognos, and MathAnd around.
Why is Sameer gmailing me? (Probably to brag about his Wii that he's going to get.)  :mad:
Soo Hyun, I heard from Sameer what happened at that particular concert.
He constantly said that he felt that Connie and you were chatting too much.

There's soooo many things to tell about Sameer; wish he had an AoPS account.[/quote]

gmail him back and FORCE HIM TO MAKE AN AOPS MWAHAHAHA!!!! :evilgrin:  :evilgrin:  :evilgrin:

and who is MathAnd? He is not on the memberlist....

[quote="dynamo729"][email] him back and FORCE HIM TO MAKE AN AOPS MWAHAHAHA!!!! :evilgrin:  :evilgrin:  :evilgrin:

and who is MathAnd? He is not on the memberlist....[/quote]Why should he make an account here?  (Also, MathAnd is short for MathAndKnowledge)

[quote="MustItAlwaysBeMe"][quote="dynamo729"][b]e[/b]mail him back and FORCE HIM TO MAKE AN AOPS MWAHAHAHA!!!! :evilgrin:  :evilgrin:  :evilgrin:

and who is MathAnd? He is not on the memberlist....[/quote]Why should he make an account here?  (Also, MathAnd is short for MathAndKnowledge)[/quote]

why did you alter my quote??? It is now a misquote :(

[quote="logic812"]He tends to follow me, cognos, and MathAnd around.[/quote]

acutally, it is you who tends to follow sameer, mustit, mathand, and i around. this thread is getting too spammy so let me post a problem to make it seem less spammy.

Find the Value of:

$ (1 \plus{} i)^{17}$ + $ (1 \minus{} i)^{17}$

[hide="hint"]use de Moivre's theorem[/hide]

[hide="answer"]$ 512$[/hide]

[quote="dynamo729"]why did you alter my quote??? It is now a misquote :([/quote]

Sorry about that :)

Uhh yeah, let's not let this turn into a hate thread.  (Ok, I'll admit I went crazy with that post :) )

Also, cognos: you are NOT bad at math.  At all.  You never were. (maybe when you were 3)  Don't say that.  Please. :D

[quote="logic812"]seriously, cognos, if you plan to turn this into a problem solving thread, make one that is not titled under sAMEER MISHRA.  :lol:
and for the above comments, sooey, I am in Korea... (unless i'm not) :huh:
MustIt, speak for yourself . and (I understand that gmail is an online email service offered by Google)
Inferior? I wasn't trying to sound like it, but I see your point.
Rephrase: Sameer has many friends he sits at lunch with, some of them bieng, me, MustIt, MathAnd, and cognos. (happy now?) :wink:
and cognos, stop insulting me in various threads.....it's getting lengthy.  (pm me about this)
Sameer is(was)  more interested in spelling (no more spelling bee for him :( ) than math, in fact he's
not the math nerd. (Not yet anyways...)[/quote]

sorry 'bout that.

yeah, but has anybody come into actual contact, (see him) with him lately?
(sorry about my hypothetical hypocritical speech)
Also, if cognos is BAD at math, where would that put me?
(seriously, I'd say you're (very) good)

Didn't you know he was on jimmy kimmel live (which apparently isn't live) yesterday (along with magic johnson-lol he was on abc in BOSTON for the first game of the finals- and shaq).  Well, I didn't watch it; I better see if it's on youtube.

[quote="logic812"]cognos, since when was sameer *nerdy*? (MathAnd is the real nerd)
[/quote]

Ok, I'm not goint to deny that I'm a nerd, but compared to you, that is the [i]most hypocritical statement[/i] that I've heard from you.  Logic, calling [i]me[/i] a nerd?...  I will now give a few quotes to demonstrate my point.

"Qing (a nice person in our school, but was in PreAlgebra at the time of this question), solve these logarithm problems.  If you can't, you're an idiot."
"Mrs. (censored name of an 8th grade English teacher), can you solve these USAMO problems?"
"(Directed to me), solve this logarithm problem from AoPS Volume 2! (And he follows me around the whole day and the next day, asking me to solve the problem. BTW, the answer is 2.)"

And I think the AoPS books are great, but carrying them around from class to class in school isn't very practical, and it especially doesn't make others think that you're not a nerd.

I'm done with that rant.  I'll now employ cognos's strategy, post one thing about math!

What is $ 1 \plus{} 1$?
[hide="Hint 1"]It's a number.[/hide]
[hide="Hint 2"]It's a real number.[/hide]
[hide="Hint 3"]It's a rational number.[/hide]
[hide="Hint 4"]It's an integer.[/hide]
[hide="Hint 5"]It's a whole number.[/hide]
[hide="Hint 6"]It's a natural number.[/hide]
[hide="Hint 7"]It's either $ 1$, $ 2$, or $ 3$.[/hide]
[hide="Hint 8"]It's not $ 1$.[/hide]
[hide="Hint 9"]It's not $ 3$.[/hide]
[hide="Hint 10"]It's $ 2$.[/hide]

Hint 10 should help out a lot.

Sigh, ... the post above is a self explanatory statement. (no offense) :lol:

[quote]the most hypocritical statement that I've heard from you. Logic, calling me a nerd?[/quote]

omg MathAndKnowlege get it right!! the dictionary.com definition of "nerd" is an [size=150][b]intelligent[/b][/size] but single-minded person obsessed with a nonsocial hobby or pursuit.

how can that apply to logic. although you could mean the other definition:

a stupid, irritating, ineffectual, or unattractive person.

that applies to logic but not you. JAYKAY.  :rotfl:

[size=200]PROBLEM 3 [/size](my favorite USAMO problem)

Prove that:

$ \frac {1}{\cos{0}\cos{1}} \plus{} \frac {1}{\cos{1}\cos{2}} \plus{} \frac {1}{\cos{2}\cos{3}} \plus{} .... \plus{} \frac {1}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin^2{1}}$

[hide="hint"]Multiply by $ \sin{1}$. We get:

$ \frac {\sin{1}}{\cos{0}\cos{1}} \plus{} \frac {\sin{1}}{\cos{1}\cos{2}} \plus{} ... \plus{} \frac {\sin{1}}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin{1}}$

we can right this as:

$ \frac {\sin{1 \minus{} 0}}{\cos{0}\cos{1}} \plus{} \frac {\sin{2 \minus{} 1}}{\cos{1}\cos{2}} \plus{} ... \plus{} \frac {\sin{89 \minus{} 88}}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin{1}}$

This is an identity $ \tan{a} \minus{} \tan{b} \equal{} \frac {\sin{(a \minus{} b)}}{\cos{a}\cos{b}}$

Therefore;

$ \displaystyle\sum_{i \equal{} 1}^{89}[\tan{k} \minus{} \tan{(k \minus{} 1)}] \equal{} \tan{89} \minus{} \tan{0} \equal{} \cot{1}$, because of telescoping.

but since we multiplied $ \sin{1}$ in the beginning, we need to divide by $ \sin{1}$. So we get that:

$ \frac {1}{\cos{0}\cos{1}} \plus{} \frac {1}{\cos{1}\cos{2}} \plus{} \frac {1}{\cos{2}\cos{3}} \plus{} .... \plus{} \frac {1}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin^2{1}}$
as desired. QED[/hide]

lol wow you guys up there in west lafayette

anyways, i was watching the spelling bee for the first time on friday (was at arml), and to be honest, some of those championship words were actually quite easy

the kulturkampf was a word from AP euro so i had that right.

also had introuvable, if you just know that "trouve" is find in french it will help alot and "in" could mean none

and also the one that ended in "poeia" (that was the hard part as the rest of the word you can sound it out), i cant believe the other guy missed it.  it was probably nerves because if you just know onomatopoeia he wouldve gotten it right...

but then again there was a word that started with like "ap" and the sound was just "a" that i wouldve obviously missed.

oh well never met the kid, but he seemed quite normal and deserved to win.

When the Jamacain girl (earlier in the day) got hidra..... something I got that one wrong :( I would have spelled it hydra... she did too and she was my favorite :(

but yeah I got a lot of the champion words. More so even than the non-champion ones.

[quote="cognos599"]
[size=200]PROBLEM 3 [/size](my favorite USAMO problem)

Prove that:

$ \frac {1}{\cos{0}\cos{1}} \plus{} \frac {1}{\cos{1}\cos{2}} \plus{} \frac {1}{\cos{2}\cos{3}} \plus{} .... \plus{} \frac {1}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin^2{1}}$

[hide="hint"]Multiply by $ \sin{1}$. We get:

$ \frac {\sin{1}}{\cos{0}\cos{1}} \plus{} \frac {\sin{1}}{\cos{1}\cos{2}} \plus{} ... \plus{} \frac {\sin{1}}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin{1}}$

we can right this as:

$ \frac {\sin{1 \minus{} 0}}{\cos{0}\cos{1}} \plus{} \frac {\sin{2 \minus{} 1}}{\cos{1}\cos{2}} \plus{} ... \plus{} \frac {\sin{89 \minus{} 88}}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin{1}}$

This is an identity $ \tan{a} \minus{} \tan{b} \equal{} \frac {\sin{(a \minus{} b)}}{\cos{a}\cos{b}}$

Therefore;

$ \displaystyle\sum_{i \equal{} 1}^{89}[\tan{k} \minus{} \tan{(k \minus{} 1)}] \equal{} \tan{89} \minus{} \tan{0} \equal{} \cot{1}$, because of telescoping.

but since we multiplied $ \sin{1}$ in the beginning, we need to divide by $ \sin{1}$. So we get that:

$ \frac {1}{\cos{0}\cos{1}} \plus{} \frac {1}{\cos{1}\cos{2}} \plus{} \frac {1}{\cos{2}\cos{3}} \plus{} .... \plus{} \frac {1}{\cos{88}\cos{89}} \equal{} \frac {\cos{1}}{\sin^2{1}}$
as desired. QED[/hide][/quote]
that semi-officially makes you better than me at math :(

in what way??? i am doing the awesome math problems and i just finished sums/products, so i am a total beast at sums and products but not at anything else. i am about to be a beast at using the law of sines and cosines.  :huh: even though i have already finished pre-calculus and i already know that stuff.

A really easy problem...for lyndon..with your mother

I just told your mother that m > 3. Your mother has n trays which hold m coins between them. A move occurs when your mother takes one coin from each of two trays and places them on a third tray. Prove (or disprove) that your mother can always transfer all the coins to a single tray by a sequence of moves?


calculus
inequalities
inequalities proposed

Let a, b, c be three real numbers with a < b < c. Show that

  $3a\leq a+b+c-\sqrt{t}\leq 3b\leq a+b+c+\sqrt{t}\leq 3c$,

where $t=a^{2}+b^{2}+c^{2}-bc-ca-ab$.

You will understand the title of the th[i]r[/i]ead when b[i]o[/i]th so[i]l[/i]utions get posted. I wonder which one wi[i]l[/i]l be post[i]e[/i]d first!

  Darij

why would anyone take the time to write a straightforward solution if we know by now there's a 2-lined one :D

Hmm, the straightforward solution also has some advantages... e. g., for those like me, an advantage is that it doesn't use calculus (BTW another hint towards the smart solution). But I find the 2-lines solution just ingenious!

  Darij

Let :
\[ f(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2+(ab+bc+ca)x-abc \]
\[ f'(x) = 3x^2-2(a+b+c)x+(ab+bc+ca) \]
\[ f(x)=0 \leftrightarrow x=a , x=b , x=c \]
\[ f'(x)=0 \leftrightarrow x= \frac{a+b+c- \sqrt{t}}{3} , x= \frac{a+b+c+ \sqrt{t}}{3} \]
Where \[ \sqrt{t}=a^2+b^2+c^2-ab-bc-ca \]
So we have :
\[ a \leq \frac{a+b+c- \sqrt{t}}{3} \leq b \leq \frac{a+b+c+ \sqrt{t}}{3} \leq c \] (Roll's Theory )
(q.e.d)

Bravo, this was the two-lines solution! (It actually takes two lines if one doesn't write down all the calculations.) And yes, here is how the straightforward but elementary solution goes:

Since a < b < c, we can introduce the variables x = b - a > 0 and y = c - b > 0. Then, b = a + x and c = b + y = a + x + y. Thus, a + b + c = 3a + 2x + y. Now you can find that $t = x^2+y^2+xy$, and hence it remains to show that

  $3a\leq 3a+2x+y - \sqrt{x^2+y^2+xy}\leq 3a+3x\leq 3a+2x+y + \sqrt{x^2+y^2+xy}\leq 3a+3x+3y$.

Subtracting 3a, we see that this is equivalent to

  $0\leq 2x+y - \sqrt{x^2+y^2+xy}\leq 3x\leq 2x+y + \sqrt{x^2+y^2+xy}\leq 3x+3y$.

This follows from the following four inequalities:

[b]1.[/b] $\sqrt{x^2+y^2+xy}\leq 2x+y$.
[b]2.[/b] $y-x\leq\sqrt{x^2+y^2+xy}$.
[b]3.[/b] $x-y\leq\sqrt{x^2+y^2+xy}$.
[b]4.[/b] $\sqrt{x^2+y^2+xy}\leq 2y+x$.

We will prove only [b]1.[/b] and [b]2.[/b] (since the other two are analogous). In fact, inequality [b]1.[/b] follows from $x^2+y^2+xy\leq \left( 2x+y\right) ^2$, since $\left( 2x+y\right) ^2 -\left( x^2+y^2+xy\right) = 3x^2 + 3xy >0$, and the second one is trivial for y < x (since the left hand side is < 0 and the right hand side is > 0), while for y > x it can be proved by squaring: $\left( y-x\right) ^2 \leq x^2 + y^2 + xy$ directly follows from $-2xy < xy$, what is trivial. So the proof is complete.

  Darij


calculus
integration
logarithms
real analysis
real analysis unsolved

Is it true or false ?
I can prove it, can you help me, maybe it's easy.

$ \int_{0}^{1}\ln \left( 1\minus{}x\right) \ln \left( x\right) dx\equal{}2\minus{}\dfrac{1}{6}\pi
^{2}$ (true)

$ \iint\nolimits_{\left[ 0,1\right] ^{2}}\ln \left( 1\minus{}xy\right) \ln \left(
x\right) \ln \left( y\right) dxdy\equal{}\minus{}4\plus{}\zeta \left( 2\right) \plus{}\zeta \left(
3\right) \plus{}\zeta \left( 4\right)$ (think true)

$ \idotsint\nolimits_{[0,1]^{n}}\ln \left(1\minus{}\prod\limits_{i\equal{}1}^{n}x_{i}\right) \prod\limits_{i\equal{}1}^{n}\ln \left(x_{i}\right) dx_{1}\cdots dx_{n}$=$ \left( \minus{}1\right) ^{n\minus{}1}\times \left(\minus{}2n\plus{}\sum_{k\equal{}1}^{2n}\zeta \left( k\right) \right)$ (true or false ?)

$ \zeta$ is Riemann Zeta function.

:lol:
Oh, yeah !
so easy for you  misan !
thanks

Sorry if I ask a silly question.
What are MCT and PF? :oops:

I think that
PF= partial fraction
MCT = monotone convergence theorem
I'm right  misan ?


ειναι γνωστη και ως "εις ατοπον απαγωγη"

Υπαρχουν κανονες της λογικης που να δικαιολογουν την παραπανω αποδεικτικη μεθοδο???

ΟΧΙ(ΟΧΙ(Ρ))=Ρ

[quote="maniopas"]ΟΧΙ(ΟΧΙ(Ρ))=Ρ[/quote]

Εγω δεν μπορω να δω την σχεση,ισως καποιος αλλος απο το φορουμ??

Ισως ενα παραδειγμα θα βοηθουσε να δουμε πως νομιζεις οτι ο κανονας της διπλης αρνησεως στηριζει την αποδειξη δια της αντιφασεως

[quote="Demetres"][quote="τρικλινο"][quote="Demetres"]Θέλουμε να δείξουμε το Ρ. Δείχνουμε  $ (\neg P) \Rightarrow \bot$, δηλαδή $ \neg \neg P$. Από τον κανόνα $ \neg \neg P \Rightarrow P$ (και modus ponens) παίρνουμε το Ρ.[/quote]

Δηλαδη ,υποθετουμε ~P Κατοπιν καταληγουμε σε αντιφαση $ \bot$ .Τωρα απο την αντιφαση πως  παιρνουμε $ \neg\neg P$

Εκτος και εαν εννοεις οτι η αντιφαση ειναι αρνηση[/quote]

Εξ'ορισμού το $ \neg P$ είναι συντομογραφία του $ P \Rightarrow \bot$.[/quote]

Μα ποιοι ειναι οι κανονες της λογικης που δικαιολογουν τετοιο ορισμο???

[quote="τρικλινο"][quote="Demetres"]
Εξ'ορισμού το $ \neg P$ είναι συντομογραφία του $ P \Rightarrow \bot$.[/quote]

Μα ποιοι ειναι οι κανονες της λογικης που δικαιολογουν τετοιο ορισμο???[/quote]

Δεν καταλαβαίνω τι εννοείς. Αυτό θέλουμε να σημαίνει και αυτό διαλέγουμε να σημαίνει. Θέλουμε η έκφραση $ A \Rightarrow B$ να είναι λανθασμένη αν και μόνο αν Α ορθό και Β λανθασμένο. Στην περίπτωσή μας αν $ P \Rightarrow \bot$ θέλουμε το P να είναι λανθασμένο. Αυτην ακριβώς την ιδιότητα θέλουμε να έχει και η άρνηση του P και γι'αυτό επιλέγουμε αυτόν τον ορισμό. Αν δεν σου αρέσει σημασιολογικά, δες το συντακτικά. Σε όποια πρόταση υπάρχει το σύμβολο $ \neg P$ αντικατέστησέ το με το  $ P \Rightarrow \bot$. Ο κανόνας που έγραψα πιο πάνω γίνεται $ ((P \Rightarrow \bot) \Rightarrow \bot) \Rightarrow P$.

[quote="Demetres"]
. Στην περίπτωσή μας αν $ P \Rightarrow \bot$ θέλουμε το P να είναι λανθασμένο..[/quote]

Γιατι αν $ P \Rightarrow \bot$ [b]ειναι[/b] το P λανθασμενο.

Δηλαδη απο  $ P \Rightarrow \bot$ πως αποδεικνυουμε ~P

[quote="Demetres"][quote="τρικλινο"]
Δηλαδη απο  $ P \Rightarrow \bot$ πως αποδεικνυουμε ~P[/quote]

Έχω πει ότι δεν το αποδεικύουμε αλλά το ορίζουμε έτσι. Πως αλλιώς θα όριζες το ~Ρ;[/quote]

Σε ενα αλλο thread με τον τιτλο βασικα μαθηματικα ( ποια η διαφορα μεταξυ ορισμου και αξιωματος) στο ποστ #4 εγραψες.

"Γιατί θέλουμε να ορίσουμε την τετραγωνική ρίζα; Διότι αποδείξαμε ένα θεώρημα που λέει ότι για κάθε μη αρνητικό αριθμό  υπάρχει μοναδικός μη αρνητικός αριθμός  έτσι ώστε . Αν δεν είχαμε αυτό το θεώρημα, τότε δεν θα είχαμε κανένα λόγο να ορίσουμε την τετραγωνική ρίζα. Και αν από την αρχή προσθέταμε την τετραγωνική ρίζα σαν primitive symbol με τα "γνωστά" αξιώματα, αλλά το παραπάνω θεώρημα δεν ίσχυε, τότε η θεωρία θα κατέρρεε"

Επομενως ποιο ειναι το θεωρημα που δικαιολογει τον ορισμο της αντιφασης??   Οριζουμε το ~P Ως   $ P \Rightarrow \bot$

[quote="τρικλινο"]
Σε ενα αλλο thread με τον τιτλο βασικα μαθηματικα ( ποια η διαφορα μεταξυ ορισμου και αξιωματος) στο ποστ #4 εγραψες.

"Γιατί θέλουμε να ορίσουμε την τετραγωνική ρίζα; Διότι αποδείξαμε ένα θεώρημα που λέει ότι για κάθε μη αρνητικό αριθμό  υπάρχει μοναδικός μη αρνητικός αριθμός  έτσι ώστε . Αν δεν είχαμε αυτό το θεώρημα, τότε δεν θα είχαμε κανένα λόγο να ορίσουμε την τετραγωνική ρίζα. Και αν από την αρχή προσθέταμε την τετραγωνική ρίζα σαν primitive symbol με τα "γνωστά" αξιώματα, αλλά το παραπάνω θεώρημα δεν ίσχυε, τότε η θεωρία θα κατέρρεε"

Επομενως ποιο ειναι το θεωρημα που δικαιολογει τον ορισμο της αντιφασης??   Οριζουμε το ~P Ως   $ P \Rightarrow \bot$[/quote]

Είχα όμως πει πως έχουμε δύο ειδών ορισμούς. Ο ορισμός του ~P πέφτει στην κατηγορία της συντόμευσης. Ξέχνα προς το παρόν την ερμηνεία "~P = άρνηση" και "$ P \Rightarrow \bot$ = αντίφαση". Επιλέγουμε οπουδήποτε υπάρχει το $ P \Rightarrow \bot$ να γράφουμε ~P στην θέση του. Δεν υπάρχει κανένα απολύτως πρόβλημα με αυτό.

Όταν λοιπόν έγραψα ότι ο λογικός κανόνας της εις άτοπο απαγωγής είναι ο ~~Ρ ==> Ρ, θα έπρεπε να έγραφα ((Ρ ==> (λαθος))==>(λάθος)) ==> Ρ. Πρόσεξε ότι εδώ δεν εμφανίζεται πουθενά η άρνηση του P.

Επιλέγουμε να ερμηνεύουμε το ~Ρ ώς η άρνηση του Ρ επειδή έχει όλες τις ιδιότητες που θέλουμε να έχει η άρνηση. (Έχουμε ήδη ορίσει τις ιδιότητες του $ \bot$ και του $ \Rightarrow$.) Δεν γνωρίζω πως αλλιώς θα μπορούσαμε να μιλήσουμε για την άρνηση του Ρ.

[quote="Demetres"]
Είχα όμως πει πως έχουμε δύο ειδών ορισμούς. Ο ορισμός του ~P πέφτει στην κατηγορία της συντόμευσης. Ξέχνα προς το παρόν την ερμηνεία "~P = άρνηση" και "$ P \Rightarrow \bot$ = αντίφαση". Επιλέγουμε οπουδήποτε υπάρχει το $ P \Rightarrow \bot$ να γράφουμε ~P στην θέση του. Δεν υπάρχει κανένα απολύτως πρόβλημα με αυτό.

Όταν λοιπόν έγραψα ότι ο λογικός κανόνας της εις άτοπο απαγωγής είναι ο ~~Ρ ==> Ρ, θα έπρεπε να έγραφα ((Ρ ==> (λαθος))==>(λάθος)) ==> Ρ. Πρόσεξε ότι εδώ δεν εμφανίζεται πουθενά η άρνηση του P.

Επιλέγουμε να ερμηνεύουμε το ~Ρ ώς η άρνηση του Ρ επειδή έχει όλες τις ιδιότητες που θέλουμε να έχει η άρνηση. (Έχουμε ήδη ορίσει τις ιδιότητες του $ \bot$ και του $ \Rightarrow$.) Δεν γνωρίζω πως αλλιώς θα μπορούσαμε να μιλήσουμε για την άρνηση του Ρ.[/quote]

Τι εννοεις (λαθος) εις το ((P====>(λαθος))====>(λαθος))===>P??