analytic geometry
graphing lines
slope
geometry
3D geometry
sphere
trigonometry

1. Evaluate cos(Arc cot(5/12))-sin(Arc cos(4/5))
[hide]-14/65[/hide]

2. With his eyes 75 feet above water, an observer spots 2 ships in the distance. The 2 ships and
the observer are all coplanar. The angle of depression to one ship is 5° and to the other it is 10°.
Find the distance between the two ships rounded to the nearest foot.
[hide]432 ft[/hide]

3. Find the slope of the tangent line to the curve x^2+4xy =5+ y^2 at the point (1, 2).
[hide]No answer given, but if you know how to solve it, it would be helpful[/hide]

4. Find the shortest distance from the line with parametric equations x =3t −2, y= −2t, and
z=4t+1 to the point (3, -1, 4).
[hide]radical 6[/hide]

5. Find the volume of the largest right circular cylinder that can be inscribed in a sphere with
diameter 20. Round the final answer to the nearest tenth of a cubic unit.
[hide]2418.4[/hide]

Any sort of help will be much appreciated, and will be rewarded with high cube points :)

You should post this in the high school forum. The high schoolers will be able to evaluate trig and stuffs.

Us middle schoolers are not that smart yet (with a few exceptions).

[hide="Hint for trig question"]
Pythagorean Triples
[/hide]

[hide="Solution"]
Well, just split it into two parts. When considering cos(Arc cot(5/12)), first look at the inside. You need to find an angle which will give a cotangent of 5/12. If you draw the unit circle, you will know that this angle will have a hypotenuse of 13. So now you have all the information you need to take the cosine. Repeat the process for the other part of the problem and you're done.

You end up getting 5/13 - 3/5 = -14/65
 [/hide]

#3
The first thought that comes to mind is simply to differentiate implicitly and solve. However, this would then become a trivial calculus exercise. I wonder if their is a more insightful way...

[hide="Solution to 2"]
We will form a right triangle.  Leg $ L_{1}$ will be the height of the observer above the water.  Leg $ L_{2}$ will be the distance from the water directly underneath the observer to the ship. The hypotenuse $ H$ will be the distance from the observer directly to the ship. With each ship, a different triangle will be formed. We want to find the distance between the different $ L_2$s.

Height of observer above water $ \equal{} 75$
Angle between $ L_{1}$ and $ {H} \equal{} (90\minus{}10\equal{}80)$ or $ (90\minus{}5\equal{}85)$.

$ \tan 85\equal{} \frac{L_{21}}{75}$

$ \tan 80\equal{} \frac{L_{22}}{75}$

$ L_{21}\minus{}L_{22}\equal{}75\tan 85 \minus{} 75\tan 80\approx 432$

[/hide]
[hide="Solution to 3"]
Implicitly differentiate.
$ 2x\plus{}4y\plus{}4x\frac{\mathrm{d} y}{\mathrm{d} x}\equal{}2y\frac{\mathrm{d} y}{\mathrm{d} x}$

$ \frac{\mathrm{d} y}{\mathrm{d} x}\equal{}\frac{2x\plus{}4y}{2y\minus{}4x}$

$ \frac{\mathrm{d} y}{\mathrm{d} x}$ at $ (1,2)$ is undefined $ (\frac{10}{0})$
[/hide]

[hide="Problem 1"]Draw a triangle with opposite side 12 and adjacent side 5. Therefore, the hypotenuse is 13. The cosine of this triangle is therefore $ \frac{5}{13}$.

Similarly, for the second part, a triangle with adjacent side 4 and hypotenuse 5 has opposite side 3, so the sine is $ \frac{3}{5}$.

Then, compute $ \frac{5}{13} \minus{} \frac{3}{5} \equal{} \boxed{\minus{}\frac{14}{65}}$.

[/hide]

We already answered your questions...theres no need to double post this http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315216


geometry
conics
ellipse
geometric transformation
reflection
parabola
analytic geometry

In an ellipse, a ray of light is emitted from one focus. If the inner surface of the ellipse is a mirror, prove that the reflected ray will pass through the other focus.

This property is called the reflection property of ellipse(similar property exists for parabola also)

To prove this,prove that the tangents and normals to the ellipse$ \frac{x^2}{a^2}\plus{}\frac {y^2}{b^2}\equal{}1$ at point $ P$ are the angle bisectors of line joining the foci to the point $ P$.

There are many ways to prove this result.

[hide="some sketches"]
1.Taking any point on ellipse in parametric form$ P\equiv(acos\theta,bsin\theta)$,write the equation of focal chords.Then write the eqn of their angle bisectors(which can be reduced in the form of their tangents and normals at $ P$.

2.Let the point of intersection of normal at $ P$ with the $ X\minus{}axis$ be $ Q$.Using the eqn of normal,find coordinate of $ Q$.
Let the foci be $ S_1,S_2$.Then use the converse of angle bisector theorem.

There must be other ways but i can't think of any more at this point of time.[/hide]

I am sorry, but I forgot to mention that I was hoping for a proof that did not involve co-ordinate geometry. Is it possible to get a proof in some other way?

[quote="Akashnil"]Since you're asking for a non-coordinate proof, I'd assume you have a non-coordinate definition for ellipse as well. Since you've not provided one, I'd assume one myself:

An ellipse is the set of points in the plane from which the sum of distances of 2 fixed points on the plane is equal to some fixed constant. The two fixed points are called the foci of the ellipse. And the fixed length the focal constant.

[b]Lemma 1:[/b]
If a point lies outside an ellipse then the sum of distances from the point to the foci of the ellipse is greater than the focal constant.

Proof:
Consider the point to be $ P$ and the 2 foci $ F_1,F_2$ let $ F_1P$ intersect the ellipse at $ P'$.
$ F_1P \plus{} F_2P \equal{} F_1P' \plus{} P'P \plus{} F_2P > F_1P' \plus{} P'F_2 \equal{}$ the focal constant.
(the inequality step comes from triangle inequality).

[b]Lemma 2:[/b]
given a line $ XY$ and 2 points $ P_1,P_2$ in a plane in the same side of $ XY$, then for a point $ Q\in XY$, $ \angle P_1QX \equal{} \angle P_2QY$ iff $ QP_1 \plus{} QP_2 < Q'P_1 \plus{} Q'P_2$ for all other points $ Q'\in XY$

Proof:
Let the foot of the perpendicular from $ P_2$ to $ XY$ be $ T$. Let $ P_2'$ be a point on $ P_2'T$ extended such that $ P_2'T \equal{} P_2T$.
$ P_2TQ\cong P_2'TQ$
$ P_2TQ'\cong P_2'TQ'$
From first congruence:
$ \angle P_1QP_2' \equal{} \angle P_1QP_2 \plus{} 2\angle P_2QY \equal{} \angle P_1QP_2 \plus{} \angle P_2QY \plus{} \angle P_2QY \equal{} 180^0$
So $ P_1QP_2$ collinear.
Also from the second congruence:
$ P_1Q' \plus{} P_2Q' \equal{} P_1Q' \plus{} P_2'Q' > P_1P_2' \equal{} P_1Q \plus{} QP_2' \equal{} P_1Q \plus{} P_2Q$
(here too the inequality step comes from triangle inequality)

So let a ray from $ F_1$ reflect at $ P$ on the ellipse. We shall show that the reflected ray passes through $ F_2$.
Let the tangent at the point $ P$ to the ellipse be $ l$.
except $ P$, all other points on $ l$ lies outside the ellipse. By lemma 1 the sum of distance from the two foci to a point on line $ l$ is minimized at $ P$. By lemma 2 $ F_1P$ and $ F_2P$ make equal angles with $ l$. Hence the reflected ray will go through $ F_2$.[/quote]

And with this Akashnil is back in business......... :)  :lol:

How do u think so meticulously ?? :?:

Thanks a lot, Akashnil.
Not only was your proof excellent, the way you presented it was also an important lesson for me.

[hide="@akashnil"]
are you a school student ?
If yes, which class do you study in ?[/hide]

:omighty: brilliant one akashnil  :D

i never thought about any geometry solution to it

@ keyree
Akashnil gave 10 boards this year from rahara ramkrishna mission.........Wish he follows the footsteps of riddhipratim basu

@UKO:
Is Riddhipratim Basu the IMO contestant from Kolkata (in 2005 & 2006[bronze medal])? :omighty:
btw, r u also from the same school as Akashnil? (during class X)

yes,u r correct

No I am not from rahara


geometry
rectangle
parallelogram
trigonometry
analytic geometry
graphing lines
slope

Let $ABC$ be a right triangle $\angle ACB=90^{o}$ and $AC \ge BC$. Point $M$ is on $AC$ and $AM=BC$, $N$ is on $BC$ and $BN=MC$. Lines $AN$ and $BM$ intersect in $K$. Find $\angle AKM$.

Yes, that right but I'm looking for a synthetical solution :lol:

See the attached diagram.

Construct points $P,Q$ such that $ACBQ$ is a rectangle and $PB=BC=AM=AQ\implies PQ=MC=BN$. Then $\triangle CBM\cong\triangle BPN\cong\triangle QAP$, which gives $PA=PN$ and $\angle APN=90^\circ$. Quadrilateral $AMBP$ is a parallelogram, hence $\angle MKA=\angle KAP=45^\circ$

Farenhajt's solution is very nice. I think it can enter the proof without words contest! :lol:
Here is mine synthetical solution i liked before seeing Farenhajt's :blush: [hide]
Let $BC=a, AC=b$ then $AM=a, MC=BN=b-a, CN=2b-a$
From Menelaus $\frac{NK}{KA}\frac{AM}{MC}\frac{CB}{BN}= 1$ so $\frac{NK}{KA}= \frac{(b-a)^{2}}{a^{2}}$ and $\frac{NK}{NA}= \frac{(b-a)^{2}}{2a^{2}-2ab+b^{2}}$. As $NA^{2}= 2(2a^{2}-2ab+b^{2})$ (From Pythagoras in $\Delta ANC$) we have $NK.NA=2(b-a)^{2}$.
Let $P$ be on line $BC$ and $BP=BN=b-a$. $\Delta ACP$ is isosceles right triangle so $\angle APC=45^{o}$.
We have $NK.NA=2(b-a)^{2}=NB.NP$ so by power of point theoreme $AKBN$ is circumscribed and $\angle AKB= 180^{o}-\angle APB = 135^{o}$[/hide]

[quote="delta"][color=red]Let $ABC$ be a right triangle, $CA\perp CB$ and $a<b<2a$. Given are the points $M\in AC$ and $N\in BC$, $N$ so that $AM=BC$ and $BN=MC$.
Denote the intersection $K\in AN\cap BM$. Find the value of the angle $\widehat{AKM}\ .$[/color][/quote]
[color=darkblue][size=117][b]A preliminary metrical study.[/b][/size] Prove easily that : $AM=a$ ; $NB=MC=b-a$ ; $CN=2a-b$ ;

$BM^{2}=$  $a^{2}+(b-a)^{2}$ ; $AN=\sqrt 2\cdot BM\ .$ I'll establish the position of the point $K$ on the segments $[BM]$ and $[AN]$.

Apply the [b][i]Menelaus' theorem[/i][/b] to the transversals $\overline{AKN}$, $\overline{BKM}$ for the triangles $BCM$, $ACN$ respectively :

$\blacktriangleright\frac{AM}{AC}\cdot\frac{NC}{NB}\cdot\frac{KB}{KM}=1$ $\Longrightarrow$ $\frac{a}{b}\cdot\frac{2a-b}{b-a}\cdot\frac{KB}{KM}=1$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \frac{KB}{b(b-a)}=\frac{KM}{a(2a-b)}=\frac{1}{BM}\ }\ .$
$\blacktriangleright\frac{BN}{BC}\cdot \frac{MC}{MA}\cdot\frac{KA}{KN}=1$ $\Longrightarrow$ $\frac{b-a}{a}\cdot\frac{b-a}{a}\cdot\frac{KA}{KN}=1$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \frac{KA}{a^{2}}=\frac{KN}{(a-b)^{2}}=\frac{\sqrt 2}{BM}\ }\ .$

If denote $m(\widehat{AKM})=\phi$, $m(\widehat{CBM})=u$, $m(\widehat{ANC})=v$, then $\phi =v-u\ .$

[size=117][b]Proof 1.1 (metric).[/b][/size] Apply the [b][i]generalized Pythagoras' theorem[/i][/b] to the side $AM$ in the triangle $AKM$ :

$2\cdot KA\cdot KM\cdot\cos \phi =KM^{2}+KA^{2}-AM^{2}=$ $2\cdot\frac{a(2a-b)}{BM}\cdot \frac{a^{2}\sqrt 2}{BM}\cdot\cos\phi =\frac{a^{2}(2a-b)^{2}}{BM^{2}}+\frac{2a^{4}}{BM^{2}}-a^{4}$ $\Longrightarrow$

$2\sqrt 2\cdot a(2a-b)\cos\phi=$ $(2a-b)^{2}+2a^{2}-BM^{2}=$ $2a(2a-b)$ $\Longrightarrow$ $\cos\phi =\frac{\sqrt 2}{2}$ $\Longrightarrow$ $\boxed{\ \phi=45^{\circ}\ }\ .$

[size=117][b]Proof 1.2 (slight metric).[/b][/size] Denote the intersection $P$ between the bisector of the angle $\widehat{ACN}$ and the line $AN$.

Apply the [b][i]theorem of the angle bisector[/i][/b] to the ray $[CP$ in the triangle $ACN$ :

$\frac{PN}{CN}=\frac{PA}{CA}$ $\Longrightarrow$ $\frac{PN}{2a-b}=\frac{PA}{b}=\frac{AN}{2a}$ $\Longrightarrow$ $\boxed{\ PA=\frac{b}{2a}\cdot NA\ }\ .$ Therefore, $\boxed{\ \phi=45^{\circ}\ }$ $\Longleftrightarrow$ the quadrilateral  $CKPM$ is cyclically

 $\Longleftrightarrow$ $AP\cdot AK=AM\cdot AC$ $\Longleftrightarrow$ $\frac{b}{2a}\cdot NA\cdot \frac{a^{2}\sqrt 2}{BM}=ab$ $\Longleftrightarrow$ $\frac{b}{2a}\cdot BM\sqrt 2\cdot \frac{a^{2}\sqrt 2}{BM}=ab$, O.K.

[b]Remark.[/b] Let $R$ be the middlepoint of the segment $[AN]$. Then $\triangle ACN\sim\triangle BRM$, where $\frac{AN}{BM}=\sqrt 2\ .$ Particularly, $\boxed{\ RB\perp RM\ \ \wedge\ \ CP\parallel MR\ }\ !!$

[size=117][b]Proof 2.1 (slight trigonometric).[/b] Apply the [b][i]Sinus' theorem[/i][/b] in the triangle $AMK$ :

$\frac{a}{\sin \phi }=\frac{AM}{\sin\phi }=\frac{AK}{\sin (90^{\circ}+u)}=\frac{a^{2}\sqrt 2}{BM\cos u}=\frac{a^{2}\sqrt 2}{a}=a\sqrt 2$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \phi=45^{\circ}\ }\ .$ [/size]

[size=117][b]Proof 2.2 (trigonometric).[/b][/size] $\tan u=\frac{CM}{CB}=\frac{b-a}{a}$, $\tan v=\frac{CA}{CN}=\frac{b}{2a-b}$ and $\tan \phi=\tan (v-u)=$

$\frac{\tan v-\tan u}{1+\tan u\tan v}=$ $\frac{\frac{b}{2a-b}-\frac{b-a}{a}}{1+\frac{b}{2a-b}\cdot\frac{b-a}{a}}=$ $\frac{ab+2a^{2}-3ab+b^{2}}{2a^{2}-2ab+b^{2}}=1\ .$  Therefore, $\boxed{\ \phi =45^{\circ}\ }\ .$ [b]In the my opinion, this proof is [u]very quickly ![/u][/b]

[size=117][b]Proof 2.3 (trigonometric).[/b][/size] $\sin u=\frac{b-a}{BM}$, $\cos u=\frac{a}{BM}$, $\sin v=\frac{b}{AN}$, $\cos v=\frac{2a-b}{AN}\ .$ Thus, $\sin \phi =\sin (v-u)=$

$\sin v\cos u-\sin u\cos v=$ $\frac{b}{AN}\cdot\frac{a}{BM}-\frac{b-a}{BM}\cdot \frac{2a-b}{AN}=$ $\frac{2a^{2}-2ab+b^{2}}{AN\cdot BM}=\frac{BM^{2}}{AN\cdot BM}=\frac{BM}{AN}=\frac{\sqrt 2}{2}$, i.e. $\boxed{\ \phi =45^{\circ}\ }\ .$

[size=117][b]Proof 3 (analytic).[/b][/size] In an analytical system denote $C(0,0)$, $B(a,0)$, $A(0,b)$, where $a<b<2a\ .$ Thus, the slope of the line $BM$ is

 $m=\frac{a-b}{a}$ and the slope of the line $AN$ is $n=\frac{b}{b-2a}$. Therefore, $\tan \phi =\left|\frac{m-n}{1+mn}\right|=$ $\left|\frac{\frac{a-b}{a}-\frac{b}{b-2a}}{1+\frac{a-b}{a}\cdot\frac{b}{b-2a}}\right|=1$, i.e. $\boxed{\ \phi =45^{\circ}\ }\ .$

[size=142][b]Remark.[/b] The previous [b][i]Farenhajt's proof[/i][/b] is naturally, very nicely and simply ! It is the [b]"proof 0 (without words)".[/b][/color][/size]

I had posted this problem again in the past. If I remember well , Fahrehajt had sent me two solutions.I proposed this problem to the Greek mathematical society , but they put the proposers solution.
 [color=red]An easy solution can be done also using coordinates and the slopes of the two lines.[/color][u] Babis[/u]

[quote="Virgil Nicula"][quote="delta"][color=red]Let $ABC$ be a right triangle, $CA\perp CB$ and $a<b<2a$. Given are the points $M\in AC$ and $N\in BC$, $N$ so that $AM=BC$ and $BN=MC$.
Denote the intersection $K\in AN\cap BM$. Find the value of the angle $\widehat{AKM}\ .$[/color][/quote]
[color=darkblue][size=117][b]A preliminary metrical study.[/b][/size] Prove easily that : $AM=a$ ; $NB=MC=b-a$ ; $CN=2a-b$ ;

$BM^{2}=$  $a^{2}+(b-a)^{2}$ ; $AN=\sqrt 2\cdot BM\ .$ I'll establish the position of the point $K$ on the segments $[BM]$ and $[AN]$.

Apply the [b][i]Menelaus' theorem[/i][/b] to the transversals $\overline{AKN}$, $\overline{BKM}$ for the triangles $BCM$, $ACN$ respectively :

$\blacktriangleright\frac{AM}{AC}\cdot\frac{NC}{NB}\cdot\frac{KB}{KM}=1$ $\Longrightarrow$ $\frac{a}{b}\cdot\frac{2a-b}{b-a}\cdot\frac{KB}{KM}=1$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \frac{KB}{b(b-a)}=\frac{KM}{a(2a-b)}=\frac{1}{BM}\ }\ \ (1)\ .$
$\blacktriangleright\frac{BN}{BC}\cdot \frac{MC}{MA}\cdot\frac{KA}{KN}=1$ $\Longrightarrow$ $\frac{b-a}{a}\cdot\frac{b-a}{a}\cdot\frac{KA}{KN}=1$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \frac{KA}{a^{2}}=\frac{KN}{(a-b)^{2}}=\frac{\sqrt 2}{BM}\ }\ \ (2)\ .$

If denote $m(\widehat{AKM})=\phi$, $m(\widehat{CBM})=u$, $m(\widehat{ANC})=v$, then $\phi =v-u\ .$

[size=117][b]Proof 1.1 (metric).[/b][/size] Apply the [b][i]generalized Pythagoras' theorem[/i][/b] to the side $AM$ in the triangle $AKM$ :

$2\cdot KA\cdot KM\cdot\cos \phi =KM^{2}+KA^{2}-AM^{2}=$ $2\cdot\frac{a(2a-b)}{BM}\cdot \frac{a^{2}\sqrt 2}{BM}\cdot\cos\phi =\frac{a^{2}(2a-b)^{2}}{BM^{2}}+\frac{2a^{4}}{BM^{2}}-a^{4}$ $\Longrightarrow$

$2\sqrt 2\cdot a(2a-b)\cos\phi=$ $(2a-b)^{2}+2a^{2}-BM^{2}=$ $2a(2a-b)$ $\Longrightarrow$ $\cos\phi =\frac{\sqrt 2}{2}$ $\Longrightarrow$ $\boxed{\ \phi=45^{\circ}\ }\ .$

[size=117][b]Proof 1.2 (slight metric).[/b][/size] Denote the intersection $P$ between the bisector of the angle $\widehat{BCM}$ and the line $BM$.

Apply the [b][i]theorem of the angle bisector[/i][/b] to the ray $[CP$ in the triangle $BCM$ : $\frac{PB}{CB}=\frac{PM}{CM}$ $\Longrightarrow$

$\frac{PB}{a}=\frac{PM}{b-a}=\frac{BM}{b}$ $\Longrightarrow$ $\boxed{\ BP=\frac{a}{b}\cdot BM\ }\ .$ Therefore, $\boxed{\ \phi=45^{\circ}\ }$ $\Longleftrightarrow$ the quadrilateral  $CKPN$ is cyclically $\Longleftrightarrow$

$BP\cdot BK=BN\cdot BC$ $\Longleftrightarrow$ $\frac{a}{b}\cdot BM\cdot BK=a(b-a)$ $\Longleftrightarrow$ $BM\cdot BK=b(b-a)$, what is truly - see the relation $(1)\ .$

[b]Remark.[/b] Let $R$ be the middlepoint of the segment $[AN]$. Then $\triangle ACN\sim\triangle BRM$, where $\frac{AN}{BM}=\sqrt 2\ .$ Particularly,

$\boxed{\ RB\perp RM\ \ \wedge\ \ CP\parallel MR\ }\ !!$  Let $S\in BC$ and $T\in BM$ so that $C\in (BS)$, $CS=BN$ and $ST\perp BM$. Then $\boxed{\ BK=ST\ }\ !!$

[size=117][b]Proof 2.1 (slight trigonometric).[/b] Apply the [b][i]Sinus' theorem[/i][/b] in the triangle $AMK$ :

$\frac{a}{\sin \phi }=\frac{AM}{\sin\phi }=\frac{AK}{\sin (90^{\circ}+u)}=\frac{a^{2}\sqrt 2}{BM\cos u}=\frac{a^{2}\sqrt 2}{a}=a\sqrt 2$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \phi=45^{\circ}\ }\ .$ [/size]

[size=117][b]Proof 2.2 (trigonometric).[/b][/size] $\tan u=\frac{CM}{CB}=\frac{b-a}{a}$, $\tan v=\frac{CA}{CN}=\frac{b}{2a-b}$ and $\tan \phi=\tan (v-u)=$

$\frac{\tan v-\tan u}{1+\tan u\tan v}=$ $\frac{\frac{b}{2a-b}-\frac{b-a}{a}}{1+\frac{b}{2a-b}\cdot\frac{b-a}{a}}=$ $\frac{ab+2a^{2}-3ab+b^{2}}{2a^{2}-2ab+b^{2}}=1\ .$  Therefore, $\boxed{\ \phi =45^{\circ}\ }\ .$ [b]In the my opinion, this proof is [u]very quickly ![/u][/b]

[size=117][b]Proof 2.3 (trigonometric).[/b][/size] $\sin u=\frac{b-a}{BM}$, $\cos u=\frac{a}{BM}$, $\sin v=\frac{b}{AN}$, $\cos v=\frac{2a-b}{AN}\ .$ Thus, $\sin \phi =\sin (v-u)=$

$\sin v\cos u-\sin u\cos v=$ $\frac{b}{AN}\cdot\frac{a}{BM}-\frac{b-a}{BM}\cdot \frac{2a-b}{AN}=$ $\frac{2a^{2}-2ab+b^{2}}{AN\cdot BM}=\frac{BM^{2}}{AN\cdot BM}=\frac{BM}{AN}=\frac{\sqrt 2}{2}$, i.e. $\boxed{\ \phi =45^{\circ}\ }\ .$

[size=117][b]Proof 3 (analytic).[/b][/size] In an analytical system denote $C(0,0)$, $B(a,0)$, $A(0,b)$, where $a<b<2a\ .$ Thus, the slope of the line $BM$ is

 $m=\frac{a-b}{a}$ and the slope of the line $AN$ is $n=\frac{b}{b-2a}$. Therefore, $\tan \phi =\left|\frac{m-n}{1+mn}\right|=$ $\left|\frac{\frac{a-b}{a}-\frac{b}{b-2a}}{1+\frac{a-b}{a}\cdot\frac{b}{b-2a}}\right|=1$, i.e. $\boxed{\ \phi =45^{\circ}\ }\ .$

[size=142][b]Remark.[/b] The previous [b][i]Farenhajt's proof[/i][/b] is naturally, very nicely and simply ! It is the [b]"proof 0 (without words)".[/b][/color][/size][/quote]


trigonometry
geometry
inequalities
geometric transformation
rotation
complex numbers
triangle inequality

Let the diagonals of a quad be m,n and the sides be a,b,c,d in an order.
The angles are A,B,C,D as per convention..

Prove that:
(mn)^2 = (ac)^2 + (bd)^2 - 2(abcd)cos(A+C)

[hide]Please dont use complex numbers as it becomes a next-to-nothing prob with that....
Use pure geometry (which I am too lazy to pursue) and thus, did it by complex geometry...[/hide]

yeah indeed by cosine rule and ptolemy theorem it comes with a lil trig juggling.

[quote="gauravpatil"]just use the ptolemys proof that involves construction of a similar triangle
and use cosine rule instead of the triangle inequality :D  :D
just uses minor trig but most of it is pure
i personelly feel that this proof is far more simple than the complex one[/quote]

Its for you to decide..

My proof was sth like...

Let the four vertices be 0,Z1,Z2,Z3...

We just have to prove that the complex numbers a*b,c*d,m*n form a triangle and angle between a*b,c*d is (A+C) which can be trivially done by using the fact that rotation means (e^(ix)) for anticlockwise direction....

i'm sorry i hurt you
it was just a personel opinion
take that easy :wink:   :wink:

We can also derive Brahmagupta's expression for the area of a quadrilateral. That too resembles this form.

[quote="gauravpatil"]i'm sorry i hurt you
it was just a personel opinion
take that easy    :wink:[/quote]

You hurt me..??
You are surely joking.... :rotfl:  :rotfl:

[quote="Euclidean Geometer"]and to add to ashwath, we can even derive expressions for the lengths of the diagonals of a cyclic quad.  :D[/quote]

And to add to that we could write a whole treatise on geometry called Soumyashant's Elements starting from this theorem as the axiom.... :rotfl:  :rotfl:  :rotfl:


geometry
3D geometry
FTW
frustum
Pythagorean Theorem

I don't get this problem because i fail at geometry
 [geogebra]780827281f68b00c502d5bbc58b2db98f12fe623[/geogebra]
pretend that is a megaphone or cone without a top

the height is 18cm
and radius of the bottom is 10cm
and the radius of the top is 1cm
how do you get the lateral area?
please explain how
thanks!  :)

hmmmm im not that good at geometry either but if u just ask pl on ftw they should be able to explain it.

I know how to do the volume, which is $ 594\pi$ cubic centimeters, but, sorry, can't help with the lateral area.

[hide]The formula for lateral area of a cone is $ \pi rl$, where $ r$ is the radius and $ l$ is the slant height.

[img]http://img212.imageshack.us/img212/9206/coneui9.png[/img]

For this question, you can find the lateral area you want by finding the lateral area of the larger cone with radius 10 and subtracting the lateral area of the smaller cone with radius 1.

To find the slant height, set up a right triangle with the radius and the height as the two legs and the slant height as the hypotenuse.

Since the height of the smaller cone isn't given, we can find it using ratios.

Let $ x$=the height of the smaller cone.

$ \dfrac{x}{x \plus{} 18} \equal{} \dfrac{1}{10}$
$ 10x \equal{} x \plus{} 18$
$ 9x \equal{} 18$
$ x \equal{} 2$

The height of the smaller cone is 2, so the height of the larger cone is 18+2=20.

Now, to find the slant height of the larger cone, form a right triangle using the height and radius. A right triangle with legs 20 and 10 is formed. Using the Pythagorean Theorem, the slant height of the larger cone is found to be
$ 20^2 \plus{} 10^2 \equal{} l^2$
$ 400 \plus{} 100 \equal{} l^2$
$ 500 \equal{} l^2$
$ l \equal{} \sqrt {500}$
$ l \equal{} 10\sqrt {5}$

Using the same thing with the smaller cone, the smaller cone is found to have a slant height of
$ 2^2 \plus{} 1^2 \equal{} l^2$
$ 4 \plus{} 1 \equal{} l^2$
$ 5 \equal{} l^2$
$ l \equal{} \sqrt {5}$

Now, we plug the slant height and radius into the formula for lateral area of a cone, $ \pi rl$.

For the larger cone,
$ LA \equal{} \pi * 10 * 10\sqrt {5}$
$ \equal{} 100\pi \sqrt {5}$

For the smaller cone,
$ LA \equal{} \pi * 1 * \sqrt {5}$
$ \equal{} \pi \sqrt {5}$

To find the lateral area of the frustum, subtract the two.
$ 100\pi \sqrt {5} \minus{} \pi \sqrt {5}$
$ \equal{} \boxed{99\pi \sqrt {5}}$
[/hide]

thank you!
thank you!
thank you!!!!!
 ^_^ :D


linear algebra
matrix
superior algebra
superior algebra unsolved

Find all zero divisors, units and nilpotents in $ \mathbb{Z}_{12}[i]$.

Ehm, sorry, I can only state the nilpotent elements of $ {Z}_{12}[i]$
[hide="Hint"]
Note that the ring given is isomorphic to the set of matrix $ [{m}_{ij}],{m}_{ij} \in {Z}_{12}$ with $ {m}_{11} = {m}_{22}$ and $ {m}_{21} = - {m}_{12}$.
We will prove the following lemma : If $ R$ is isomorphic to $ R'$, then the amount of nilpotent elements in $ R$ is equal to the amount of nilpotent elements in $ R'$
Proof : Supposed $ f$ is the isomorphism between $ R$ and $ R'$. Supposed $ x$ is a nilpotent elements in $ R$, then $ x^n = 0$ for some natural number $ n$, then $ f(x^n) = 0$. Conversely, if $ x'$ is a nilpotent elements in $ R$, then there exist natural numbers $ n$ such that $ {x'}^{n} = 0$, then $ x' = f(x)$ for some $ x \in R$, then $ 0 = f(x^n)$ and $ x^n = 0$. Now, note that, if $ M$ is the nilpotent elements in the set of matrix I have stated above, then $ (det(M))^n = 0$ for some natural number $ n$. The condition above can only satisfied when $ det(M) = 0$ or $ det(M) = 6 \in {Z}_{12}$.  The entries of the matrix can only be come from the set {$ 9,3$} or $ {6,0}$} (In other words, if $ 6$ is an entry of the matrix, the other entry can only be $ 6$ or $ 0$. I have obtained that $ 3 + 9i$ and $ 9 + 3i$ and $ 3 + 3i$ and $ 9 + 9i$ and $ 6i,6,6 + 6i$ and $ 0$ are nilpotent elements, so we have only $ 8$ elements.
[/hide]


In a game, you roll a standard six-sided die until you roll a $1$, $2$, or a $6$. If the game ends with a $6$, you win no money. If the game ends with a $1$ or a $2$ after $n$ rolls, you win $\$n$. Per game, what is the expected number of dollars you win?

[hide]

Is the answer 4/3?

My work is very complicated, and I don't want to post it unless it's right.

[/hide]

EDIT:  I didn't see minsoens' post when I posted.  But I did basically the same thing.

[hide]
You see there's a $\frac13$ chance that you get 1 dollar, $\frac13 \cdot \frac12$ chance you get 2 dollars (since there's a $\frac12$ chance you don't get 1 2 or 6 the first time), and $\frac13 \cdot \frac12 \cdot \frac12$ chance you get 3 dollars, etc.
So we can write the expected value as $\left( \frac13 \right) \left( 1 + 2 \left( \frac12 \right) + 3 \left( \frac14 \right) + 4 \left( \frac18 \right) + \cdots \right)$
In general here's how we can solve the sum $1 + 2n + 3n^2 + 4n^3 + \cdots$
Set it equal to $S$.  Notice that multipling $S$ by $n$ makes the coefficients of each degree one less than it is in $S$.  Therefore $S - Sn = 1 + n + n^2 + n^3 + \cdots \implies S = \frac{1}{(1-n)^2}$
Using this, we find the solution is $\left( \frac13 \right) \left( \frac{1}{(\frac12)^2} \right) = \frac43$
[/hide]

Nice solutions. Much nicer than the one I got. :D


If $ A$,$ B$,$ C$ are angles of a triangle,prove that:
$ 2(sinAsin\dfrac{A}{2}\plus{}sinBsin\dfrac{B}{2}\plus{}sinCsin\dfrac{C}{2})\geq sinA\plus{}sinB\plus{}sinC$ :(


geometry
inequalities
inequalities proposed

[quote="xzlbq"][size=150]Let $ ABC$ is a triangle,$ w_a, w_b, w_c$ are bisectors of angles,$ h_a, h_b, h_c$ are altitudes respectively, $ r$ is radius of the incircle

$ h_a \plus{} h_b \plus{} h_c \geq 3(\frac {(s \minus{} b)(s \minus{} c)w_a}{w_bw_c} \plus{} \frac {(s \minus{} c)(s \minus{} a)w_b}{w_cw_a} \plus{} \frac {(s \minus{} a)(s \minus{} b)w_c}{w_aw_b}) \geq 9r$

BQ[/size][/quote]

i can prove it by Muirhead theorem :lol:


Hi, I'm in a bit of a pickle here. I need help stating the degree of the following cubic function:

p(x) = 3x^5 - 6x^4 + 2x^3 - x^2 + 10

Thanking you in advance,

QuadraticFormula :lol:

Edit: Paste my location into your run box to see where I live!

It is in the 5th degree? Not a cubic function?

I worked it out.

Its just the highest power which in this case is 5 from 3x^5.

Can't beleive I didn't see it.


find all function$f:N\rightarrow N$ which that:
$(m+n)f(m^2+n^2)=mf(m)+nf(n)$


The Rainbow Game is played by a team of seven. Each player gets a hat, which can be any one of the seven colors red, orange, yellow, green, blue, indigo, and violet. The colors of the hats are independent of each other and repetitions are allowed: for instance, it may happen that all the hats are green. Each player can see only the colors of the six hats worn by the rest of the team; no player can see the color of his or her own hat. The players are to guess the colors of their own hats, and if at least one player guesses correctly then the team as a whole wins.

The players may not communicate in any way during the game, and they must all announce their guesses simultaneously. They are, however, allowed to plan out a strategy in advance, and they hope to find a strategy which will guarantee them success for every possible arrangement of hats. Is there such a strategy?

A very nice (and difficult) problem! Don't be too quick to say no. For you skeptics, [hide] Consider smaller cases. For two people, what happens if A guesses B's color and B guesses the opposite of A's color? [/hide]

[quote="SM4RT"]How would you do that with 7 people though...like would you pair them up in {2,2,3}..so the people who are paired in two's would say the colour of the hat of their partner (as Phelpedo suggested) and the remaining three (lets say A,B,C) do this:
A guesses B's, B guesses C's, C guesses A's...could that work??[/quote]
You couldn't be farther from the solution. Obviously that could not work. First of all, with two people, the trick was for person A to guess B's color, and for person B to guess [i]the opposite[/i] of A's color. Secondly, that could only work if there are only two colors. With seven colors, two people cannot do it.

[hide]
name the people from A to G, and name the colors from 0 to 6. From here on, all calculations are mod 7.
The trick is to see that each guess is an equation. If B guesses 5, this forms the equation B=5. We are looking for seven equations, such that [i]always at least one[/i] equation is correct. This is actually very easy to find:
$A+B+C+D+E+F+G=0$ or
$A+B+C+D+E+F+G=1$ or
$A+B+C+D+E+F+G=2$ or
$A+B+C+D+E+F+G=3$ or
$A+B+C+D+E+F+G=4$ or
$A+B+C+D+E+F+G=5$ or
$A+B+C+D+E+F+G=6$ mod 7.
rearranging, we get
$A=0-B-C-D-E-F-G$ or
$B=1-A-C-D-E-F-G$ or
$C=2-A-B-D-E-F-G$ or
$D=3-A-B-C-E-F-G$ or
$E=4-A-B-C-D-F-G$ or
$F=5-A-B-C-D-E-G$ or
$G=6-A-B-C-D-E-F$.
Then A guesses $0-B-C-D-E-F-G mod (7)$, B guesses $1-A-C-D-E-F-G mod (7)$, C guesses $2-A-B-D-E-F-G mod (7)$, D guesses $3-A-B-C-E-F-G mod (7)$, E guesses $4-A-B-C-D-F-G mod (7)$, F guesses $5-A-B-C-D-E-G mod (7)$, and G guesses $6-A-B-C-D-E-F mod (7)$. Always at least one of the people will be right.[/hide]

I got basically the same solution as shalevbd.

[hide]Assign each color a number mod 7. The sum of the 7 numbers is a number mod 7. Assign each person a number mod 7. When they see the hats of the others, sum up the colors of those hats mod 7. Have each person announce the difference of the number he was assigned and the number he summed.

Each person's answer would result in a different total sum mod 7. There are 7 possible sums. Therefore, one of them must have been working with the correct total sum and would give the correct color of his own hat.[/hide]

ya sorry i thought the team got more then one guess...so my bad..

For reference, the source for this problem was the Mathcamp 2006 Qualifying Quiz.  There are quite a few other "fun" problems on there, though make sure you don't discuss the 2007 qualifying quiz on this forum when it comes out.  The rainbow game one was definitely my favorite, but try numbers 6 and 7 too.

http://www.mathcamp.org/quiz/quiz06.php


algorithm
rotation
number theory
divisibility tests

Here are some easy number theory and counting questions, which I am having trouble on. Can someone please help me?

1. Can a number written with one hundred 0's, one hundred 1's, and one hundred 2's be a perfect square?

The answer to this question does not seem to answer the question. It says: "The sum of the digits of any number such as described in the problem is 100(0 + 1 + 2) = 300, which is divisible by 3 but not 9. This, then, must be true of the number in the problem, regardless of the order in which its digits appear.

2. Using Euclid's algorithm,

A.) Find gcd(2^100 - 1, 2^120 - 1)

B.) Find gcd(111...111,11...11), (where there are one hundred 1's in the decimal rep of the first number and sixty 1's in the decimal rep of the second number.

Perhaps I am dividing them wrong, because I don't get the answers 2^20 - 1 and 111...11 (twenty 1's) respectively. Can someone show me the division?

3. How many wyas are there to put 12 white and 12 black checkers on the black squares of a chessboard?

The answer is (32 C 12)(20 C 12). However, shouldn't the answer be (1/4) that because you can rotate the board?

4. A person has 6 friends. Each evening, for 5 days, he or she invites 3 of them so that the same group is never invited twice. How many ways are there to do this?

Can someone verify if the answer should be P((6 C 3),5)?

5. How many ways are there to put seven white and two black billiard balls in nine pockets? Some of the pockets may remain empty and the pockets are considered indistinguishable.

My answer is (15 C 8)(10 C 8); the correct answer is supposed to be (16 C 8)(10 C 8). Can anyone tell me why?

6. How many ways are there to choose 6 cards form a complete deck of 52 cards so that all four suits will be present?

What is wrong with my answer: (1/6!)[(13 C 1)^4](48 C 4)?

[quote="mathnovice"]Here are some easy number theory and counting questions, which I am having trouble on. Can someone please help me?

1. Can a number written with one hundred 0's, one hundred 1's, and one hundred 2's be a perfect square?

The answer to this question does not seem to answer the question. It says: "The sum of the digits of any number such as described in the problem is 100(0 + 1 + 2) = 300, which is divisible by 3 but not 9. This, then, must be true of the number in the problem, regardless of the order in which its digits appear.

[/quote]

Yes.  Remember, a number which has the sum of its digits divisible by 3 IS divisible by 3 (I can prove this if you want).  However, it then must be divisible by 9 to be a perfect square, but the condition for divisibility by 9 is ALSO sum of digits divisible by nine.  Hence such a number is necessarily not a perfect square.

Where did these questions come from by the way?


Given $a\neq 0$, find all function $f: \mathbb{R}\to\mathbb{R}$ such that: $f(x)+f\left(\frac{a^{2}}{a-x}\right)=x$  $\forall x\neq a$

There is no such function, unfortunately.  The approach is the standard one for cyclic functions; in this case, if $g(x) = \frac{a^{2}}{a-x}$, we have $g(g(g(x))) = x$, and this allows us to make two substitutions to get a linear system in $f(x), f(g(x))$ and $f(g(g(x)))$ which we can solve.  However, the unique resulting function is not defined at $x = 0$.

[quote="JBL"]if $g(x) = \frac{a^{2}}{a-x}$, we have $g(g(g(x))) = x$, and this allows us to make two substitutions to get a linear system in $f(x), f(g(x))$ and $f(g(g(x)))$ which we can solve.[/quote]

I really don't understand what's you say. Can you explain more detail?

I think actually that my final deduction (no such function exists) may be wrong.  But the method of solution to find candidate functions is fine:  For any $x \neq 0, a$, substitute in $x \mapsto \frac{a^{2}}{a-x}$, which gives you another good functional equation.  We can then make the same substitution again.  This will give us three linear equations in the unknowns $f(x), f(g(x))$ and $f(g(g(x)))$ (where again $g(x) = \frac{a^{2}}{a-x}$) which we can solve for $f(x)$.  Then, after that, we have to figure out what $f$ can do when $x= 0, a$.

Yes, that is a right subtituing. I will rewrite solution more detail:
From the condition of problem: $f(x)+f\left(\frac{a^{2}}{a-x}\right)=x$ (1)
Subtitute $x$ by $\frac{a^{2}}{a-x}$ into above relation, we get:
$f\left(\frac{a^{2}}{a-x}\right)+f\left(\frac{ax-a^{2}}{x}\right)=\frac{a^{2}}{a-x}$ $\forall x\neq 0$ (2)
Put $x$ by $\frac{a^{2}}{a-x}$ in the place of (2), we have:
$f\left(\frac{ax-a^{2}}{x}\right)+f(x)=\frac{ax-a^{2}}{x}$  $\forall x\neq 0$(3)
Combine (1), (2) with (3) infering:
$f(x)+f\left(\frac{a^{2}}{a-x}\right)+f\left(\frac{ax-a^{2}}{x}\right)=\frac{1}{2}.\left(x+\frac{a^{2}}{a-x}+\frac{ax-a^{2}}{x}\right)$
Therefore: $f(x)=\frac{1}{2}\left(x-\frac{a^{2}}{a-x}+\frac{ax-a^{2}}{x}\right)=\frac{1}{2}.\frac{x^{3}-a^{2}x+a^{3}}{x(x-a)}$ $\forall x\neq 0$ and $f(0)=-f(a)$


Εστω ακολουθία που ορίζεται από $ a_1\equal{}1, a_2\equal{}3$ και $ a_n\cdot a_{n\plus{}2}\equal{}a^2_{n\plus{}1}\plus{}8$.
Να βρεθεί το $ a_n$ (κλειστός τύπος).

Εχει και δεύτερο ερώτημα που θα βάλω αργότερα!

Μήπως το άλλο ερώτημα είναι ότι το $ (\minus{}1)^n\plus{}a_n$ είναι τετράγωνο φυσικού άριθμού?

[quote="r_boris"]Μήπως το άλλο ερώτημα είναι ότι το $ ( \minus{} 1)^n \plus{} a_n$ είναι τετράγωνο φυσικού άριθμού?[/quote]
Αν ναι δες εδώ
http://www.mathematica.gr/forum/viewtopic.php?f=50&t=201

Ναι αυτό ήταν :wink:  Πολύ ωραίες οι αποδείξεις κ. Μπόρη!


I need help for this mathcounts problem

Square ABCD has side lengths of 13 units. Point E lies in the interior of the square such that AE = 5 units and BE =12 units. What is the distance from E to side A? Express your answer as a mixed number.
_________________
'' " Do or do not there is no try" - Yoda ''

by copy i meant for a mod to move not literally copy...


Let $ C_f$ be the mapping cone of a map $ f: X \rightarrow Y$.    Find a long exact sequence relating the reduced homology groups of $ X$, $ Y$, and $ C_f$.

[i]Hint: Use Mayer-Vietoris Theorem[/i]

I guess $ C_f$ should be considered as the space written as the union of two open subspaces in order to use Mayer-Vietoris Theorem, but I can't see which such open subspaces I should take.

Thanks

I'm pretty sure that it should work nicely if one is everything minus the "top point" of the cone, since this will def. retract to Y, and the other one is just the "cone" part, which is simply connected. The intersection will def. retract to a cylinder.

(This was intuitive, and maybe the details don't work out that nicely).

Thanks a lot for your answer, but I don't see where the homology of $ X$ shows up in your idea.

Oops. I guess if I express more formally what I meant by "just the cone part" then I would see that the intersection is homotopy equivalent to $ X$.

Take $ A\equal{}C_f\setminus \{(x,1)\}$ and $ B\equal{}CX\setminus \{(x,0) : x\in X\}$.

Then we get the long exact sequence $ \cdots \rightarrow H_{n\plus{}1}(C_f)\rightarrow H_n(X)\rightarrow H_n(Y)\oplus 1\rightarrow H_n(C_f)\rightarrow \cdots \rightarrow H_0(Y)\rightarrow H_0(C_f)\rightarrow 0$

Where the maps are the standard ones.

ooh I see. It wasn't so hard, isn't it?

Thank you  :oops:

I had to solve this problem, and I looked up my notes on it. I'd say the hard part is not only coming up with those sets, but then actually rigorously proving they are homotopy equivalent to what I said they are. Usually that is the hard part of Mayer-Vietoris. The actual application is usually easy once you have those details.

But you're right, it isn't all that hard in this case anyway. I think all of the deformation retracts can be done with "straight line homotopies."


Cea mai cunoscuta forma a teoremei de medie este urmatoarea:
Daca f:[a, b]->R este integrabila pe [a, b], atunci exista $\lambda\in[m,M]$, unde m=inf f, M = supf, astfel ca $\int^{b}_{a}f(x)dx=\lambda(b-a)$
Daca f e continua, atunci exista $c\in(a, b)$ astfel ca $f(c)=\lambda$, deci $\int^{b}_{a}f(x)dx=f(c)(b-a)$. Se pune intrebarea: Daca f are proprietatea lui Darboux(dar nu continua!), ramane valabila $\int^{b}_{a}f(x)dx=f(c)(b-a)$?? Mie mi s-ar parea logic, deoarece f ia toate valorile intre f(a) si f(b), deci ia si pe $\lambda$. Insa niciunde nu zicea chestia asta, desi pare mai generala decat cea cu continuitatea.

Cred ca nu e valabila decat pt continuitate dar nu iti pot da un contrexemplu :D

Intr-adevar, teorema de medie este adevarata intr-o ipoteza mai tare si anume: $f$ este integrabila si are proprietatea lui Darboux. De fapt nu trebuie ca $f$ sa aiba neaparat proprietatea lui Darboux, ci o functie pe care o construim in prealeabil. Oricum chestii de genul acesta pot fi gasite in cartea:  "Probleme de Analiza matematica-Calcul Integral", autor Tudorel Lupu,ed. Gil, Zalau, 1996. ;)


i think this is a part of a problem from some imo.....very elemenatry
In $ \Delta ABC$, $ m\angle A\equal{}60$.$ AP$ bisects $ \angle BAC$ and $ P$ lies on $ BC$.$ BQ$ bisects $ \angle ABC$ and $ Q$ lies on $ AC$.Extend $ ray CB$ to $ A'$ so that $ AB\equal{}A'B$.Extend $ ray BQ$ to $ A''$ so that $ AQ\equal{}A''Q$.Also, $ A''B\equal{}A'P$.Prove that $ AA''$ parallel to $ A'B$

mmmmm... I don't know what I am doing wrong:

If extend ray$ CB$ means extend $ CB$ from $ C$ to $ B$ then:
$ AA'$ is parallel to $ BQ$ and so to $ BA''$.
If $ AA''$ is parallel to $ A'B$ then $ A'BA''A$ is a parallelogram: $ c\equal{}AB\equal{}A'B\equal{}AA''\equal{}AQ\equal{}\frac{bc}{a\plus{}c}$
Thus $ a\plus{}c\equal{}b$, so there is no triangle.... :maybe:

Where is the mistake?

 :)

[quote="ElChapin"]$ AQ \equal{} \frac {bc}{a \plus{} c}$

[/quote]

 :?: How?

Because $ BQ$ bisects $ \angle ABC$ and $ Q \in AC$ : $ \frac {AQ}{QC} \equal{} \frac {AB}{BC}$ (Teorema de la bisectriz, I don't know how to say "bisectriz" in english)

 :)

From your equation it looks like you're referring to the angle bisector theorem.


A------------||--------wwwwww-------------||------------wwwwww----------B
                       6V          2 ohms                  9V                 .7 ohms

Find the potential difference between the points A and B
Here the source is a battery

a)5.1V          b)51V            c).51V           d)1.5V

[quote="phucnv87"][color=blue]I can't understand your picture[/color][/quote]

here the first point is A.It is connected to a source battery (cell) of 6V which is again

connected to a resistance of 2 ohms.It is connected to a cell of 9V which is again

connected to a resistance of .7 ohms.It is now ended at a point(terminal) B.

These all are in a straight line .Now the question

is ,find the potential difference between A and B

now i think you can understand the picture

[quote="bus"]Uhm, and what are points A and B attached to? If they're not attached to anything, shouldn't it be just 6V + 9V = 15V ?[/quote]

i think you are correct but ,will the current flow in the 2 ohm resistor i.e from
9V cell to 6V cell as the positive terminal of 9V cell is connected to negative terminal of
6V cell

will any current pass through the given circuit?

No. Unless you give it a reason to flow. The point is - current flows from points of lower potential to points of higher potential but the potentials on either ends of a battery are not fixed. If you have a 6V battery you don't know what the potentials on its terminals are, you only know that they differ by 6V. So, let's set the potential at point A to 0V. We can always choose the potential at one point, it's like choosing a frame of reference - choosing a point with zero potential energy. Then the potential right past the first battery is 6V. Resistors do nothing unless there's some current flowing through them so the potential doesn't change and is still 6V right before the second battery. The potential past the second battery is 15V and the second resistor again doesn't do anything so the potential at point B is 15V.

If there were some current flowing through the circuit the situation would be different - but electric current usually doesn't flow from nowhere to nowhere, you'd have to connect points A and B to something so that new electrons can enter and leave the circuit, then measure the resistance between points A and B and only then figure out the right answer. In this case the resistance between A nad B is infinity so there's no current.


This is probably a very simple question, but I dont know how to handle it.
Suppose you have a normal distribution with mean 100 and standard deviation 3
How do you find in which precentile 120 is?
Please dont just give me the result, but show me the method!

I think you have to look at its Gaussian distribution.  One standard deviation will correspond to some percentile, so in this case, you're looking at about $ \frac{20}{3}\approx6.67$ standard deviations.

Following up on what JRav said: $ 6.67$ standard deviations is HUGE.

For the normal distribution, if $ z$ is large, we can approximate

$ P(Z > z)\approx\frac {e^{ \minus{} z^2/2}}{\sqrt {2\pi}z}.$

Putting $ z \equal{} \frac {20}3$ into that expression, we get

$ P(Z > 6.67)\approx 3\times 10^{ \minus{} 11}.$

Hence we could say that this is deep, deep, deep into the 99th percentile - about the $ 99.999999997$th percentile.

Of course, the normal distribution in its exact form does not exist in nature - everything that looks normal is a consequence of the Central Limit Theorem. And this far away from the mean, the CLT does start to break down as a reasonable approximation. So don't trust that number very far - certainly, don't trust it with that precision.

---

Now let's suppose you had asked a more reasonable question. Say the mean is 100, the standard deviation is 10, and your point is 113. That's 1.3 standard deviations above the mean. Now we need a normal distribution table. I've got one in the "outside probability" form. It tells me that $ P(Z > 1.3)\approx .0968.$ Subtract that from 1 to get $ .903,$ round a little, and we'll call this the $ 90$th percentile.

Thanx a lot for your help!
I really appreciate that!

Kent Merryfield, I'm not JBL but you referenced me as him :lol: .

[color=green][Edited - K.M.][/color]


Lynn and Cece play tic-tac-toe on a standard 3 by 3 board. They each make 3 moves before getting bored with the game and stopping. If they move completely at random (taking each turn in a random unused square), what's the probability that there is no winning set (of three in a row horizontally, vertically, or diagonally) when they stop playing?

Um.... the numbers don't seem to be very pretty. I got $ \displaystyle \frac{5468}{6561}$, which is almost guaranteed to be wrong, but...

[quote="leoxnlin"]Um.... the numbers don't seem to be very pretty. I got $ \displaystyle \frac {5468}{6561}$, which is almost guaranteed to be wrong, but...[/quote]
A solution would be appreciated.

Here is my attempt. Sorry if its not very clear.
There are 9!/6!3!=84 ways to place an X, and 6!/3!3!=20 ways to then place an O. 84*20=1680 ways to place 3 X's and 3O's. There cannot be 3 X's in a row and three O's in a row because somebody would have won on the previous turn and there would have been only 5 X's and O's on the board. There are 12 ways to have two winners, so 1680-12=1668. To count the number of times that have no winner, we subtract the number of times when there is a winner from the total amount of possibilities. If there is a line of three, there are 6!/3!3!=20 ways to fill the rest in. For columns and rows though, there are two ways that produce two winners which cannot happen. 20-2=18*6=108. If the three X's or O's are on a diagonal, the other three cannot be in a winning formation, so 20*2=40. 108+40=148*2(To cover both X's and O's winning)=296. 1668-296=1372. 1372/1668=[b]343/417[/b]
I could be wrong though.

[quote="Bacteria"]Here is my attempt. Sorry if its not very clear.
There are 9!/6!3!=84 ways to place an X, and 6!/3!3!=20 ways to then place an O. 84*20=1680 ways to place 3 X's and 3O's.[/quote]
Sorry? I don't think that part is quite right.

EDIT: By the way, the answer I got is 4/21. I used essentially the same method as bacteria, but counted differently.

I assumed that if somebody won before the other person placed his third letter the arrangement was not valid. I did it again allowing this and got a 320/1680 probability of someone winning, or a 1360/1680=17/21 probability of neither of them winning.

Yes I did it like Bacteria and got 17/21.Was temperal counting the probablity someone wins?

I think Temperal meant to take the complement and then forgot later. That would explain the solutions being 4/21 and 17/21.


[quote="hjoon0125"]uh, whats the point of this thread?????[/quote]
the point of this thread is to trick you into unneccissarily posting here.

oh darn i fell for it too!!!

O NO ITS A USELESS THREAD!!!!!
 :|  :spam: :D

maybe we can make use of it by turning it into a discussion thread of something?

Since the name of this topic is ................ Lets talk about DOTS!

Dots. This word is a favorite for many people. What comes to your mind when you think of dots? When I think of dots many things come to my mind. From the game Dots, to the tasty Ice Cream Dots! Mmmm...The dots ice cream...tasty..icy...creamy....huh? Oh wow..I was day dreaming about icecream. The game dots is extremely fun and addicting. Many people play it. (If you dont know what I'm talking about its the game where you connect dots to make squares, and you play against any number of opponents..but the objective is to have the most number of squares.) Now..speaking of the game dots...another fun thing comes to my mind. We all know the classic "Connect the Dots" game. You just connect the dots to make a figure! It may be boring now, but we all had fun playing it when we were kids. If you really think about it...dots are a big thing in all of our lives. We all need dots. For example, I need a dot now because I have to end this sentence with a dot.

Or we could try to write posts without dots. :)

FORBIDDEN:

.?!;:ij
all smilies


For $i = 1,2, \ldots, 1991$, we choose $n_i$ points and write number $i$ on them (each point has only written one number on it). A set of chords are drawn such that:

(i) They are pairwise non-intersecting.
(ii) The endpoints of each chord have distinct numbers.

If for all possible assignments of numbers the operation can always be done, find the necessary and sufficient condition the numbers $n_1, n_2, \ldots, n_{1991}$ must satisfy for this to be possible.


I was trying to write a school essay when I came up with this game.  As I thought about it, I wasn't sure how to solve it, or what level difficulty it was, or what's a good question to ask about it.  Here is the problem/game:

You are given a $100$ x $100$ grid and a 7 letter word, such as $Peanuts$.  Only one letter can fit per square.  Also, each letter in the word must be in a square on the board and must be in the correct order $Peanuts$.  A move is defined as randomly selecting a letter from the word $Peanuts$ and rotating the entire word about that letter (that letter acts as a fulcrum or pivot point and hence, does not change squares) either $90$ degrees clockwise or counter-clockwise.  You win the game by placing the letter $P$ in $Peanuts$ on any of the four corners.  How many possible starting positions are possible where it is possible to win the game?

What are the requirements for the initial position and orientation?

Hopefully, this answers your question.

The word must stay in the same order, though it can be reversed.  Example

P e a n u t s

If I select $n$ and rotate the word 90 degrees to the right, then I get...

- - - P - - -
- - - e - - -
- - - a - - -
- - - n - - -
- - - u - - -
- - - t - - -
- - - s - - -

Select $u$ now and rotate 90 degrees to the right again

- - - - - - - -
- - - - - - - -
- - - - - - - -
- - - - - - - -
- s t u n a e P
- - - - - - - -
- - - - - - - -

The inital position can be anything and orientated anyway, however, the word must be either vertical or horizontal at all times (no diagonal) and the word cannot start with the first letter in a corner.

Does this answer your question?


Suppose that $f_{1},...,f_{n}$ are $n$ real positive numbers and $g_{1},...,g_{n}$ are $n$ positve integers which are constant and $x_{1},...,x_{n}$ are $n$ real positive numbers. Also assume that  $(x_{1}^{g_{1}})...(x_{n}^{g_{n}})$ is constant. Find the minimum amount of $f_{1}x_{1}+...+f_{n}x_{n}$.

This problem and a similar problem posted on [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=405955#405955[/url] are both solved using the same inequality; I shall prove that inequality:
First of all because for every $i\in{\{1,2,...,n}\}, g_{i}$ is a positive integer, we have:

$\sum_{i=1}^{n}f_{i}x_{i}=\sum_{i=1}^{n}\sum_{j=1}^{g_{i}}\left(\frac{f_{i}}{g_{i}}x_{i}\right)$

Now by applying the AM-GM inequality to the right hand side, we get:

$\sum_{i=1}^{n}f_{i}x_{i}\geq \left({\sum_{i=1}^{n}g_{i}}\right) \left(\prod_{i=1}^{n}{\left(\frac{f_{i}}{g_{i}}x_{i}\right)}^{g_{i}}\right)^{1/({\sum_{i=1}^{n}g_{i}})}$


inequalities
function
trigonometry
calculus
derivative
Putnam
college contests

These are from Larsen 7.4. Again, just give hints if you can.

1) Prove that

$ \left(\frac {a \plus{} 1}{b \plus{} 1}\right)^{b \plus{} 1} > \left(\frac {a}{b}\right)^b$

for $ a,b > 0$ and $ a \neq b$.

2) For positive numbers a,b,c such that $ 1/a \plus{} 1/b \plus{} 1/c \equal{} 1$, show that $ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1) \geq 8$.

3) For all x>0, (2+cos x) x > 3 sin x

a) Prove this inequality by considering the function $ F(x) \equal{} x \minus{} (3 \sin x)/(2 \plus{} \cos x)$.

b) Prove this inequality by considering the function $ F(x) \equal{} (2 \plus{} \cos x) x \minus{} 3\sin x$.

[b]Attempt at solution:[/b]

1 is just AM-GM when b is rational and I can get  for real b using the density of the rationals but we need strict inequality and I am not sure how to get that.

2 is clearly an application of AM-GM, but I am not sure how exactly.

For 3, it is not hard to show that we need only consider 0<x<3.

[hide="2"]For 2:

Note that $ abc \equal{} ab \plus{} bc \plus{} ca$
Use AM-HM[/hide]

[quote="iniquitus"]For 2:

Note that $ abc \equal{} ab \plus{} bc \plus{} ca$
Use AM-HM[/quote]

OK, so I get that the inequality in the problem statement is equivalent to

$ a\plus{}b\plus{}c \geq 9$

and I still can't figure out how to apply AM-GM.

not AM-GM, AM-HM:

$ \frac{a\plus{}b\plus{}c}{3}\geq GM \geq \frac{3}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}}$  :D

Hint for 1
[hide]
Use calculus. Question to everyone, how do you prove this without resorting to calculus?
[/hide]

Further hint.
[hide]
This is how I thought: I bet $ 1$ could be replaced by any positive constant. If it is, then the function $ \left(\frac{a\plus{}x}{b\plus{}x}\right)^{b\plus{}x}$ would be strictly increasing. Now, that means the derivative must be... (at this step, it helps to set $ a\equal{}b\plus{}k$)
[/hide]

[hide=" Alternate Solution for 2"]

Let $ x \equal{} \frac {1}{a}, y \equal{} \frac {1}{b}, z \equal{} \frac {1}{c}$

We are given x+y+z = 1

Then y+z = 1-x
        z+x = 1-y
        x+y = 1-z

Thus $ (1 \minus{} x)(1 \minus{} y)(1 \minus{} z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x) \ge 8xyz$

Substituting back for x, y and z, we get

$ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1) \ge 8$
[/hide]

[quote="iniquitus"]not AM-GM, AM-HM:

$ \frac {a \plus{} b \plus{} c}{3}\geq GM \geq \frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}$  :D[/quote]

Haha I thought that was a typo.

[quote="Alfred"]
This is how I thought: I bet $ 1$ could be replaced by any positive constant. If it is, then the function $ \left(\frac {a \plus{} x}{b \plus{} x}\right)^{b \plus{} x}$ would be strictly increasing. Now, that means the derivative must be... (at this step, it helps to set $ a \equal{} b \plus{} k$)
[/quote]

I am a little confused about your logic.

Let $ f(x) \equal{} \left(\frac {a \plus{} x}{b \plus{} x}\right)^{b \plus{} x}$. I don't think the problem statement implies that f is increasing on ANY interval. If you can show that f is strictly increasing on [0,1], then that will imply the problem statement, but that would also prove much more and it seems to me like that will be harder.

Using calculus is a good idea though because this problem is in the calculus section of chapter 7. But we to figure out what function to consider.

First an apology: My original idea kind of works, but not as I intended. I must have messed up my algebra the first try. There must be a (much) slicker solution...

Set $ a\equal{}b\plus{}k$. Then $ f(x) \equal{} \left(\frac {b \plus{} x\plus{}k}{b \plus{} x}\right)^{b \plus{} x}\equal{} \left(1\plus{}\frac {k}{b \plus{} x}\right)^{b \plus{} x}\equal{}e^{\log(1\plus{}\frac {k}{b \plus{} x})(b\plus{}x)}$. $ f(x)$ is increasing if and only if $ \log(1\plus{}\frac {k}{b \plus{} x})(b\plus{}x)$ is increasing. The derivative of this is $ \log(1\plus{}\frac{k}{b \plus{} x})\plus{}(\minus{}\frac{k}{(b\plus{}x)^2})\frac{1}{1\plus{}\frac{k}{b\plus{}x}}(b\plus{}x)\equal{}\log(1\plus{}\frac{k}{b \plus{} x})\minus{}\frac{k}{b\plus{}x\plus{}k}$.

Now, we first note that $ b\plus{}k\equal{}a>0$. Hence $ b\plus{}x\plus{}k>0$.

How do we show that the above derivative is always positive? Take the derivative with respect to $ k$ (for $ k\equal{}0$, there is an equality).

We get the following derivative (again, wrt to $ k$):
$ \frac{1}{b\plus{}x}\frac{1}{1\plus{}\frac{k}{b\plus{}x}}\minus{}\frac{1}{b\plus{}x\plus{}k}\plus{}\frac{k}{(b\plus{}x\plus{}k)^2}\equal{}\frac{k}{(b\plus{}x\plus{}k)^2}$. Hence for $ k\equal{}0$, the derivative of $ f$ achieves its minimum (note that $ b\plus{}x\plus{}k>0$ by above).

So if the derivative of $ f$ achieves its minimum (with respect to $ k$) when $ k\equal{}0$ (which implies equality), we must have that the derivative of $ f$ is greater than zero.

I like how you differentiate w.r.t k! Anyway you did prove much more than was necessary so there probably is a "slicker" way. But your proof definitely works. One suggestion: you should say that f is defined on $ [0,\infty)$  because that is where you proved it was increasing. Now what about question 3...

[hide="Quick solution to 1"]
For $ b\leq1$ we have the following:
\[ \frac{a\plus{}1}{b\plus{}1}>1\plus{}b\frac{a\minus{}b}{ab\plus{}b}\geq\left(1\plus{}\frac{a\minus{}b}{ab\plus{}b}\right)^b\]
Where the first $ >$ is equivalent to $ (a\minus{}b)^2>0$, and the second is Bernoulli's inequality.
Then, \[ \frac{a\plus{}1}{b\plus{}1}>\left(1\plus{}\frac{a\minus{}b}{ab\plus{}b}\right)^b\iff
\left(\frac {a \plus{} 1}{b \plus{} 1}\right)^{b \plus{} 1} > \left(\frac {a}{b}\right)^b\]

Otherwise, let $ c\equal{}\frac{1}{b}$. Then we must show:
\[ \left(c\frac {a \plus{} 1}{c \plus{} 1}\right)^{c \plus{} 1} > ac\]

for $ 0<c<1$, which is even more straightforward than the other case:
\[ \left(c\frac {a \plus{} 1}{c \plus{} 1}\right)^{c \plus{} 1}\equal{}\left(1\plus{}\frac {ac \minus{}1}{c \plus{} 1}\right)^{c \plus{} 1}>1\plus{}(c\plus{}1)\frac{ac\minus{}1}{c\plus{}1}\equal{}ac\]
Notice we have strict inequality because $ c\plus{}1>1$ and $ ac\not\equal{}1$.

Basically, the two inequalities used were those found on http://en.wikipedia.org/wiki/Bernoulli_inequality under "generalization"
[/hide]


[quote="mathgeniuse^ln(x)"]For any integer n greater than 1, show that $n^{2}-2n+1$ is a factor of $n^{n-1}-1$.

Please Help...[/quote]

$n^{n-1}-1$ can be factored into $(n-1)(1+n+n^{2}+...+n^{n-2})$, and $n^{2}-2n+1$ can be factored into $(n-1)^{2}$, meaning this problem can be reduced to the following:

For any integer n greater than 1, show that $n-1$ is a factor of $(n^{0}+n^{1}+n^{2}+...+n^{n-2})$.

$1$ is equal to $1$ mod $n-1$
$n$ is equal to $1$ mod $n-1$
All of the terms up to $n^{n-2}$ are also equal to $1$ mod $n-1$.

There are $n-1$ terms from $n^{0}$ to $n^{n-2}$, and each of them are equal to 1 mod n - 1, meaning that $(n^{0}+n^{1}+n^{2}+...+n^{n-2})$ is divisible by $n-1$.

[hide="Solution"]Since $n^{2}-2n+1=(n-1)^{2}$, we get

${n^{n-1}-1\over (n-1)^{2}}={n^{n-2}+n^{n-3}+\dots+n+1\over n-1}$

Now look at the denominator modulo $n-1$: All the terms are $\equiv 1\pmod{n-1}$, and as there are $n-1$ of them, the sum is $\equiv n-1\equiv 0\pmod{n-1}$. QED[/hide]

Thanks So much...I was going to approach this using Treething's way, but I could'nt figure out what to do from the step of n-1 being a factor of n^(n-2)+n^(n-3)+...1

For the last step, you can also use the Division Algorithim. For this, you can just plug in 1 for n (not in the exponent). This sum condenses into $n-1$, so the geometric series is divisible by 1 when $n\ne1$.


let $ x,y,z \geq0$ such that $ x\plus{}y\plus{}z\equal{}1$ prove that:

$ \frac{x^3}{x^2\plus{}y^2}\plus{}\frac{y^3}{y^2\plus{}z^2}\plus{}\frac{z^3}{z^2\plus{}x^2}\geq \frac{1}{2}$

it can be proved that  :

$ \frac{a^4}{a^3\plus{}b^3}\plus{}\frac{b^4}{b^3\plus{}c^3}\plus{}\frac{c^4}{c^3\plus{}a^3}\ge\frac{a\plus{}b\plus{}c}{2}$ , also easy .

[quote="alex2008"]it can be proved that  :

$ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\ge\frac {a \plus{} b \plus{} c}{2}$ , also easy .[/quote]
i think use SOS . :roll:

[quote="iqtimo"]
i think use SOS . :roll:[/quote]

Show how .


I remember "Walter" showing up in AHSME problems a few times. Walter Mientka was the director of the AMC. Of course he probably wasn't the person inserting that name; it doesn't really matter. Your friend Gene might have been the inspiration without knowing anything about it himself.

I once taught a freshman algebra course in which the course director wrote the word problems, and he always used "Olaf" and "Anneke." Whatever.

[quote="ZzZzZzZzZzZz"]Does it really matter?[/quote]

No, not really. I'm just curious as to why Gene has been used twice in the same year.

[quote="le00327146"]I'm just curious as to why Gene has been used twice in the same year.[/quote]

I was thinking the same thing when I saw Gene again.

[quote="Erin J. Schram"][quote="le00327146"]This is just curiosity...How do you (the creators of the problems) choose names in the problems? Is it just a coincidence that "Gene" has been used twice already?[/quote]
When I write a math problem that needs a name, I tend to use either my own name, the name of the person who inspired the problem, or a collection of names in alphabetical order (so that the problem-solver may abbreviate them as A, B, C, etc.).

When I spotted the name "Gene" in Problem 1/1/19, I asked Gene Berg, one of the former directors and current graders of the USAMTS, whether he had written that problem. He had not. I haven't asked him about Problem 2/3/19 yet.

Erin Schram
USAMTS grader[/quote]

That was indeed in honor of Gene Berg.  Forgot I'd already used his name once this year.  I'll have to switch the next one to Blair.  Or Erin :)


You are given a closed curve(can be absolutely any closed curve). Prove that if you take any arbitrary point inside the curve, there always exists a chord of the curve with that point as its mid-point.

Note that a chord is a line segment joining two points on a curve.
??

[hide]
This isn't as much an actual proof as just reasoning. I don't know if it works for all closed curves, either.

Let $P =$ the arbitrary point. Now locate the point on the curve that is closest to $P$; call it point $A$. Draw the chord through $P$ and $A$, and call the point on the other side of the chord point $B$. Clearly, since $A$ is the point closest to $P$, $AP \leq BP$. Now here's the tricky part. What if we redifined point $A$, "sliding" it along the curve and redifining $B$ in the process? This will make $AP$ decrease, until finally, $A$ and $B$ are switched so that $AP \geq BP$. But since the curve is continuous, in the process of going from $AP \leq BP$ to $AP \geq BP$ there must be at least one instance where $AP = BP$.
[/hide]

what if at such an instance you intersect the curve more than once??? He didn't mention that the curve is "convex'".


calculus
integration
real analysis
real analysis unsolved

[quote=misan]can you solve :D
$ \int_{0}^{\infty}\;\;\frac{x\cdot e^{x}}{\left(\sqrt{\;4\,e^{x}\minus{}3\;}\right)\cdot\left(1\plus{}2\,e^{x}\minus{}\sqrt{\;4\,e^{x}\minus{}3\;}\right)}\;\;\textbf dx$[/quote]

\begin{align*}
&\int_{0}^{+\infty}\;\frac{x\,e^{x}}{\left(1+2\,e^{x}-\sqrt{\;4\,e^{x}-3\;}\right)\sqrt{\;4\,e^{x}-3\;}}\;\mathrm{d}x\\
\\
&\qquad\qquad\qquad\quad\dfrac{2G}{3}+\dfrac{\pi\ln(2+\sqrt{3})}{6}
\end{align*}


.........{x-3           if x<= -2
f(x) =..{ -x^2        if -2<x<1
.........{-x+4         if x>=1

I undertand most of this question in terms of graphing but I am having trouble with the -x^2 part.

I understand that the basic function of y = x^2 is a parabola centered at the origin.

I also understand that a negative in front of -x^2 means an upside down parabola picture at the origin.

However, given the side condition on the left side for -x^2, I came up with two points:

(-2, -4) and (1, -1).  When I tried to graph this piece of the piecewise defined function, the picture does not make sense.

Can anyone help me?

The function is not continuous.

It's like 3 separate pictures.

You got your line, it stops abruptly switches to an upside-down parabola, which stops abruptly and starts another line.

[hide="Pixelized graph of function"][img]http://img387.imageshack.us/img387/1466/graphjl2.png[/img][/hide]


function
integration
calculus
real analysis
real analysis solved

Let A be the set of all functions f:[a,b]->[0,00), f=/=0 (f isn't identically null), f(a)=0, f'(x)<=M for a certain M>0 (M is the same for all f in A.

Find max {f in A} [( \int {a->b} (f(x))^3 dx)^1/2]/[ \int {a->b} f(x)dx]

K(f)=( \int  {a->b} (f(x))^3 dx)^1/2]/[  \int  {a->b} f(x)dx
If there is a function g \in A for wich max_{f\in A}K(f) is reach
for any t real near 0, K(g+tf) \leq K(g)
K(g+tf)=K(g)+t(term)+t^2(term involving f,g )+..
g is the maximum of the functional K coefficient of t must be 0,
coefficient of t^2 must be negatif, these coefficients seems complicate.
----------------------------------------------------------------------------
Maybe there is a more elementary proof
f'(x) \leq M=> f(x) \leq M(x-a)
f(a)=0, f take its value in R+ => f'(a) \geq 0
 \int  {a->b} (f(x))^3 dx \leq M^3 \int  {a->b} (x-a)^3 dx=M^3/4(b-a)^4
..

It took me some time to remember this one:

[color=white]

Let F(x)= \int {b->x} f(x)dx. We can see that F(b)=0 and, since f>=0, F is increasing, so F(x)<=0 for all x in [a,b], and F(a)<0 because F isn't always 0.

By part integration formula we have  \int {a->b} f(x) 3 dx = F(x)f(x) 2 | b  a - 2 \int {a->b} F(x)f(x)f'(x)dx = -2 \int {a->b} F(x)f(x)f'(x)dx because f(a)=0 and F(b)=0. Because F has values <=0, it means that -2 \int {a->b} F(x)f(x)f'(x)dx <= -2M \int {a->b} F(x)f(x)dx = -M[F(b) 2 -F(a) 2] = M*F(a)  2 , which means that  \int {a->b} f(x) 3 dx <= M*F(a) (#), with equality iff f'(x)=M for all x in [a,b].

 \int {a->b}f(x)dx = F(b)-F(a)=-F(a), and by combining this with (#) we get that expression in the text of the problem <=  \sqrt M, with equality iff f'(x)=M for all x in [a,b].

[/color]

Hope it's correct...


[quote="nebula42"][quote="bpms"][hide="there is probably a better way but oh well"]If 5 is the last digit then you have 8 choices for the first digit, 10, 10 again, then 1.
800.
For the other way around you have 1,10,10,9=900
1700 numbers...[/hide][/quote]
Sorry, that's not correct.
By the way, I think you misread the problem. It isn't first digit and last digit; it's [i]tens[/i] digit and last (units) digit.[/quote]
Oh, you are right.
[hide]For the tens digit being 5
9x10x1x9=810
For the ones
9x10x9x1=810
1620.[/hide]
How odd that anirudh, who never ever posts a solution at all no matter how simple the solution is, got the exact same wrong answer as I did after me and had no work for his wrong answer...

[quote="anirudh"][quote="nebula42"]How many 4-digit numbers are there such that 5 is either the tens digit or the units digit, but not both?[/quote]

[hide="this problem was somehwre in mathcounts"]
1700---I have seen it somehwere[/hide][/quote]

See, this is why you don't just look at other's people's answer post it without a solution in all the topics just to get your post count up.

why can't 5 be in the thousands place?  :maybe:

[hide]$9\cdot10\cdot9\cdot2=\boxed{1620}$[/hide]

[quote="ajai"]for some reason, i get something different.....correct me if i'm wrong....

[hide]
$8*9*1*9=648 8*9*9*1=648$

FIRST
because 0 or 5 can't be in the thousandths digit, and 5 can't be in the hundredths or units digit

SECOND
same. but 5 can't be in the tenths digit...
[/hide][/quote]
He never said that you have to couldn't have 5 anywhere, just not both...


[b][size=100][color=DarkBlue]Δίνεται τρίγωνο $ \bigtriangleup ABC$ και έστω $ D,$ μεταβλητό σημείο μεταξύ των $ B,\ C.$ Γράφουμε τους κύκλους $ (O_{1}),\ (O_{2}),$ τους περιγεγραμμένους περί τα τρίγωνα $ \bigtriangleup ADB,$ $ \bigtriangleup ADC$ αντιστοίχως, οι οποίοι τέμνουν τις ευθείες $ AC,\ AB,$ τα σημεία $ E$ και $ F.$ Εάν $ M$ είναι το σημείο τομής των $ BE,\ CF,$ αποδείξτε ότι η ευθεία $ MD,$ περνάει από σταθερό σημείο [/color][/size][/b].

Κώστας Βήττας.

K. Βήττα, θα μπορούσατε σας παρακαλώ να βάλετε μια λύση σε αυτό το πρόβλημα? Εχει πολύ ενδιαφέρον σχήμα...(Miquel configuration) κι είναι κρίμα να μείνει άλυτη

[quote="bilot4"]...(Miquel configuration) κι είναι κρίμα να μείνει άλυτη. [/quote]
Δημήτρη, δεν έβαλα αυτό το ''hind'', γιατί νομίζω ότι οδηγεί άμεσα στη λύση.

Συμφωνώ ότι δεν πρέπει να μένουν άλυτες οι ασκήσεις στο φόρουμ αυτό, που για οποιοδήποτε λόγο δεν συζητήθηκαν.

Ίσως αν ορίζονταν ( από τους moderators ? ), ότι μετά από κάποιο χρονικό διάστημα και πέρα, να πρέπει υποχρεωτικά να βάζουν τη λύση τους ( αν έχουν ), όσοι πρότειναν κάποιο πρόβλημα, να βοηθούσε στο ξεπέρασμα των όποιων αναστολών μπορεί να έχουν μερικοί ( κι’ εγώ είμαι ένας από αυτούς ) για κάτι τέτοιο, όταν δεν έχει υπάρξει ανταπόκριση στην πρότασή τους.

Κώστας Βήττας.

Αφείστε σας παρακαλώ την άσκηση μέχρι τη δευτέρα (Βέβαια μην το πείτε σε κανέναν θα με σφάξει η καθηγήτρια της έκθεσης  :P )

**Όσο αφορά τις άλυτες ασκήσεις , μην στενοχωρίστε θα τις ξεθάψω  :wink:  το καλοκαίρι . Καμμιά δεν θα μείνει άλυτη εκτός από την υπόθεση Riemmann που είχε προτείνει κάποιος μυστήριος παλιά

[quote="silouan"]Καμμιά δεν θα μείνει άλυτη [b]εκτός από την υπόθεση Riemmann που είχε προτείνει κάποιος μυστήριος παλιά[/b][/quote]
:what?:  :w00t: Ωραίος ο τύπος :rotfl:

Αυτή η άσκηση είναι πολύ απλή ρε mim.
$ \angle MBD\equal{}\angle DAC\equal{}\angle MFD$ άρα $ BFMD$ είναι εγγράψιμο. Όμοια $ DCEM$ είναι εγγράψιμο. Άρα $ \angle MDC\equal{}\angle MEA\equal{}\angle BDA$.
Φέρνουμε το ύψος $ AZ$ του τριγώνου $ ABC$ με $ Z\in BC$ και θεωρούμε $ X$ το σημείο τομής της $ AZ$ με την $ MD$.
Τότε η $ DZ$ είναι διχοτόμος και ύψος στο τρίγωνο $ ADX$ άρα $ XZ\equal{}ZA$ άρα $ X$ είναι το σταθερό σημείο.

[quote="Vaf"]....
$ \angle MBD \equal{} \angle DAC \equal{} \angle MFD$ άρα $ BFMD$ είναι εγγράψιμο. Όμοια $ DCEM$ είναι εγγράψιμο.

Άρα $ \angle MDC \equal{} \angle MEA \equal{} \angle BDA$.

Φέρνουμε το ύψος $ AZ$ του τριγώνου $ ABC$ με $ Z\in BC$ και θεωρούμε $ X$ το σημείο τομής της $ AZ$ με την $ MD$.

Τότε η $ DZ$ είναι διχοτόμος και ύψος στο τρίγωνο $ ADX$ άρα $ XZ \equal{} ZA$ άρα $ X$ είναι το σταθερό σημείο.[/quote]
Τάσο σε ευχαριστώ και χαιρετίσματα στον αγαπητό φίλο Νίκο Ιωσηφίδη του ''ΑΠΟΛΛΩΝΙΟΥ'', από τη Βέροια.

Κώστας Βήττας.

:wallbash: ...έχουν αρχίσει να με πειράζουν οι εξετάσεις μου στο κεφάλι :P

Πράγματι ήταν εύκολο... για τα εγγράψιμα αρκεί να πεις ότι $ D$ το σημείο Miquel του πλήρες τετραπλέυρου... (κι εγώ έψαχνα με αντιστροφές και άλλα τέτοια  :blush: )

Κωστα να συπληρωσω ενα ερωτημα ακομη στην ωραια ασκηση
Αν οι περιγεγραμενοι κυκλοι των τριγωνων $ ABC$ και $ FBD$ τεμνονται ακριβως στα $ B$ και $ M$ ν.δ.ο. $ \angle{O_2MB} \equal{} 90^o$.
Θυμιζει κατι;

[quote="xp"]Κωστα να συπληρωσω ενα ερωτημα ακομη στην ωραια ασκηση
Αν οι περιγεγραμενοι κυκλοι των τριγωνων $ ABC$ και $ FBD$ τεμνονται ακριβως στα $ B$ και $ M$ ν.δ.ο. $ O_2MB \equal{} 90^o$.
Θυμιζει κατι;[/quote]
Χρήστο, νομίζω πως το σημείο $ M\equiv BE\cap CF,$ δεν μπορεί να ανήκει στον περιγεγραμμένο κύκλο $ (O)$ του $ \bigtriangleup ABC$ και ίσως εννοείς τη γωνία $ \angle O_{2}NB \equal{} 90^{o}$ αντί $ \angle O_{2}MB \equal{} 90^{o},$ όπου $ N\equiv (O)\cap (O_{3})$ και όπου $ (O),\ (O_{3}),$ είναι αντιστοίχως, οι περίκυκλοι των τριγώνων $ \bigtriangleup ABC$ και $ \bigtriangleup FBD.$

Αν είναι έτσι, πράγματι μου θυμίζει κάτι. Είναι το κλασσικό θεώρημα που λέει ότι σε κάθε εγγράψιμο τετράπλευρο, η ευθεία που συνδέει το κέντρο του περιγεγραμμένου κύκλου, με το σημείο τομής των διαγωνίων του, είναι κάθετη στην ευθεία που συνδέει τα σημεία τομής των απέναντι πλευρών του.

Εδώ έχουμε την ευθεία $ O_{2}N,$ κάθετη στην ευθεία που συνδέει τα σημεία $ B\equiv AF\cap CD$ και $ K\equiv AC\cap DF,$ τομής των απέναντι πλευρών του εγγραψίμου τετραπλεύρου $ ACDF$ και περνάει από το σημείο $ P\equiv AD\cap CF,$ τομής των διαγωνίων του.

Κώστας Βήττας.

Κωστα οντως δεν ενοουσα το $ M$ της πρωτης εκφωνησης , αλλα Αυτο που εσυ ονομασες $ N$
Ηταν θεμα στην ΙΜΟ '85 (Φιλανδια)

Δείτε και την άσκηση 101 από εδώ
http://www.mathlinks.ro/viewtopic.php?t=165293&start=40

[quote="bilot4"]Αυτό είναι βασική εφαρμογή στις πολικές! ($ O_2$ ορθόκεντρο του $ KPB$). Ομως να ρωτήσω κατί... πώς αποδεικνύεται ότι τα $ N,P,O_2$ είναι συνευθειακά? (συγχωρέστε με αν είναι trivial και απλά δεν το βλέπω :oops:)[/quote]
Στο σχήμα [b][size=100]t=202726(α)[/size][/b], το σημείο $ N$ είναι το [b][size=100]σημείο Miquel[/size][/b] του εγγραψίμου τετραπλεύρου $ ACDF$ και ύπάρχει μία στοιχειώδης (=elementary) απόδειξη για την καθετότητα $ O_{2}N\perp BK,$ αλλά και για το ότι τα σημεία $ O_{2},\ P,\ N$ είναι συνευθειακά, στο http://www.mathlinks.ro/Forum/viewtopic.php?t=110887 .

Κώστας Βήττας.


Princeton college calculus derivative email

Right now I'm a sophomore at Princeton planning to major in economics/applied math. I am wondering how much education is required to be a derivatives trader? Will an undergrad degree suffice, or should I go on to pursue a MS degree in say statistics, applied math, or financial engineering? What is the typical career path (in terms of education) of a derivatives trader? There's not much info. on the internet on this. Any help would be greatly appreciated. Thanks in advance.

Nowadays at many firms all you need is an undergraduate degree. Most people I've interviewed for the trader position at Jane Street are either undergrads or did graduate work because their previous education had little to do with finance and they wanted to switch over. A big reason to do graduate work is to signal that you're qualified, but since you're at Princeton you don't need to do anything extra to signal that.

As a side note, top trading firms know that getting a degree in econ is not super hardcore. If you want to stand out, you should make sure to do something hardcore -- advanced math/ORFE coursework, doing cool research, winning competitions, etc. Try MCM -- if you haven't heard about it already, just send an email to the Math Club and they'll tell you about it.


geometry
ratio
analytic geometry
similar triangles
AMC

If $ABCD$ is 2 x 2 square, $E$ is the midpoint of $\overline{AB}$, $F$ is the midpoint of $\overline{BC}$, $\overline{AF}$ and $\overline{DE}$ intersect at $I,$ and $\overline{BD}$ and $\overline{AF}$ intersect at $H$, then the area of the quadrilateral $BEIH$ is

$\text{(A)} \ \frac{1}{3} \qquad \text{(B)} \ \frac{2}{5} \qquad \text{(C)} \ \frac{7}{15} \qquad \text{(D)} \ \frac{8}{15} \qquad \text{(E)} \ \frac{3}{5}$

I wasted so much of my time on this because I didn't realize $\triangle DAE$ and $\triangle ABF$ are exactly same!

[hide="View"]

Okay.. If you know the above, the problem is pretty trivial. I still don't believe how I missed that.. Blah, I can be bit dumb sometimes in geometry.

It's easy to see that $\triangle AHD$ and $\triangle FHB$ are similar. Why? $AD$ and $BF$ are parallel and the middle angles are vertical angles. Since they're similar, we know that their altitudes have to be in same ratio as their sides. The side ratio is 2:1 so their altitude (from AHD to FHB) is in 2:1, namely 4/3:2/3 (they add up to 2). So, the area of HBF is 1*2/3*1/2 = 1/3. Since ABF is a right triangle, this now tells that [ABH] = 1-1/3 = 2/3. Now, time to bring out [AEI]. Using my stupid mistake of not realizing DAE and ABF being exactly same, we know $\angle AED = \angle AFB$. Also, $\angle BAF$ is in both of triangles $AEI$ and $AFB$. By AA, they're similar and $\triangle AEI$ is right as well. Using similar triangle ratio, we find that $AI = 2/\sqrt{5}$ and $EI = 1/\sqrt{5}$. Thus, the area of AEI is $2/\sqrt{5} \cdot 1/\sqrt{5} \cdot 1/2 = 1/5$. Now, using our above information, we can conclude that [BEIH] is 2/3-1/5 = 7/15, which is C. [/hide]

This solution used no skill of drawing any extra lines (you can.. but you don't need to do). At least, I learned a lesson to look carefully in geometry problems again. Blah.

Yah...I was about to type the solution but you posted already....

I think the hardest part is seeing that $\triangle AIE$ is similar to $\triangle ABF$ or $\triangle ADE$ (unlike you, I see the two are eqivalent in the start), the rest can be obtained by similar triangles.

I did not read your soution thoroughly, but I'll outline mine.

Strategy:  Since only lengths are given, and squares have lots of right angles, we have to apply some Pythagorean, and maybe stewarts and similar triangles.  Since there are no cevians, we are left with similar triangles and Pythagorean.

Also, it will be hard to prve quad. $EIHB$ is cyclic, so we want to find area of $\triangle BCD$, $\triangle AHD$, and $\triangle AEI$ and subtract it from 4 to find the area.  Of course, seeing that we should go after these triangles take some insights and experience.

After getting the $AF = DE = \sqrt{5}$ and $DB = \sqrt{2}$, we run out of useful right angles, so the rest must be similar triangles.  Since opposite sides of a square are parallel and we are not given any angles, we go after the transversals for the alternate interior angles for similar triangles.

by going after $\triangle ADH$'s height (since we want the area), by comparing it with $\triangle BFH$, we can start eliminating areas from the square.  After seeing $\triangle AEI$ has one-fifth area because it shares angles with other triangles, we can finish it by concluding the Quad. has $\frac{7}{15}$ area.

I set up a coordinate system with A(0,0), B(0,2), C(2,2), and D(2,0).  By finding the points of intersection of the different lines, you can find I(2/5, 4/5) and H(2/3, 4/3).

I have a dummy question. Where can I find the practice sheet?


The product of the digits of many positive four-digit integers is equal to $ 5!$. What is
the sum of the largest and smallest such four-digit numbers?

The smallest such number is $ 1358$. The largest such number is $ 8531$. $ 8531\plus{}1358\equal{}\boxed{9889}$.

SRY for the bump, but I just realized that I solved this question [b][u]exactly[/u][/b] like the AoPS answer.

Edit (July 17, 2021): Why was this relevant?

[hide=Solution]
$5!=5\times4\times3\times2\times1$.

We need to put the possible digits in smallest to largest in order to create the smallest four-digit number possible, and vice versa for the largest.

Smallest digit $\implies 1$ and largest digit $\implies4\times2=8$. So we have $1\textunderscore\textunderscore8$. After looking at $5!$, we get the smallest to be $1358$ and largest to be flipped, which is $8531$.

Thus, the sum is $1358+8531=\boxed{9889}$.
[/hide]


geometry
number theory unsolved
number theory

We say (a, b, c) is a primitive Heronian triple if a, b, and c are positive integers with no common factors (other than 1), and if the area of the triangle whose sides measure a, b, and c is also an integer. Prove that if a = 96, then b and c must both be odd.

If m is the area of the triangle, we may write

[tex]\displaystyle (a+b+c)(-a+b+c)(a-b+c)(a+b-c)=16m^2[/tex], i.e.

[tex]\displaystyle 4b^2c^2-(b^2+c^2-a^2)^2=16m^2[/tex].

It is now clear that 2 divides b^2+c^2-a^2. Since a is even, b and c must have the same pairity. But b and c cant both be even, since then a, b, and c would have the factor 2 in common. Therefore, b and c are odd.


let $ L$ be the union of the $ z$-axis and the unit circle in the xy-plane, compute $ \pi_{1}(\mathbb{R}^{3} \minus{} L)$ ? thanks

There's a deformation retract to a torus; realize the torus as $ z^2\plus{}(r\minus{}1)^2\equal{}\frac14$, and just move each point toward or away from the nearest point on the circle until we reach distance $ \frac12$. Deformation retracts preserve $ \pi_1$.

I kind of want to use Van Kampen's theorem on this, but I can't really think of any good decompositions on this space. I would be curious to see if anyone else has suggestions.

On the torus? You can do it, but its not necessary; you should be able to do $ \pi_1(S^1 \times S^1)$ directly since it's a product of spaces.

I don't see why you wouldn't just do it to the torus -- it's standard practice to take weird spaces and replace them with simpler, homotopy-equivalent ones, whenever possible. And the torus is about as simple as it gets. Doing Van-Kampen on this directly would probably be one of the more difficult computations you could do -- though on the torus, it's a reasonable exercise to get familiar with Van-Kampen... but even there, you usually look at the Torus as a quotient space of the square $ I \times I$, instead of looking at its embedding in $ \mathbb{R}^3$.


I've been slowly (like 1-2 hours a day) going through AoPS Vol I; I'm still not quite 1/3 of the way done, but I took the 2008 AMC 10B today to see what kind of progress I had made - I was pleasantly surprised when I found out I had scored a 108 :D !  (That's up from a 60 something on last spring's AMC 10).  Anyways, I was wondering: what's the best way to take "practice" tests?  I only got through 17 problems in $ 1 \frac{1}{2}$ hours, so obviously I need to build up my speed, but should I just forgot the clock and finish the whole test?  Or should I take it like a "practice test", stopping when the time expires (and then going over the questions I didn't get to later) ?

Well, it depends on what you're trying to do. If you're trying to get used to the format of the AMC, then do it under timed conditions, and then after time's up, score yourself and continue with the rest of the test.

This is the best way to do it, this way, when it's competition time, you'll be ready, and won't need to spend time figuring out how fast you need to go, what order to do the problems in etc.

This is the way I do it. Also, make sure you understand the solutions of every problem in the competition, not just the ones you get wrong (in case of guesses). You'll be surprised to find out that your score will be at around 140 after you finsish AoPS volume 1 if you're using the book the way you should.  :lol:

Evolved, I must say, you are almost as the same situation as me. As for me, i'm chapter 7-8 on Volume 1 and my average is about 120-130 since i had a good amount of practice just practicing for MC last year. As of now, i recommend you follow the same plan as me, as it seems to be helping a lot. What I do is one chapter a day on AOPS V1 on weekdays. As for weekends, i take 1 recent(some test in this Millennium) AMC10 on Saturday and on Sunday and review the problems, understanding their solutions. The understanding what you got wrong part is probably more important than even taking the test :maybe: . And if i have extra time, go over what chapter was hard.

[quote="Mewto55555"]HiDN, while it may interest you that both you and the OP are in the same boat, you did not answer his question.

What I personally would do is take it like a real test, that is, check what you have done if you feel like you can't get the rest, and then when time runs out, make a note of where you were (circle problems that you had done or something), and then finish the test before looking at the answers. That way the benefits are twofold. You get used to taking it in test conditions, with accurate feedback over how you would have done, and you do all of the problems.

Of course, don't do all of the tests before you finish AoPS 1. Once you've seen all of the problems on a test once, you don't learn as much or get as accurate of results if you go through it a 2nd time.[/quote]

I sooo agree with the last paragraph (I haven't read the rest :P). You'll run out of AMC 10's in 2 weeks. I suggest you do AIMEs, HMMT (available on their website), ARML and Mandelbrot problems :)


algebra
system of equations
Pythagorean Theorem
geometry
IMO Shortlist
IMO 1984

Let $a,b,c$ be positive numbers with $\sqrt{a}+\sqrt{b}+\sqrt{c}= \frac{\sqrt{3}}{2}$
Prove that the system of equations
\[\sqrt{y-a}+\sqrt{z-a}=1\]
\[\sqrt{z-b}+\sqrt{x-b}=1\]
\[\sqrt{x-c}+\sqrt{y-c}=1\]
has exactly one solution $(x,y,z)$ in real numbers.

It was proposed by Poland. Have fun!  :lol:

I don't think IMO problems belong in the PRE-OLYMPIAD forum.

The ones I post certainly do.   :lol:

Very old IMO problems (Like this one) are relatively easy, and they are comparable with many other problems in this forum. So many Shortlist problems are OK for Pre-Olympiad, though I admit this is a hard one, and I doubted to post it here  :wink:

We'll see...

[hide="Solution"] We make the following substitutions, since all the numbers involved are positive.  Let

$p = \sqrt{a}, q = \sqrt{b}, r = \sqrt{c}$

And

$u = \sqrt{x}, v = \sqrt{y}, w = \sqrt{z}$

Then our condition becomes $p+q+r = \frac{ \sqrt{3}}{2}$, and our equations become

$\sqrt{v^{2}-p^{2}}+\sqrt{w^{2}-p^{2}}= 1$, etc.

Notice that if we let $p, q, r$ be the altitudes to some triangles and let $v, w$, etc. be their side lengths, then by Pythagorean Theorem we have produced the following triangles:

$(v, w, 1)$
$(w, u, 1)$
$(u, v, 1)$

And now the real geometric meaning behind the problem is apparent!

Let there be a point $P$ inside an equilateral triangle of side length $1$.  Let $p, q, r$ be the length of the perpendiculars from $P$ to each side.  It's well-known that $p+q+r = \frac{ \sqrt{3}}{2}$ (we can sum up some areas).

This tells us that $u, v, w$ are simply the distances from $P$ to the vertices, and then the proof is obvious:

Consider the locus of points that are a distance $p$ away from one of the given sides.  They form a line within the triangle parallel to that side.  Consider the intersection of that locus and the locus of points a distance $q$ away from another side.  There is precisely one point which satisfies our conditions ($P$), which, by the above argument, must be a distance $r$ from the third side.  Then $u, v, w$ exist and are unique.

QED. [/hide]


Hi everyone!
  Could anyone help me with this puzzle: (I am still thinking about it).

There are N devices; some of them work correctly and some of them are broken. These devices may be tested two at a time telling us the status of each other. A device working correctly always tells the truth:" Good" if the device it is testing works correctly and "Bad" if it is broken. In contrast, a broken device can tell us "Good" or "Bad" about the device it is testing regardless of its actual status.
1. Let the number of devices that work correctly be more than N/2. Find an algorithm that separates devices working correctly from the broken ones.
2.Under the same condition, find the linear algorithm (It means that the maximum number of testing should be not more than some linear function of N) doing the same.
3.Prove that if the number of devices that work correctly less than or equal to N/2 then there is no such an algorithm, linear or not.

Here's an algorithm for #1:
Test all pairs once each. For each machine, take the majority opinion.


function
calculus
derivative
real analysis
real analysis unsolved

Let $ f: [a,b]\to\mathbb{R}$ be a differentiable function such that
$ (f'(a)-m)(f'(b)-m)>0$
, where $ m=\frac{f(b)-f(a)}{b-a}$.
 Prove that there is $ c\in (0,1)$ such that
$ \frac{f(c)-f(a)}{c-a}=f'(c).$

The proof of [b]Flett[/b]'s Theorem works here too. It has been discussed several times, but I think I'll write it again.

We want to find $ \xi \in (a,b)$ such that $ g^\prime \left( \xi \right) = 0$, where $ g(x) = \frac{f(x)-f(a)}{x-a}$.

A derivative has the intermediate value property, so if the above doesn't happen, then WLOG $ g^\prime > 0$, i.e. $ g$ is strictly increasing. By using algebraic manipulations that remind us of convex functions we see that $ f^\prime$ is also increasing. From here we get that \[ f^\prime (a) < f^\prime (k) = \frac{f(b)-f(a)}{b-a}< f^\prime (b) ,\] contradiction.


inequalities calculus integration function

If there are n positive quantities a,b,c,....k then
$\frac{a^m+b^m+c^m....+k^m}{n}>(\frac{a+b+c....k}{n})^m$ (where m is integral)

Well known relationship between the arithmetic mean and the power mean of the m-th order.

Jensen's Inequality:

For a convex function $f(x)$, given values $x_1, x_2, ... x_n$ and non-negative weights $w_1, w_2, ... w_n$, we have (where each $\sum$ is understood to be $\sum_{k=1}^{n}$):

$\frac{ \sum w_k f(x_k) }{ \sum w_k} \ge f \left( \frac{ \sum w_k x_k}{ \sum w_k} \right)$

For a concave function, this becomes

$\frac{ \sum w_k f(x_k) }{ \sum w_k} \le f \left( \frac{ \sum w_k x_k}{ \sum w_k} \right)$

The desired result is attainable by setting $w_i = \frac{1}{n}$ and $f(x) = x^m$ (a convex function).

Edit:  The desired result seems to be attainable through Chebyshev.  Much easier solution  :rotfl:

[quote="gotztahbeazn"]haha a lil deep for the intermediate forum aint it? XP[/quote]

Yeah, I'm bumping it.

[quote="gotztahbeazn"]actually... try writing out a solution using jensens... for some reason i dun think it works o.O[/quote]

It works.

[quote="gotztahbeazn"]actually... try writing out a solution using jensens... for some reason i dun think it works o.O[/quote]

[hide="Solution"]
$w_1 = w_2 = ... = w_n = \frac{1}{n} \implies \sum w_k = 1$
$f(x) = x^m$

Then Jensen's tells us that $\frac{a^m + b^m + c^m + ... }{n} \ge \left( \frac{a + b + c + ... }{n} \right)^m$, precisely the desired result.
[/hide]

[hide="Solution without Jensens"]
The result follows directly from an extension of Chebyshev, which can be proved by Rearrangement.
[/hide]

If i'm rigth,this can be done without any help from another inequality.

[hide="Procedure"]value of a+b+c....k does not change if e substitute 2 values as (a+b)/2,but value of $a^m+b^m....$ does change bcos of the AM-GM thing.Now manipulate geniuses :lol: [/hide]

[quote="shadysaysurspammed"]If i'm rigth,this can be done without any help from another inequality.
...
but value of $a^m+b^m....$ does change bcos of the AM-GM thing.[/quote]

Great job not using another inequality  :rotfl:

Besides, there's not much point in not using another inequality if it's as simple as Rearrangement.

[quote="t0rajir0u"][quote="shadysaysurspammed"]If i'm rigth,this can be done without any help from another inequality.
...
but value of $a^m+b^m....$ does change bcos of the AM-GM thing.[/quote][/quote]

oh yeah...sorry :D


Let $n$ be natural number. Each of the numbers $\in\{1,2,\ldots ,n\}$ is coloured in black or white. When we choose a number, we flip it's colour and the colour of all the numbers which have at least one common divider with the chosen number. At the beginning all the numbers were coloured white. For which $n$ are all the numbers black after a finite number of changes?

The following problem has been discussed, in various formulations, several times on the forum (and it has received several solutions; I remember giving an algebraic one, but there were combinatorial proofs as well):

Given a graph $G$ (in which eac vertex is considered to be connected to itself) there is a subset $V$ of vertices such that every vertex of $G$ is connected to an odd number of vertices in $V$.

Apply this to the graph having $1,\ldots,n$ as vertices and in which $i$ is connected to $j$ iff they are not coprime. This should prove that every positive integer $n$ works: choose precisely the numbers belonging to the set $V$ provided by the problem above, and perform the changes on those; this will change the color of all the numbers, as can easily be seen.

Thank you I will look more  :)

P.S What is  "vertices" ?? You can tell in Rumanian maybe.. lol

Well, since you seem to speak french, I'll tell it in french : sommets. :)

I prove this for any graph (I prove the result grobber stated).

We use induction. Base case is trivial.

We say a graph "works" if we can touch vertices such that from a configuration of all-white, we reach a configuration of all-black.

Now assume any graph with $n-1$ vertices works. Then take a graph $G$ with $n$ vertices. "Forget" about one of $G$'s vertices, and apply the inductive hypothesis. If the vertex you forgot about turns black, we're done. So this vertex stayed white. So we now have a method for changing the color of all the vertices except for one, and we can choose this one.

If $n$ is even, apply this method for all vertices and we're done.

If $n$ is odd, apply this method for all vertices except two. We now have a method to change the color of all vertices except for two, and we can choose these two. We call this method, method $X$. So, since there are an odd number of vertices in $G$, there must be a vertex $V$ with an even degree. Touch this vertex $V$. Change the color of the remaining vertices using method $X$ $(n-d(V)-1)/2$ times. Now we have reached the all-black configuration.

So we're done.


$ \sqrt {2x^2 \plus{} x \plus{} 5} \plus{} \sqrt {x^2 \plus{} x \plus{} 1} \equal{} \sqrt {x^2 \minus{} 3x \plus{} 13}$
solve for real x

[hide="A Solution"]
$ \sqrt{2x^{2}\plus{}x\plus{}5}\plus{}\sqrt{x^{2}\plus{}x\plus{}1}\equal{}\sqrt{x^{2}\minus{}3x\plus{}13}$

$ 2x^{2}\plus{}x\plus{}5\plus{}x^{2}\plus{}x\plus{}1\plus{}2\sqrt{(2x^{2}\plus{}x\plus{}5)(x^{2}\plus{}x\plus{}1)} \equal{} x^{2}\minus{}3x\plus{}13$

Rearrange equation, with square root on one side, then square again.
$ (2x^{2}\plus{}5x\minus{}7)^{2}\equal{}4{(2x^{2}\plus{}x\plus{}5)(x^{2}\plus{}x\plus{}1)}$

Solve, using a calculator maybe...make sure to check solutions in equation as squaring often produces some extra solutions.
I cannot seem to think of a more elegant solution to this problem.
[/hide]

[hide="MySolution"]
Lets multiply the first root with 2, and the second with $ \sqrt {7}$ Thus we get:
$ \frac {\sqrt {8x^2 \plus{} 4x \plus{} 20}}{2} \plus{} \frac {\sqrt {7x^2 \plus{} 7x \plus{} 7}}{\sqrt {7}} \equal{} \sqrt {x^2 \minus{} 3x \plus{} 13}$
Now after substituting $ \sqrt {8x^2 \plus{} 4x \plus{} 20} \equal{} a \; ; \sqrt {7x^2 \plus{} 7x \plus{} 7} \equal{} b$ (notice that $ a^2 \minus{} b^2 \equal{} x^2 \minus{} 3x \plus{} 13$ ) we get: $ \frac {a}{2} \plus{} \frac {b}{\sqrt {7}} \equal{} \sqrt {a^2 \minus{} b^2} \Leftrightarrow \frac {a^2}{4} \plus{} \frac {b^2}{7} \plus{} \frac {ab}{\sqrt {7}} \equal{} a^2 \minus{} b^2 \Leftrightarrow \frac {3a^2}{4} \minus{} \frac {ab}{\sqrt {7}} \minus{} \frac {8b^2}{7} \equal{} 0$
Which can be solved easily, but I'm just too lazy to do it :ninja:
[/hide]


One question

What exactly does the angle formed of the opposite edges of a tetrahedron look like?

I can't visualize it.

Thank

[url]http://mathworld.wolfram.com/Tetrahedron.html[/url]
It helped me, at least.  You can drag around all the tetrahedrons.

But wouldn't the opposite edges not intersect?

Apparently I am missing something too in the problem, but I saw a problem and it said Let alpha be the angle formed by the pair of opposite edges in a tetrahedron.

I was like what?


I don't think this is the right forum for that type of question. See [url]http://www.mathlinks.ro/Forum/index.php?f=360[/url]

Plenty of chess topics there  :)

Well, first I'd like to read on a particular topic. Then I would get online and try the variations (or consult with my chess teacher - and we'd go over the thing I am working on). Eventually, when Im more confident I will try it in some tournaments to see how people respond to it and modify my playing from there.

Moved.

well first read some endgame stuff from books
then read some openings... it really helps

[quote="bos1234"]How do you guys actually study chess. I have some e-books and they have moves which I put on winboard. But they have so many variations it gets confusing. DO you guys analyse variations too?[/quote]
As D3m0n said, you should try some of what you're learning in several real games to see how people react and change your tactics depending on their moves. I would develop my openings and endgames before going into the middle game because it is the most complex and probably hardest to study. After learning how to begin and how to finish off, you just need a lot of practice in real games to learn how to play middlegame.

Good luck! :)

Pay a coach to teach you once a week and get tons of books, first study the basic moves of openings and the basic ideas, than move on to studying in detail the openings you like. Modern Chess Openings is an amazing book, buy it!


linear algebra
matrix
vector
linear algebra unsolved

I was told that this place is the best to ask for help with it:
Let $||x||=\sqrt{x^{H}x}$ that is $l_2 - norm$
Let $M \in C^{nxn}$ be a hermitian positive definite, and consider the vector norm:
$||x||_{M}=\sqrt{x^{H}Mx}$.
Find an expression for the matrix norm $||A||_M$ induced by this vector norm, show also that
$||A||_M \le \sqrt{K(M)}||A||$, where $||K||=||M||||M^{-1}||$

I can't see beyond this..
$||A||_M= \sup_{x\neq0}\frac{||Ax||_M}{||x||_M}=\sup_{x\neq0}\frac{\sqrt{(Ax)^{H}M(Ax)}}{\sqrt{x^{H}Mx}}$

I don't know what to do with the M..


geometry
circumcircle
geometric transformation
reflection
trigonometry
geometry proposed

Here your solution>>>>>>
Proof:
Because CO ís perpendicular to A1B1 .This easy to prove it.
Then C,O,C2 is colinear because that CC2 is perpendicular to A1B1.
We have Angle CAO=Angle OCA( because OC=OA so triangle OAC is scalen) and Angle ACC2 =Angle CC2B1 because that B1C=C2B1 by reflection)
So Angle CAO =Angle CC2B1  then OB1AC2 is cyclic.
Then CO.CC2=CB1.CA. But HA1AB1 is cyclic, so CH.CA1 =CB1.CA.Then CO.CC2=CH.CA1 .
So O,H,A1,C2 are concyclic

Consider the inversion through pole $ C$, power $ \overline {CA_1}\cdot \overline {CB}\equal{}\overline {CB_1}\cdot \overline {CA}\equal{}\overline {CH}\cdot \overline {CC_1}\equal{}k^2$. We have $ \mathcal {I}(C,k): A_1\mapsto B$, $ B_1\mapsto A$, hence $ A_1B_1\mapsto (O)$ Therefore $ O$ will be the image of $ C_2$, which is the reflection of $ C$ through $ A_1B_1$, through $ \mathcal {I}(C,k)$. Hence $ \overline {CC_2}\cdot\overline {CO}\equal{}k^2\equal{}\overline {CH}\cdot \overline {CC_1}\Longrightarrow C_2,O,H,C_1$ are concyclic.
Our proof is conpleted then.

[quote="April"]Suppose $ H$ and $ O$ are the orthocenter and the circumcenter of acute triangle $ ABC$; $ AA_1$, $ BB_1$ and $ CC_1$ are the altitudes of the triangle. Point $ C_2$ is the reflection of $ C$ in $ A_1B_1$. Prove that $ H$, $ O$, $ C_1$ and $ C_2$ are concyclic.[/quote]
Let $ CC_2 \cap A_1B_1\equal{}H_1$. we know that $ O \in CH_1$. So, $ CO\cdot CC_2\equal{}R\cdot 2 CH_1\equal{}2R\cdot CB_1\sin B\equal{}b\cdot CB_1\equal{}CH\cdot CC_1$. So, $ C_1$ and $ C_2$ are concyclic.

Dear Mathlinkers,
in order to have a proof with a minimum of calcul, we can involve the Monge theorem

Sincerely
Jean-Louis

[hide=Storage]Notice that $$\measuredangle OAB_1=\measuredangle OAC=\measuredangle OCA=\measuredangle C_2CB_1=\measuredangle CC_2B_1=\measuredangle OC_2B_1$$It follows that $\{C_2,O,B_1,A\}$ lie on a circle. Consequently, $$CO\cdot CC_2=CB_1\cdot CA=CH\cdot CC_1$$ which finishes the problem. $\blacksquare$[/hide]

Let CC2 meet A1B1 at P. ∠B1CP = 90 - ∠PB1C = 90 - ∠B and ∠B1CO = 90 - ∠B so C,O,C2 are collinear So we can prove CO.CC2 = CH.CC1 . Note that CH.CC1 = CB1.CA so we instead can prove AB1OC2 is cyclic. ∠B1AO = 90 - ∠B = ∠B1CC2 = ∠B1C2O so AB1OC2 is cyclic.
we're Done.