Prove that the Simson LInes of triangles with the same incircles and circumcircles for a fixed point on the circumference are concurrent.

According to the [url=http://mathworld.wolfram.com/FonteneTheorems.html]2nd Fontene theorem[/url], the orthopole of a line passing through the circumcenter $(O)$ of a triangle $\triangle ABC$ lies on the circumcircle the pedal triangle $\triangle XYZ$ of any point $P$ on this line. For a proof, see [url=http://de.geocities.com/darij_grinberg/]Darij Grinberg, Generalization of the Feuerbach point  1)[/url]. The pedal triangle of the circumcenter $O$ is the medial triangle $\triangle A'B'C'$ and its circumcircle is the 9-point circle $(N)$ of the triangle $\triangle ABC$. Hence, the orthopoles of all lines passing through the circumcenter lie on the 9-point circle. If the point $P$ is the intersection of a line through the circumcenter with the circuncircle $(O)$, the pedal triangle degenerates into a line segment and its circumcircle into a Simson line with the pole $P$, which passes through the orthopole of the line $OP$.
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The pedal triangle of the incenter $I$ is the contact triangle and its circumcircle is the incircle $(I)$ of the triangle $\triangle ABC$. The orthopole of the diacentral line $OI$ lies both on the 9-point circle and on the incircle, hence, it is identical with their tangency point $F \equiv (I) \cap (N)$ (Feuerbach point). Let the diacentral line $OI$ intersect the circumcircle and incircle at points $P, P'$ and $Q, Q'$, so that the points $P, Q, I, Q', P'$ and the points $P, O, I, P'$ follow on the line $OI$ in these orders. The orthocenter $H$ is the external homothety center of the circumcircle and the 9-point circle. The Feuerbach point $F$ is the external homothety center of the incircle and the 9-point circle. Thus the line $HF$ is the external homothety axis of the circumcircle, 9-point circle, and the incircle passing through the external homothety center $E$ of the incircle and the circumcircle, the intersection of the diacentral line $OI$ with the homothety axis $HF$. Let the line $HF$ intersect the circumcircle $(O)$ at a point $F'$. Because of the homothety between the incircle and circumcircle, the right angle triangles $\triangle QFQ' \sim \triangle PF'P'$ are cenrally similar with homothety cemter $E$. Because of the homothety between the 9-point circle and circumcircle with homothety center $H$ and homothety coefficient $\frac 1 2$, the 9-point circle cuts the segments $PH, P'H$ at their midpoints $R, R'$, the lines $PP' \equiv QQ' \parallel RR'$ are parallel and the segment $RR'$ is a diameter of the 9-point circle. Consequently, the right angle triangles $\triangle PFP' \sim \triangle RFR'$ are centrally similar with homothety center $H$ and the right angle triangles $\triangle QFQ' \sim \triangle RFR'$ are centrally similar with homothety center $F$. The Simson lines of the poles $P, P'$, perpendicular to each other, pass through these midpoints $R, R'$ and they meet at the orthopole $F$ of the diacentral line $PP' \equiv OI$. Hence, the Simson lines of the poles $P, P'$ are the lines $RF, R'F$. Because of the homothety of the triangles $\triangle QFQ' \sim \triangle RFR'$, the Simson lines $RF, R'F$ pass through the points $Q, Q'$. But these 2 points are completely independent of the "rotation" of the triangle $\triangle ABC$ around the fixed circumcircle $(O)$ and the incircle $(I)$ according to Poncelet's porism. As a result, the intersections $Q, Q'$ of the incircle with the diacentral line $OI$ are the fixed points of Simson lines with the poles $P, P'$.
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Let $P, S$ be 2 points of the circumcircle with the central angle $0 < \measuredangle POR = \phi \le 180^o$. Then the Simson lines $s_P, s_S$ of the poles $P, S$ form an angle of $\frac \phi 2$. This easily follows from the usual construction of a Simson line in the desired direction. Let the lines parallel to the Simson lines $s_P, s_S$ and passing through one of the triangle vertices, say $A$, itersect the circumcircle at points $K, L$ different from the vertex $A$. Thus the angle between the Simson lines $s_P, s_S$ is equal to the angle $\measuredangle KAL$. The poles $P, S$ are intersections of the lines throught the points $K, L$ perpendicular to the triangle side $BC$ with the circumcircle $(O)$. Hence, the quadrilateral $KLSP$ is a cyclic trapezoid, which means that it is isosceles with $KL = PS$. Consequently, the central angles $\measuredangle KOL = \measuredangle POS = \phi$ are equal and the angle formed by the 2 Simson lines is equal to $\measuredangle KAL = \frac \phi 2$. As a special case, if the poles $P, S$ are diametrally opposite points of the circumcircle $(O)$, the angle $\phi = \measuredangle POS = 180^o$ and the Simson lines of the poles $P, S$ are perpendicular to each other.
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Reflections of any 2 circumcircle diameters $PP', SS'$ in any side line of the medial triangle $\triangle A'B'C'$ meet on the 9-point circle. This is because the circumcenter $O$ of the triangle $\triangle ABC$ is the orthocenter of the medial triangle and the 9-point circle $(N)$ is its circumcircle. The reflection of the orthocenter in any side of a triangle lies on its circumcircle and the reflection of any line passing through the orthocenter in any triangle side must pass through the corresponding orthocenter reflection on the circumcircle. Let the reflections of the diameters $PP', SS'$ in the side line $B'C'$ of the medial triangle meet at a point $W$ on the 9-point circle $(N)$. These reflected lines form the same angle $\phi = \measuredangle POS$ as the diameters $PP', SS'$. According to the above reference [url=http://de.geocities.com/darij_grinberg/]1)[/url], the reflections of the diameters $PP', SS'$ in the sides $B'C', C'A', A'B'$ of the medial triangle concur at the orthopoles $U, V$ of these diameters. Thus the angle $\measuredangle UWV$ spanning the arc $UV$ of the 9-point circle $(N)$ is equal to $\measuredangle UWV = \measuredangle POS = \phi$ and the central angle of the orthopoles $U, V$ on the 9-point circle is equal to $\measuredangle UNV = 2 \measuredangle UWV = 2 \phi$. As a special case, if the diameters $PP', SS'$ are perpendicular to each other, i.e. $\phi = \measuredangle POS = 90^o$, their orthopoles $U, V$ are diametrally opposite points of the 9-point circle.
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Let $R, r$ be the fixed circumradius and inradius, $d = OI$ the fixed distance from the circumcenter to the incenter and $\triangle ABC$ an arbitrary triangle with the fixed circumcircle $(O)$ and incircle $(I)$. The 9-point circle $(N)$ is the circumcircle of the medial triangle $\triangle A'B'C'$, hence, its radius is $n = \frac R 2$ is also fixed. The 9-point circle is tangent to the fixed incircle at a point $F$ (Feuerbach point). Thus the locus of the 9-point circle centers $N$ of the triangles $\triangle ABC$ "rotated" according to Poncelet's porism is a circle $(I, m)$ centered at the incenter $I$ and with radius $m =  n - r = \frac{R - 2r}{2}$. Consider the circumcircle diameter $PP'$, which coincides with the diacentral line $OI$, and an arbitrary circumcircle diameter $SS'$ forming the central angle $\phi = \measuredangle POS \le 90^o$ with the diacentral diameter $PP'$. If the angle $\phi = 0$, i.e., if the diameters $SS' \equiv PP'$ are identical, the locus of orthopoles $F$ of this diameter (Feuerbach points) is the incircle $(I)$. If the angle $\phi = 90^o$, i.e., if the diameters $SS' \perp PP'$ are perpendicular, the orthopole $V$ is the diametrally opposite point of the Feuerbach point on the 9-point circle. Hence, the locus of the orthopoles $V$ diametrally opposite to the Feuerbach points $F$ on the 9-point circles is a circle $(I, k)$ centered at the incenter $I$ and with radius $k = m + n = \frac{R - 2r}{2} + \frac R 2 = R - r$. If the angle $0 < \phi < 90^o$, consider the triangle $\triangle INV$. The side $IN = m = \frac{R - 2r}{2}$ is fixed, the side $NV = n = \frac R 2$ is fixed and the angle $\measuredangle INV \equiv \measuredangle FNV$ is also fixed, because the Feuerbach point $F$ lies on the center line $IN$ of the incircle and the 9-point circle and the 9-point circle central angle of the orthopoles $F, V$ of the circumcircle diameters $PP', SS'$ is equal to $\measuredangle FNV = 2 \measuredangle POS = 2 \phi$. Hence, the remaining side of this triangle is fixed as well and given by the cosine theorem:

$IV = \sqrt{NV^2 + IV^2 - 2\ NV \cdot IV \cos{\widehat{INV}}} = \sqrt{\left(\frac R 2\right)^2 + \left(\frac{R - 2r}{2}\right)^2 - 2\ \frac R 2 \cdot \frac{R - 2r}{2} \cos{2 \phi}}\ =$

$= \sqrt{r^2 + (R^2 - 2rR)\ \frac{1 - \cos{2 \phi}}{2}} = \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}} = \sqrt{(R - r)^2 - (R^2 - 2rR) \cos^2{\phi}}$

Since the incenter $I$ is fixed and the distance $IV$ is independent of the "rotation" of the triangle $\triangle ABC$ according to Poncelet's porism, the locus of orthopoles $V$ of the circumcircle diameter $SS'$ is a circle $(I, v)$ centered at the incenter $I$ and with radius $v = IV$. From the above formula, it is clear that the radius of this circle is limited by $r \le v \le R - r$ for $0 \le \phi \le 90^o$.
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Since the angle $\measuredangle FNV  = 2 \phi$ is fixed and the triangle $\triangle INV$ just rotates around the incenter $I$, the angle $\measuredangle FNV = 2 \psi$ is also fixed. Using the sine theorem for the triangle $\triangle INV$,

$\frac{R}{2 \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}}} = \frac{NV}{IV} = \frac{\sin{\widehat{NIV}}}{\sin{\widehat{INV}}} = \frac{\sin{\widehat{FIV}}}{\sin{\widehat{INV}}} = \frac{\sin{2 \psi}}{\sin{2 \phi}}$

$\sin{2 \psi} = \frac{R \sin{2 \phi}}{2 \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}}}$

From the above formula, it is clear that $0 \le 2 \psi < 180^o$ for $0 \le 2 \phi \le 180^o$ and that $\sin{2 \psi} = 1,\ 2 \psi = 90^o$ for

$4R^2 \sin^2{\phi}\ \cos^2{\phi} = 4\left[r^2 + (R^2 - 2rR) \sin^2{\phi}\right]$

$R^2 \sin^4{\phi} - 2rR\ \sin^2{\phi} + r^2 = (R\ \sin^2{\phi} - r)^2 = 0$

$\sin^2{\phi} = \frac r R,\ \ \cos{2 \phi} = \frac{R - 2r}{R}$

The Simson lines $s_P \equiv QF \perp s_P' \equiv Q'F$ with the poles $P, P'$ pass through the orthopole $F \in (I, r)$ (Feuerbach point) of the diacentral line $OI \equiv PP'$ and they are perpendicular to each other. Likewise, the Simson lines $s_S \perp s_S'$ with the poles $S, S'$ pass through the orthopole $V \in (I, v)$ of the line $SS'$ and they are also perpendicular to each other. In addition, the Simson line pairs $s_P, s_S$ and $s_P', s_S'$ both form a constant angle equal to $\frac \phi 2$. Let $V' \equiv IV \cap (I)$ be the intersection of the line $IV$ with the incircle $(I)$ and $p \parallel s_S, p' \parallel s_S'$ 2 lines through the point $V'$ parallel to the Simson lines $s_S, s_S'$. The concentric circles $(I, v), (I, r)$ are centrally similar with homothety center $I$ and so are the parallel lines $s_S \parallel p, s_S' \parallel p'$ passing through the corresponding points $V, V'$ in this homothety. It is clear that the mutually perpendicular Simson lines $s_S \perp s_S'$ pass through fixed points of the circle $(I, v)$ iff their mutually perpendicular parallels $p \perp p'$ pass through fixed points of the circle $(I, r)$ in a rotation of the central angle $\measuredangle FIV \equiv \measuredangle FIV'$ around the incenter $I$ by an arbitrary angle $\theta$. Therefore, to complete the solution of the problem, we have to prove the following lemma:
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Let $(I)$ be an arbitrary circle with a fixed diameter $Q, Q'$ and $F$ an arbitrary point on this circle, forming a right angle triangle $\triangle QFQ'$ with the fixed hypotenuse $QQ'$. Let $0 \le \measuredangle FIV = 2 \psi \le 180^o$ be an arbitrary central angle of this circle with a fixed measure, the point $V$ also lying on the circle $(I)$. Let $p \perp p'$ be 2 perpendicular lines through the point $V$, both forming the same arbitrary fixed angle $\frac \phi 2 \le 90^o$ with the perpendicular lines $QF \perp Q'F$. In a rotation of the point $F$ around the circle center $I$ by an arbitrary angle $\theta$, the perpendicular lines $p \perp p'$ pass through fixed points $X, X'$ of the circle $(I)$, forming a right angle triangle $\triangle XVX'$ with the fixed hypotenuse $XX'$.
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The proof of this lemma is easy: Let the perpendicular lines $p \perp p'$ through the point $V$ intersect the circle $(I)$ at points $X, X'$. Since the angle $\measuredangle XVX' = 90^o$ is right, $XX'$ is a diameter of the circle $(I)$. The triangles $\triangle FIQ, \triangle VIX$ are both isosceles with $IF = IQ = IV = IX = r$, where $r$ is radius of the circle $(I)$. Hence, the angles $\measuredangle IFQ = \measuredangle IQF$ are equal and the angles $\measuredangle IVX = \measuredangle IXV$ are also equal. Let the lines $FQ, VX$ intersect at a point $Z$. From the concave quadrilateral $IFZV$, sum of the angles $\measuredangle IFQ + \measuredangle IVX$ is equal to

$\measuredangle IFQ + \measuredangle IVX = 360^o - (\measuredangle FZV + 360^o - \measuredangle FIV) = \measuredangle FIV -\measuredangle FZV = 2 \psi - \frac \phi 2$

From the convex quadrilateral $QIXZ$, the angle $\measuredangle QIX$ is equal to

$\measuredangle QIX = 360^o - (\measuredangle QZX + 180^o - \measuredangle IQF + 180^o - \measuredangle IXV) =$

$= \measuredangle IQF + \measuredangle IXV - \measuredangle QZX = 2 \psi - \frac \phi 2 - \frac \phi 2 = 2 \psi - \phi$

Since the measures of the angles $2 \psi, \frac \phi 2$ are both fixed and the point $Q$ is also fixed, the point $X$ is necessarily fixed as well. Since $XX'$ is a diameter of the circle $(I)$, both the points $X, X'$ are fixed and they do not depend on the arbitrary angle of rotation $\theta$.
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This concludes the proof of the problem proposition. Wow !!!  :cool:  And look mom, no cubics !

[b]Related problem:[/b] Instead of "rotating" the triangle $\triangle ABC$ around he fixed circumcircle $(O)$ and incircle $(I)$ according to Poncelet's porism, we can keep the triangle fixed and rotate the diametrally opposite poles $S, S' \in (O)$ around the circumcenter. When the poles $S, S'$ are identical with the intersections $P, P'$ of the diacentral line $OI$ with the circumcircle, i.e., $\phi = \measuredangle POS = 0$, the Simson line pivot points $X, X'$ are identical with the intersections $Q, Q'$ of the incircle with the diacentral line $OI$, i.e., $IQ = IQ' = r$. When the diameter $SS'$ is perpendicular to the diacentral line $OI$, i.e., $\phi = \measuredangle POS = 90^o$, the Simson line pivot points $X, X'$ are on a line perpendicular to the diacentral line $OI$ and $IX = IX' = R - r$. From this, we may conjecture that the locus of the Simson line pivot points is an ellipse centered at the incenter $I$, with the semiminor axis $IQ = r$, semimajor axis $IX = R - r$ and the focal length (using Euler's formula for $R, r, d = OI)$

$\sqrt{(R - r)^2 - r^2} = \sqrt{R^2 - 2rR} = d = OI$

I am 100 % sure that this is true, even though I have not performed the calculation yet.

Yetti

[quote="mecrazywong"]Impressive work, yetti![/quote]
Thank you. I was thinking about this problem on and off for days, trying different approaches. At first, I thought that some cubics would be involved, because the first thing one notices is that when "rotating" the triangle around the fixed circumcircle and incircle according to Poncelet's porism, the locus of the feet of normals from a fixed pole on the circumcircle to the triangle sides (i.e., to the tangents to the fixed incircle) is by definition a limacon with both the inner and outer loops tangent to the incircle and that the Simson line pivot point necessarily lies on this limacon. The solution really surprised me.

Yetti

[quote="pestich"]Every time I come accross one of Yetti's proofs I know I have to take extra care
   to avoid the 'pencil' of potential harmful side effects on the brain (the one inside the
   head, the one on the shoulders) which it definitly may cause inside an unsuspecting
   fresh unspoiled mind like mine.  Simply put, it is like he tries to  braid the however few
   brain curves I have and keep them from floating free,  as for  the solid formations
   up there, he tries to split them up and extract crumb by crumb. And he knows pretty
   well that it is exactly like this, or well within epsilon of his true intentions.[/quote]
Huh???  :?

[quote="pestich"]
   My main complaint though is that it takes too long to upload topics infected by Yetti.[/quote]
The advice for your atypical case is simple:
(a) You can get a faster computer.
(b) You can complain at the JFK airport.

Yetti

[quote="yetti"][b]Related problem:[/b] Instead of "rotating" the triangle $\triangle ABC$ around he fixed circumcircle $(O)$ and incircle $(I)$ according to Poncelet's porism, we can keep the triangle fixed and rotate the diametrally opposite poles $S, S' \in (O)$ around the circumcenter. When the poles $S, S'$ are identical with the intersections $P, P'$ of the diacentral line $OI$ with the circumcircle, i.e., $\phi = \measuredangle POS = 0$, the Simson line pivot points $X, X'$ are identical with the intersections $Q, Q'$ of the incircle with the diacentral line $OI$, i.e., $IQ = IQ' = r$. When the diameter $SS'$ is perpendicular to the diacentral line $OI$, i.e., $\phi = \measuredangle POS = 90^o$, the Simson line pivot points $X, X'$ are on a line perpendicular to the diacentral line $OI$ and $IX = IX' = R - r$. From this, we may conjecture that the locus of the Simson line pivot points is an ellipse centered at the incenter $I$, with the semiminor axis $IQ = r$, semimajor axis $IX = R - r$ and the focal length (using Euler's formula for $R, r, d = OI)$

$\sqrt{(R - r)^2 - r^2} = \sqrt{R^2 - 2rR} = d = OI$
[/quote]

And here is the proof, by a special request:

Both the orthopole $V$ of the diameter $SS'$ and the pivot point $X$ of the Simson lines with the fixed pole $S$ lie on the circle $(I, v = IV)$ centered at the fixed incenter $I$ of the variable triangle $\triangle ABC$. In addition, the angle $\theta = \angle QIX = 2 \psi - \phi$, where $Q$ is the intersection of the diacentral line $OI$ with the incircle $(I)$ closer to the fixed circumcenter $O$.

$v = IV = \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}} = \sqrt{r^2 + d^2 \sin^2{\phi}}$

$\sin{2\psi} = \frac{R \sin{2 \phi}}{2v} = \frac{R \sin \phi \cos \phi}{v}$

$\cos{2\psi} = \sqrt{1 - \sin^2{2\psi}} = \frac{\sqrt{v^2 - R^2 \sin^2{\phi} \cos^2{\phi}}}{v} =$

$= \frac{\sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi} - R^2 \sin^2{\phi} (1 - \sin^2{\phi})}}{v} =$

$= \frac{\sqrt{r^2 -2rR \sin^2{\phi} + R^2 \sin^4{\phi}}}{v} = \frac{r - R \sin^2{\phi}}{v}$

We have to show that the point $X$ lies on an ellipse with center $I$, semimajor axis $R- r$ and semiminor axis $r$ coincident with the diacentral line $OI$, i.e.,

$(?)\ \ \frac{v^2 \cos^2{\theta}}{r^2} + \frac{v^2 \sin^2{\theta}}{(R - r)^2} = 1$

regardless of the angle $\theta = 2\psi - \phi$. Using the trigonometric formula for $\sin{(2\psi - \phi)}$ and the relation $OI^2 = d^2 = R^2 - 2rR$,

$\frac{v^2 \cos^2{\theta}}{r^2} + \frac{v^2 \sin ^2{\theta}}{(R - r)^2} = \frac{v^2}{r^2} - \frac{v^2 \sin^2{\theta}}{r^2} + \frac{v^2 \sin ^2{\theta}}{(R - r)^2} = \frac{v^2}{r^2} - \frac{d^2 v^2 \sin^2{\theta}}{r^2(R - r)^2} =$

$= \frac{v^2}{r^2} - \frac{d^2 [ R^2 \sin^2{\phi} \cos^4{\phi} - 2R(r - R \sin^2{\phi}) \sin^2{\phi} \cos^2{\phi} + (r - R \sin^2{\phi})^2 \sin^2{\phi} ]}{r^2(R - r)^2} =$

$= \frac{v^2}{r^2} - \frac{d^2 \sin^2{\phi}}{r^2(R - r)^2}\ [R^2 (1 - 2 \sin^2{\phi} + \sin^4{\phi}) - 2rR(1 - \sin^2{\phi})\ +$

$+ 2R^2 (\sin^2{\phi} - \sin^4{\phi}) + r^2 - 2rR \sin^2{\phi} + R^2 \sin^4{\phi} ]\ =$

$= \frac{v^2}{r^2} - \frac{d^2 \sin^2{\phi} (R^2 - 2rR + r^2)}{r^2(R - r)^2} = \frac{v^2 - d^2 \sin^2{\phi}}{r^2} = \frac{r^2}{r^2} = 1$

[quote="pestich"]Let me say a couple of words related  to the problem of Simson lines in Poncelet set-up.

   It is known that the angle between  Simsons of two points P1 and P2 on circumcircle is
   1/2 of the central angle P1-O-P2.  Now, if we have a third point P3 on the circumcircle
   then the triangle XYZ made by Simsons is similar to P1P2P3  and independent of the
   reference  triangle ABC.[/quote]
Right. For example, it was the simplest step in the above proof. But I would not venture to say "it is known" without putting forward some argument, for I did not see it in geometry textbooks dealing with Simson lines.

[quote="pestich"]
   Poncelet  configuration just provides an anker - the fixed
   point  I- incenter of ABC.[/quote]
Huh ?? :?
There are a lot of other fixed points, besides the incenter: the circumcenter, the entire incircle and circumcircle, the entire diacentral line $OI$ including its intersections with the fixed incircle and circumcircle and of course, the fixed Simson line poles. What is so special about the incenter to be an anchor? Do you know any theorems about geometric anchors?

[quote="pestich"]
   As triangle ABC rotates around incircle a la Poncelet, for the XYZ
   to have fixed angles it's vertices have to move along some 3 concurrent circles
   (concurrency is for the case when the 3 Simsons are concurrent) and the
   sides of XYZ have to pass  thru the 3 points of pairwise circles intersections.
   These 3 points correspond to  the points P1,P2,P3.   So all Simsons for point Pi pass
   thru it's circle intersection.[/quote]
How do you know that the vertices of the triangle $\triangle XYZ$ move on circles, besides finding their locus experimentally? Rest assured that the Sketchpad does it in no time and I looked for the loci of all points of interest. It is as if I said: The Simson lines of diametrally opposite points are perpendicular to each other and intersect at their orthopole. Therefore, the orthopole moves on a fixed circle and the 2 Simson lines intersect this circle at a fixed diameter. It is certainly possible, but until you show that it is also necessary, it is just possible. You would have to find either the fixed center and fixed radius of one of these circles or the common circle of similarity of the rotation centers of the spiral similarities of all triangles $\triangle XYZ$. The second possibility would show that the 3 circles are concurrent. You did neither.

[quote="pestich"]
   The triangle of centers of these 3 circles is also similar to P1P2P3.

   When P1P2P3 slides along the circumcircle,  the 3 circles
   move along so that their centers  have circular orbits around the incenter of ABC.
   According to Lissajus,  the 3 circle intersections points are on an ellipse centered at
   incenter of ABC.[/quote]
You did not even show that you have 3 circles, much less that the Simson lines pass through their their intersections. And now you come up with another totally unproven claim - their centers, which you did not find at all, move on a circle concentric with the incircle. If these circles had constant radii (which they do not have), their intersections would move on (different) circles. To move on an ellipse, these circles have to - well - pulsate and on top of that, pulsate in the right way. Nothing you have said makes it possible to use the parametric equation of a closed Lissajous' curve, not to mention the amplitudes being different, the frequency ratio equal to one and the phase shift non-zero.

[quote="pestich"]
    It seems that every time we look at a problem of 2 intersecting circles, with lines
    thru the intersection point,  there is a 3rd circle concurrent to the pair to complete
    the picture to the fine configuration of Simson-Poncelet.

    Maj.  Pestich[/quote]
In conclusion, you did not prove a single iota of your claims. After you have seen a proof, which is indisputable. An idea is not enough, it may or it may not result in a proof. In this case, until you find the fixed intersections of the incircle with the diacentral line to be on the Simson lines with the poles identical with the endpoints of the diacentral diameter, it is like trying to nail a piece of slime to the wall. And to find them, you need not only the properties of Simson lines, but also the special case of the 2nd Fontene theorem, leading to the identity of the orthopole of the diacentral line with the Feuerbach point, and the homotheties between the circumcircle, incircle and the 9-point circle. If you commit this kind of fallacies elsewhere, I may choose to ignore them, but if you commit them in replies to me, you can expect a refutation.

Have fun.

[b]Writer:- Goutham[/b]

[b]Problem:-[/b]

Given a circle $ (O) $ and a circle $ (I) $ inside it. Consider the set of all triangles with circumcircle $ (O) $ and incircle $ (I) $(assume at least one exists). Then prove that for a fixed point $ P $ on $ (O) $, the simson's line of $ P $ with respect to these triangles are concurrent.

[b]Proof:-[/b]

Let $ABC$ be the triangle with orthocenter $H$, centroid $G$, ninepoint centre $N$, Feuerbach point $F$ and anticomplement of $F$ be $F'$. Let midpoint of $PH$ be $M$.
Suppose locus of a point $X$ is a circle. Define by movement of $X$ as the angle subtended on the locus of $x$ by the arc with endpoints as the initial and final positions of $X$ on the circle after a rotation.
Let movement of $A$ be $\angle x$ and $IA$ rotate by $\angle y$ in the direction same as $x$.
First, note that since $\frac{FI}{IN}=\text{constant}$, we have that locus of $N$ to be the locus of $F$ which is the incircle. Since $O, P$ are fixed, the locus of $H, G, M$ are each a circle.
Now, we will prove that the simson line of $P$, which passes through $M$, passes through a fixed point on the locus of $M$. For, this it is sufficient to prove that the movement of $M$ is same as the angle rotated by the simson's line.
The angle rotated by the simson line is same as the angle rotated by the line perpendicular to the simson line, which is parallel to the isogonal conjugate of $AP$. Since $A$ rotates by $\angle x$ and $IA$ rotates by $\angle y$ in the same direction, it is easily seen that the simson line rotates by angle $2y-x$.
The movement of $M$ is same as the movement of $H$, which is same as the movement of $G$ and $F$. The spiral similarity which maps the locus of $F$ onto the locus of $G$, maps the locus of $F$ to locus of $F'$. So, the movement of $F$ is same as the movement of $F'$. Since $AF'$ is isogonal to the perpendicular from $A$ onto $OI$, which is fixed, the movement of $F'$ is seen to be equal to $2y-x$ as the perpendicular from $A$ to $OI$ is fixed and $IA$ rotates by $\angle y$, but $A$ also rotates by $\angle x$ in the same direction.
We have thus proved that movement of $M$ is same as the angle rotated by the simson line and this proves the required statement.

My solution:

Let the $ \odot (O) $ and $ \odot (I) $ be the fixed circumcircle and the fixed incircle .
Let $ P $ be the fixed point on $ \odot (O) $ and $ \ell_1, \ell_2 $ be the line passing through $ P $ and tangent to $ \odot (I) $ .
Let $ P_1=\ell_1 \cap \odot (O) , P_2=\ell_2 \cap \odot (O) $ and $ \ell $ be a line tangent to $ \odot (I) $ and parallel to $ P_1P_2 $.
Let $ T $ be the projection of $ P $ on $ \ell $ and $ D=PP_1 \cap \ell , E=PP_2 \cap \ell $ .
Let $ \triangle ABC $ be an arbitrary triangle with circumcircle $ \odot (O) $ and incircle $ \odot (I) $ .
Let $ K=AP_1 \cap \ell, H=AP_2 \cap \ell  , X=AB \cap \ell  , Y=AC \cap \ell $.
Let $ G=AB \cap P_1P_2, F=AC \cap P_1P_2 $ .

Since $ \angle HDP=\angle P_2P_1P=\angle HAP $ ,
so $ A, P, D, H $ are concyclic .
Since $ \angle EKA=\angle P_2P_1A=180- \angle APE $ ,
so $ A, P, E, K $ are concyclic .
Since $ \angle XIG=\angle DIP_1=\angle FIY=\angle P_2IE=90 $ ,
so $ (K,H;X,Y)=(P_1,P_2;G,F)=(D,E;X,Y) $. i.e. $ \frac {XK \cdot XE}{XD \cdot XH}= \frac {YK \cdot YE}{YD \cdot YH} $
hence we get $ X, Y $ lie on a circle which is coaxial with $ \{ \odot (APDH), \odot (APEK) \} $ . i.e. $ X, Y, A, P $ are concyclic
Since $ P $ is the Miquel point of complete quadrilateral $ AXBCZY $ ,
so the Simson line of $ P $ WRT $ \triangle ABC $ pass through $ T $ which is a fixed point .

Q.E.D

Let $F$ be the feurerbach point, $H$ be the orthocentre of $XYZ$, $XYZ$ be the intouch triangle, $\ell$ be the simson line of $P$.

Clearly, $\dfrac{IF}{IN} = \dfrac{r}{R/2}$ is constant, so trivially the midpoint $M$ of $PH$ traces a circle tangent to $XYZ$ at say $T$, and hence $F \in TM$. We must show $\angle (\ell, TM) = \text{constant}$, which means $\ell$ and $TM$ rotate by same angle in same directions. Let the midpoint of arc $BC$ move $y$ and $A$ move $x$.

Note that if $P'$ is the isogonal conjugate of $P$ wrt $ABC$, then $AP'$ moves $x+2y$, which means $\ell$ moves $x+2y$. Also, as $X$ moves by $y$, and $AI$ moves by $x+y$, it follows $AH \cap \odot XYZ = N$ moves by $x$. But, the reflection of $OI$ on $YZ$ passes through $N, F$, and the reflection of $OI$ rotates twice the angle of $YZ$ which is $2x+2y \implies F$ rotates $x+2y$ so by homothety about $T \implies M$ rotates by $x+2y$ and we are done.


How do you go about proving that if N^2 + 14 and N is prime that N^3+4 is prime?

[hide]$N$ must greater than $2$, if $N=3k+1$ or $3k+2$ then $3|N^{2}+14$ , from this and $N^{2}+14$ is prime therefore $N^{2}+14=3$, contradiction! So $N=3k\Rightarrow N=3$ because $N$ is prime. Final $N^{2}+4=13$ is prime.[/hide]

[quote="chess64"]Actually it's $N^{3}+4 = 31$ (not $N^{2}+4=13$), which is also prime.[/quote]
 :oops: I see is $N^{2}$ and problem is $N^{3}$, but same idea :D


$\mathbb{P}$ is a n-gon with sides $l_1 ,...,l_n$ and vertices on a circle. Prove that no n-gon with this sides has area more than $\mathbb{P}$

Assume we want to maximize the area of such of a polygon. It's easy to show that from a particular $n$-gon we can obtain another one with larger area and which is convex.

Now assume we have another convex $n$-gon with the same sides as our initial one, but larger area. On the sides construct the arcs of the circle in which the first $n$-gon was inscribed (this is where I need convexity: I don't want the arcs to overlap). We get a figure with the same perimeter as that of the circle, but larger area, but it's well-known that among all shapes of the same perimeter, the circle is the one with the largest area, so we have a contradiction.

I just realized that we don't need the convexity: if the second polygon is not convex, then the perimeter and area of the figure we get after drawing the circular arcs on its sides are even smaller, and that's Ok from our point of view (we're trying to show that the area can't be larger than that of the circle with a perimeter $\ge$ that of the figure).

Is this Ok?


inequalities
trigonometry
geometry proposed
geometry

I think it is not difficult(only use some theorems),try it now! :blush:
Given triangle ABC.$ m_a,m_b,m_c$ is the length of the medians from vertex A,B,C.
Prove that $ m_a\plus{}m_b\plus{}m_c$≤4R+r

hello, let $ d_a,d_b$ and $ d_c$ be the distance of the circmcentre from the sides $ BC,CA$ and $ AB$ respectively we have
$ m_a\le R \plus{} d_a$
$ m_b\le R \plus{} d_b$
$ m_c\le R \plus{} d_c$
adding these three inequalities we have
$ m_a \plus{} m_b \plus{} m_c\le 3R \plus{} d_a \plus{} d_b \plus{} d_c$ (1)
Further we get $ d_a \plus{} d_b \plus{} d_c \equal{} R(\cos(\alpha) \plus{} \cos(\beta) \plus{} \cos(\gamma)) \equal{} R(1 \plus{} \frac {r}{R})$ (2)
Inserting (2) in (1) we get $ m_a \plus{} m_b \plus{} m_c\le 4R \plus{} r$, this was to prove.
Sonnhard.

I found another interesting result (for acute triangles ABC)

$ \sqrt {6\sqrt {3} r p \plus{} \frac{_3}{^2}(p^2 \minus{} r^2) \minus{} 6Rr} \le m_a \plus{} m_b \plus{} m_c \le 4R \plus{} r$

[quote="Dr Sonnhard Graubner"]$ d_a \plus{} d_b \plus{} d_c \equal{} R(\cos(\alpha) \plus{} \cos(\beta) \plus{} \cos(\gamma)) \equal{} R(1 \plus{} \frac {r}{R})$[/quote]
This is only true for acute triangles.


hello everybody !!

could someone help me to solve this problem ?? :blush: ...

[url=http://www.casimages.com][img]http://nsa09.casimages.com/img/2009/10/05/091005083758866948.gif[/img][/url]

Thanks ...

Hello , i found the solution , it's your turn now   :lol:

it has been 3 days when i posted this problem , so i'll  try to put my solution


linear algebra
matrix
algebra
polynomial
vector
absolute value
complex numbers

What is the interiour and adherence of complex (real) cyclic matrices?

Definitions? I've never heard "adherence" in any context, and it's not clear what you want "interior" or "cyclic" to mean either.

I was thrown off only by "cyclic matrices" :). However, I decided to think it means matrices $A$ which satisfy $A^m=I$ for some $m$ :). As for the other part, I think it's clear what he means: we have to find the interior and the closure (it's also called "aderenta" in Romanian, and the term he used is probably borrowed from French :)) of the set of cyclic matrices (whatever cyclic may mean; like I said, for now, I'll just assume it means what I said above).

The interior is empty in both cases, because in any neighbourhood of any $n\times n$ matrix there is one whose absolute value of th determinant is not $1$, meaning that not all of its eigenvalues are roots of unity.

The closure of the set consists of all matrices with all eigenvalues of absolute value $1$ in the complex case: Given a diagonal matrix with diagonal entries of absolute value $1$, we can obviously find a sequence of cyclic diagonal matrices converging towards it, so all diagonalizable matrices with eigenvalues of absolute value $1$ belong to the closure. The closure of these, in turn, contains all matrices with eigenvalues with absolute value $1$, as can easily be seen by putting such a matrix in Jordan form. Conversely, it's clear that all matrices in the closure must have eigenvalues with absolute value $1$.

As for the real case, by putting the matrices in rational canonical form, we can reduce the problem to the $2\times 2$ case: is it true that all $2\times 2$ real matrices with eigenvalues of absolute value $1$ lie in the closure of the set of real $2\times 2$ matrices?

The answer is clearly positive, since every matrix as descirbed above can be approximated arbitrarily well with a $2\times 2$ real matrix whose trace is twice the real part of a root of unity and whose determinant is $1$. This means that the answer in the real case is the same as in the complex case, with the obvious modifications: the closure of the set of real cyclic matrices is the set of real matrices with eigenvalues of absolute value $1$.

I hope it's correct.

"cyclic" matrix is an expression used in France, and perhaps only in France, to call matrices that have a minimal polynomial equal to their characteristic polynomial. (it is equivalent to say there is a vector $x$ such that $(x, Ax, A^2x,..,A^{n-1}x)$ is a basis)

Oops! Everything I did above has nothing to do with the problem harazi intended to post.. :). Back to work then! :)

Edit:

The closure is $\mathcal M_n(\mathbb C)$, since this is the case for the set of matrices with distinct eigenvalues.

The interior is the entire set (i.e. the set of cyclic matrices is open). Take any cyclic matrix $A$ and a vector $x$ as described above by alekk. Arrange the vectors $x,Ax,\ldots,A^{n-1}x$ into a non-zero determinant $D(A,x)$. If $A'$ is another matrix, close enough to $A$, then $D(A',x)$ will also be non-zero, meaning that $A'$ will be cyclic.

yep, that is.
You can also study the connectedness of this set. Or what about the set of matrices with minimal polynomial of degree $d$, where $d$ is a fixed integer ?

We can see that the set $\mathcal C$ of complex cyclic matrices is connected (and thus path connected, since these two coincide for open subsets of a Euclidean space) as follows:

$\mathcal C$ is locally connected, so the connected components are open. However, the closure of the component $\mathcal C_1$ containing all the diagonal matrices with distinct eigenvalues is the entire $\mathcal M_n(\mathbb C)$, so $\mathcal S$ can have no component different from $\mathcal C_1$ (because it would be an open set disjoint from $\mathcal C_1$).

As I recall, Moubinool posted the problem concerning the path connectedness of the set of complex matrices with minimal polynomial of degree $d$, but it was a bit trickier than this one. I'll give it some more thought.


AMC AIME email probability geometry trigonometry circumcircle

Edit- Thanks to Yetti for correcting me on a problem. On #10, MD = 39 has been changed to MD = 113.

The Rules
-------------

[b]This contest has officially ended.[/b] (A discussion of the contest can be found in the thread "Mock AIME 2 Results / Solutions / Discussion" [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=107441#107441[/url].) The rules are the same as last time. No calculators, slide rules, abaci, or any other calculation devices are permitted. You are encouraged to use scrap paper including graph paper. You will have three hours, from 2:00 - 5:00 PM EST, to complete the contest. After the contest ends, please PM you answers to me. Alternatively, you could email your answers to tmildorf@cox.net .  You need only send the final answers, no work is required, although if you have a particularly elegant or unusual solution, please submit this too. If I think that it illustrates something instructive I will add it to the solutions packet.  Credit will be received only for correct answers, so [b]check your work[/b]. There is no penalty for guessing. All problems are weighted equally, so don't spend too much time on any one question. If you have a question about any problem, please post your concern in this thread. I will try to answer you as quickly as possible and will post my response in this thread as well.

The Problems
-------------

The .pdf version of the problems can be found at [url]http://www.tjhsst.edu/~tmildorf/math/PracticeAIME2.pdf[/url]. They are reproduced here for your convenience.

$1.$ Compute the largest integer $k$ such that $2004^k$ divides $2004!$.

$2.$ $x$ is a real number with the property that $x + \frac{1}{x} = 3$. Let $S_m = x^m + \frac{1}{x^m}$. Determine the value of $S_7$.

$3.$ In a box, there are 4 green balls, 4 blue balls, 2 red balls, a brown ball, a white ball, and a black ball. These balls are randomly drawn out of the box one at a time (without replacement) until two of the same color have been removed. This process requires that at most 7 balls be removed. The probability that 7 balls are drawn can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.

$4.$ Let $S := \{5^k | k \in \mathbb{Z}, 0 \le k \le 2004\}$. Given that $5^{2004} = 5443 \cdots 0625$ has 1401 digits, how many elements of $S$ begin with the digit 1?

$5.$ Let $S$ be the set of integers $n > 1$ for which $\frac{1}{n} = 0.d_1d_2d_3d_4\dots$, an infinite decimal that has the property that $d_i = d_{i+12}$ for all positive integers $i$. Given that 9901 is prime, how many positive integers are in $S$? (The $d_i$ are digits.)

$6.$ $ABC$ is a scalene triangle. Points $D, E$, and $F$ are selected on sides $BC, CA$, and $AB$ respectively. The cevians $AD, BE$, and $CF$ concur at point $P$. If the areas of triangles $AFP, FBP$, and $CEP$ are $126, 63$, and $24$ respectively, then determine the area of triangle $ABC$.

$7.$ Anders, Po-Ru, Reid, and Aaron are playing Bridge. After one hand, they notice that all of the cards of two suits are split between Reid and Po-Ru's hands. Let $N$ denote the number of ways 13 cards can be dealt to each player such that this is the case. Determine the remainder obtained when $N$ is divided by 1000. (Bridge is a card game played with the standard 52-card deck.)

$8.$ Determine the remainder obtained when the expression

\[
\Large{2004^{2003^{2002^{2001}}}}
\]
is divided by 1000.

$9.$ Let

\[\left(1 + x^3\right)\left(1 + 2x^{3^2}\right)\cdots\left(1 + kx^{3^k}\right)\cdots\left(1 + 1997x^{3^{1997}}\right) = 1 + a_1x^{k_1} + a_2x^{k_2} + \cdots + a_mx^{k_m}\]

where $a_i \ne 0$ and $k_1 < k_2 < \cdots < k_m$. Determine the remainder obtained when $a_{1997}$ is divided by 1000.

$10.$ $ABCDE$ is a cyclic pentagon with $BC = CD = DE$. The diagonals $AC$ and $BE$ intersect at $M$. $N$ is the foot of the altitude from $M$ to $AB$. We have $MA = 25, MD = 113$, and $MN = 15$. $[ABE]$ can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Determine the remainder obtained when $m+n$ is divided by 1000.

$11.$ $\alpha, \beta$, and $\gamma$ are the roots of $x(x-200)(4x+1) = 1$. Let

\[
\omega = \tan^{-1}(\alpha) + \tan^{-1}(\beta) + \tan^{-1}(\gamma)
\]
The value of $\tan(\omega)$ can be written as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Determine the value of $m + n$.

$12.$ $ABCD$ is a cyclic quadrilateral with $AB = 8, BC = 4, CD = 1$, and $DA = 7$. Let $O$ and $P$ denote the circumcenter and intersection of $AC$ and $BD$ respectively. The value of $OP^2$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime, positive integers. Determine the remainder obtained when $m+n$ is divided by 1000.

$13.$ $P(x)$ is the polynomial of minimal degree that satisfies

\[
P(k) = \frac{1}{k(k+1)}
\]
for $k = 1, 2, 3, \dots, 10$. The value of $P(11)$ can be written as $-\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $m+n$.

$14.$ 3 Elm trees, 4 Dogwood trees, and 5 Oak trees are to be planted in a line in front of a library such that

\begin{eqnarray*}
i)\phantom{iiM}\textrm{No two Elm trees are next to each other.}\phantom{xxx} \\
ii)\phantom{iM}\textrm{No Dogwood tree is adjacent to an Oak tree.}\phantom{} \\
iii)\phantom{M}\textrm{All of the trees are planted.}\phantom{xxxxxxxxxxxxx|}
\end{eqnarray*}
How many ways can the trees be situated in this manner?

$15.$ In triangle $ABC$, we have $BC = 13, CA = 37$, and $AB = 40$. Points $D$, $E$, and $F$ are selected on $BC$, $CA$, and $AB$ respectively such that $AD$, $BE$, and $CF$ concur at the circumcenter of $ABC$. The value of

\[
\frac{1}{AD} + \frac{1}{BE} + \frac{1}{CF}
\]
can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Determine $m + n$.

[i]Moderator's note: sticky removed.  -mathfanatic[/i]

I would like to take the test early...either tonight or tomorrow afternoon. Thanks.

[quote="joml88"]Mildorf, what does the test look like?  I mean how many problems from each category did you get?  If you don't want to share that info. I ll be fine until Sunday.[/quote]

I will share it since it is sufficiently vague that it probably won't help anyone get an unfair advantage. I think it's a pretty balanced test. My Problem Distribution:

(Easy:Medium:Hard / Algebra:Combinatorics:Geometry:Number Theory)

1. E/N
2. E/A
3. E/C
4. M/N
5. M/N
6. M/G
7. M/C
8. M/N
9. M/A
10. H/G
11. H/A
12. H/G
13. H/A
14. H/C
15. H/G

Could I take it early?  Perhaps sometime on Saturday....

I do not want to take a chance that I've made a silly mistake in the answer key and score contests with a wrong answer. Since nobody is going to check over my solutions before the contest, I won't be grading your answers immediately and we will actually have a discussion of the solutions in the results / solutions / discussion thread to verify their accuracy before I compile grade submissions / compile statistics.

Also, why is there always someone who accidentally votes "Easy"? :lol:

I'm soo sad that i can't take this test - *sniff*. Stupid Washington - I'm leaving tomorrow - bleh. I couldn't find time today either. :(  :(  :(

[quote="Rep123max"]So I'll know my score by Tuesday/Wednesday right?[/quote]

Yes. I don't think it should take us that long to verify / correct the solutions since even if there is a calculation error all of the ideas are there. Hopefully I will have results posted on Sunday night or early on Monday.

I have to take the mock AIME on Saturday, which is wide open for me.

[quote="white_horse_king88"]If I miss this AIME will it affect my chances in competing in the mock USAMO?[/quote]

I guess you won't be able to qualify through this particular test, but I will be willing and able to create a third some time in the next few weeks. I am running out of good original ideas so I will need some time. :D

[quote="Mildorf"]
Also, why is there always someone who accidentally votes "Easy"? :lol:[/quote]

Wasn't me this time (I am not even going to take the test.  I got to work as an intern on Saturday).

what time is the contest in GMT? (or GMT+1 to be more precisely)

also, what are the contest rules (points for blank answer? calculator? other allowed instruments?)

could be handy to know (i didn't know last time :))

Oh, The test is on Sunday, August 8?  I thought it was on Saturday, August 7.  Well, then, I think I am going to take the test.  (or not)

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3741707_2004_mildorf_mock_aime_25]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


how would you solve a problem like:
a+bx=c^x
for x where a, b, and c are all real numbers.

example:
8.64 + 4.25x = 2^x

these problems do not have general solutions...

I do have the solutions for the problem.. i just dont know how to get there...
the soulutions for 8.64 + 4.25x = 2^x are
x=-1.97300667761 or 4.875908327

these answers are obviously not exact but they are close.
i got the answers from graphing each side of the equation and then using the intersection thing on my calculator.

so becides the calculator, is there any way to do this? thanks.

You can guess and check till the end of your life.

Just use a calculator, because I doubt they will ask you a question like this and make you do it by hand.


[quote="sagar_astakala"][hide]1/8 and 1/10 are in H.P. as 8 and 10 are in A.P.

The no. in halfway is    {(2)(1/8)(1/10)}/{(1/8)+(1/10)}

this will give the result 1/9[/hide][/quote]

Are you sure they wanted the harmonic mean? It looks like the arithmetic mean to me...

I'm pretty sure they want the arithmetic mean, so it's E.

:lol: Even the title says Arithmetic Mean  :P

[quote=Eric2009][quote=4everwise]The number halfway between $ \frac {1}{8}$ and $ \displaystyle \frac {1}{10}$ is

$ \textbf{(A)}\  \frac {1}{80} \qquad \textbf{(B)}\  \frac {1}{40} \qquad \textbf{(C)}\  \frac {1}{18} \qquad \textbf{(D)}\  \frac {1}{9} \qquad \textbf{(E)}\  \frac {9}{80}$[/quote]

$\frac{1}{8}-\frac{1}{10}=\frac{1}{40}$ So$\frac{1}{10}+\frac{1}{80}=\frac{9}{80}=\boxed{E}$[/quote]

14 year bump why

[quote=franzliszt]I would've use common denominator: $\frac{10}{80}$ and $\frac{8}{80}$ so the middle number is $\frac{9}{80}$.[/quote]

Yeah, just turned the common denominator into 80.

I think answer [hide=is]
$\frac{\frac18+\frac1{10}}2\implies{\frac{\frac{9}{40}}2}\implies\boxed{{\frac9{80}}}$

Our answer is $\boxed{\text{E}}$

[hide=Answer] Multiplying both sides by their common denominator, they both simplify to $\frac{10}{80}$ and $\frac{8}{80}$.  By simple intuition, it can be concluded that the the midpoint between the fractions is $\frac{9}{80}$, or option $\boxed{E}$. [/hide]


is it possible to solve a equation that's like this?
2^n=n^2
i can solve this graphically, getting 2, 4, and -.7666646, but is there a algebraic way to do this?

i could be wrong but.
[hide]n log 2 = n^2
log 2 = n^2 - n
0 = n^2 - n - log 2
Just use quadratic(or factor if you're into that sort of thing)from here. :)[/hide]

[quote="mickeythedunce"]n log 2 = n^2[/quote]

Obviously wrong.  $2^{n}\neq n \log 2$.

See http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm .  (People keep posting this problem!)


Έστω ένα τρίγωνο $ ABC$ και ο εγγεγραμένος του κύκλος ο οποίος εφάπτεται στις $ BC,CA,AB$ στα $ D,E,Z$.Έστω $ P$ το αντιδιαμετρικό του $ D$.Έστω οι $ AP,ZP,EP$ τέμνουν την $ BC$ στα $ M,K,N$.Νδο $ MK\equal{}MN$

[quote="Athinaios"]hint
[hide]μοιαζει παρα πολυ με την προηγουμενη του Σακη!  :wink: [/hide][/quote]
Σωστός.
Η λύση μου είναι σχεδόν ίδια με τη δικιά σου στη άσκηση του Σάκη γι'αυτό και την ανέβασα.... :P
Anyway νομίζω ότι είναι πολύ ωραία άσκηση....

Συμφωνω. Γενικα τα αρμονικα (και ολα τα συναφη) ειναι ενα πολυ ομορφο κεφαλαιο ευκλειδιας γεωμετριας!  :wink:

Ας βάλω και μια άλλη λύση που σκέφτηκα για το πρόβλημα, αυτή τη φορά χωρίς χρήση αρμονικών (και αρκετά πιο στοιχειώδης).

Απαραίτητη γνώση είναι ότι το $ M$ είναι το συμμετρικό του $ D$ ως προς το μέσο της $ BC$ (σημείο που εφάπτεται κι ο παραγγεγραμμένος).

Τώρα έχουμε εύκολα: $ \angle{EPD}\equal{}EZD\equal{}CID\equal{}90\minus{}\frac{C}{2}\iff PE// CI$
Ομοια $ PZ//BI$. (*)
Ετσι τώρα τα τρίγωνα $ PKN, BIC$ είναι ομοιόθετα με κέντρο ομοιοθεσίας το $ D$ και λόγο 2:1 (αφού η $ PI$ τέμνει την $ BC$ στο $ D$ kai $ PD\equal{}2DI$). Αυτό σημαίνει ότι το μέσο της KN είναι το συμμετρικό του D ως προς το μέσο της $ BC$ και άρα το ζητούμενο!

Υ.Γ. Απ' το (*) μπορούμε να τελειώσουμε επίοσης την άσκηση ως εξής:
$ PK//BI\iff BK\equal{}BD$. Ομοια $ DC\equal{}CN$.
Etsi tora: $ KM\equal{}BK\plus{}BM\equal{}BD\plus{}CD\equal{}BC$ και όμοια $ MN\equal{}BC$ κι έτσι αποδείξαμε το ζητούμενο χωρίς καν χρήση ομοιοθεσίας :P


function
real analysis
real analysis solved

Let $a,b\in \mathbb{R},a<b$ assume that $f:[a,b]\rightarrow [a,b]$ satisfies$|f(x)-f(y)|\leq |x-y|$ for every $x,y\in [a,b]$choose $x_1\in [a,b]$ and $x_{n+1}=\frac{x_n+f(x_n)}{2}$ show that $\{\ x_n \}\ _{n=1}^{\infty}$ converges to some of the fixed point of f.

The main idea is that the function $ g(x)=\frac{x+f(x)}{2} $ is increasing and continuous. This follows immediately from the condition of the problem. But it is well-known and easy to prove that for an increasing and continuous function the sequence of consecutive iterates converges to a fixed point of the function ( since the sequence of iterates is monotonuous). From here the conclusion follows.


[color=darkblue]Prove easily that in a triangle $ABC$ exists the inequality $\frac{\sqrt{p-a}}{a}+\frac{\sqrt{p-b}}{b}+\frac{\sqrt{p-c}}{c}\le\frac{\sqrt p}{2r}\ .$

( See http://www.mathlinks.ro/Forum/viewtopic.php?t=117311 )

Is truly and the left inequality : $\frac{\sqrt p}{R}\le \sum\frac{\sqrt{p-a}}{a}\ ?$

[b]Remark.[/b] Can use possibly the well-known inequality $\frac{1}{R^{2}}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\le\frac{1}{4r^{2}}\ .$[/color]

Are you sure about the first one?
Both inequalities are not homogenous, so they are a bit suspect... :wink:

Pierre.


does anyone have a combinatorial interpretation for this?

I just stumbled upon this thread again. Let me rewrite the question in more readable notations:

Prove that $\sum_{i=0}^{p} \dbinom{p}{i}^2 \dbinom{n+2p-i}{2p} = \dbinom{p+n}{p}^2$ for any $n \in \mathbb{Z}$ and $p \in \mathbb{N}$.

This is actually the particular case (for $a = p$, $b = p$ and $x = n$) of the following fact:

[color=green][b](1)[/b][/color] $\sum_{i=0}^{b} \dbinom{a}{i} \dbinom{b}{i} \dbinom{a+b+x-i}{a+b} = \dbinom{a+x}{a} \dbinom{b+x}{b}$ for any $x \in \mathbb{Z}$, $a \in \mathbb{N}$ and $b \in \mathbb{N}$.

This is Proposition 2.30 (g) in my [url=http://www.cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf][i]Notes on the combinatorial fundamentals of algebra[/i][/url]. The numbering will probably change, but you can find this proposition by searching for the words "an assortment of other identities" (or else, you can look at the [url=https://github.com/darijgr/detnotes/releases/tag/2016-09-04]archived version from 4 September 2016[/url], in which it is guaranteed to be Proposition 2.30 (g)).

This does not, per se, give a bijective proof, though -- my proof of [color=green][b](1)[/b][/color] uses upper negation. However, I think generalizing to [color=green][b](1)[/b][/color] is a reasonable first step, whatever kind of proof one wants to see.


trigonometry
calculus
calculus computations

$ \sum_{n\equal{}1}^{\infty} \frac {\sin{n}}{n!}$

Nice problem!

[hide="solution"]
\[ \sum_{n \equal{} 1}^\infty{\frac {\sin n}{n!}} \equal{} \frac {1}{2i}\sum _{n \equal{} 1}^\infty {\frac {e^{in} \minus{} e^{ \minus{} in}}{{n!}}}
\]
Notice that the sum is precisely the sum of two exponential series
\[ \equal{} \frac {1}{2i}\left(e^{e^{ i}} \minus{} e^{e^{ \minus{} i}}\right) \equal{} \frac {1}{2i}\left(e^{\cos 1}\left(e^{i\sin 1} \minus{} e^{ \minus{} i\sin {1}}\right)\right)
\]

\[ \equal{} e^{\cos 1}\sin \left(\sin 1\right)
\]
[/hide]


note to people I just had a new sibling.

it's a boy.

This is just basically spam...it's nothing to do with Illinois math or whatever public reason stuff...If you wanna post this, keep it to your blog, not on the forums. This forum is [b]not[/b] the right place to post this.

Guys, let's just stop this topic please....

[quote="jonathanchou711"]LOL. Is the missouri forum actually just spam?[/quote]

See for yourself.

[quote="probability1.01"]Guys, I felt obliged to post to say that this thread is spam.[/quote]

QFT

Because Zhai is my idol, I will follow in his footsteps and say
...this thread is spam...  :ninja:

[quote="gauss1181"]Same here.

*topic is now closed to an end*[/quote]

Well why is everyone here?

I was going to ignore this but I know someone else would post after your post anyway.

By *topic now closed to an end* that implies something along the lines of nobody post in this topic again.

So seriously stop spamming this forum with random junk that's not related at all.

Tytia.

[quote="probability1.01"]Oh man I would like to second that point... this topic is spam and there is no point in discussing it[/quote]

yeah man I wholeheartedly agree

what is up with this spam

Totally guys this is all spam why do you all spam... I mean my post is completely not spam since it only contains the word spam like spam number of spam times.

oh jeez the spam is getting to me

MOD MOVE TO MO Forum =D.

new idea.

proove that not all triangles can be made in to 2 congruent parts.

talk mostly in mathematics.

x=>y


trigonometry
induction
inequalities
real analysis
real analysis solved

Find the angles $\alpha$ for which $cos(2^n\alpha)\leq 0\ \forall n\in\mathbb{N}$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3495]this thread[/url] for pretty much the same problem.

\par We can assume that $\alpha\in [\frac{\pi}2,\frac{3\pi}2]$ because the addition or substraction of $2\pi$ does not influence the property in the text
of the problem. By hypothesis, for all $k\geq 0$, there exists $x_k\in\mathbb{N}$ such that $2x_k\pi+\frac{\pi}2\leq 2^k\alpha\leq 2x_k\pi+\frac{3\pi}2$.
\par Let $\beta=\frac{\alpha}{2\pi}$. Then $x_k+\frac 14\leq 2^k\beta\leq x_k+\frac 34\ \forall k\in\mathbb{N}$.\textbf{[1]}
It is easy to see that $\beta$ is not of the form $\frac{m}{2^l}$ otherwise $2^l\beta$ is an integer and this contradicts [1] which says that
$\{2^k\beta\}\in [\frac 14,\frac 34]\ \forall k\in\mathbb{N}$, where $\{x\}$ is the fractional part of $x$. So [1] can be replaced by
$x_k+\frac 14<2^k\beta<x_k+\frac 34\ \forall k\in\mathbb{N}$.\textbf{[2]} This is equivalent to
$\{2^k\beta\}\in(\frac 14,\frac 34)\ \forall k\in\mathbb{N}$. \textbf{[3]}
\par If $\frac 14<\beta<\frac 12$, then $\frac 12<2\beta<1$. By [2], $x_1+\frac 14<2\beta<x_1+\frac 34$, so $x_1=0$ and $\frac 12<2\beta<\frac 34$.
Notice that if $\beta$ verifies [3], then $2\beta$ also verifies [3], so we can assume that $\frac 12<\beta<\frac 34$. In this case, I prove by induction
on $n$, that $x_{2n}=2\frac{4^n-1}3$ and $x_{2n+1}=\frac{4^{n+1}-1}3$.
\par For $n=0$, we have $x_0+\frac 14<\beta<x_0+\frac 34\Rightarrow x_0=[\beta]=0=2\frac{4^0-1}3$, where $[x]$ is the integer part of
$x$. Notice that [1] implies that $x_k=[2^k\beta]\ \forall k\in\mathbb{N}$. By the assumption $\frac 12<\beta<\frac 34$, $x_1=[2\beta]=1=\frac{4^1-1}3$.
Thus we have checked the induction hypothesis.
\par Assume now that we have proved $2\frac{4^n-1}3+\frac 14<2^{2n}\beta<2\frac{4^n-1}3+\frac 34$ and $\frac{4^{n+1}-1}3+\frac 14<2^{2n+1}\beta<\frac{4^{n+1}-1}3+\frac 34$. By multiplying the first by 2,
$\frac{4^{n+1}-1}3-\frac 12<2^{2n+1}\beta<\frac{4^{n+1}-1}3+\frac 12$. By comparing the last two inequalities,
$\frac{4^{n+1}-1}3+\frac 14<2^{2n+1}\beta<\frac{4^{n+1}-1}3+\frac 12$. Multiplying by 2 yields
$2\frac{4^{n+1}-1}3+\frac 12<2^{2n+2}\beta<2\frac{4^{n+1}-1}3+1\Rightarrow x_{2n+2}=[2^{2n+2}\beta]=2\frac{4^{n+1}-1}3$. Considering [2], the last
inequality can be improved to $2\frac{4^{n+1}-1}3+\frac 12<2^{2n+2}\beta<2\frac{4^{n+1}-1}3+\frac 34$. If we multiply this by 2, we get
$\frac{4^{n+2}-1}3<2^{2n+3}\beta<\frac{4^{n+2}-1}3+\frac 12\Rightarrow x_{2n+3}=[2^{2n+3}\beta]=\frac{4^{n+2}-1}3$.
\par So for all $n\in\mathbb{N}$, $2\frac{4^n-1}3+\frac 14<2^{2n}\beta<2\frac{4^n-1}3+\frac 34\Rightarrow \frac 23-\frac 2{4^n3}+\frac 1{4^{n+1}}<\beta<\frac 23-\frac 2{4^n3}+\frac 3{4^{n+1}}$. When $n$ tends to infinity, we get
$\frac 23\leq\beta\leq\frac 23\Rightarrow \beta=\frac 23$.
\par Corresponding to $\frac 14<\beta<\frac 12$, we have $\frac 12<2\beta<\frac 34$, so $2\beta=\frac 23\Rightarrow \beta=\frac 13$. So the only solutions to [3] are $\frac 13$ and $\frac 23$. These correspond
in the initial problem to $\alpha\in\{\frac{2\pi}3,\frac{4\pi}3\}$.

Hmm, I thought there was a simpler solution.  Consider the base 2 expansion of $\frac{\alpha}{2 \pi}$.  To the right of the decimal point, if there are ever two 0's in a row or two 1's in a row, then we reach a contradiction quickly.  Thus the binary expansion must alternate 0's and 1's, which I think was enough to restrict $\alpha$ to the required solution set.


limit
logarithms
integration
real analysis
real analysis unsolved

Prove that $ \lim_{n\to\infty}\ \left(\frac {1^1\cdot 2^2\cdot 3^3\cdot\cdot\cdot (n\minus{}1)^{n\minus{}1}\cdot n^n}{n^{1\plus{}2\plus{}3\plus{}\cdot\cdot\cdot \plus{}(n\minus{}1)\plus{}n}}\right)^{\frac {1}{n^2}}\equal{}\frac {1}{\sqrt [4]e}\ .$

[quote="Virgil Nicula"]Prove that $ \lim_{n\to\infty}\ \left(\frac {1^1\cdot 2^2\cdot 3^3\cdot\cdot\cdot (n \minus{} 1)^{n \minus{} 1}\cdot n^n}{n^{1 \plus{} 2 \plus{} 3 \plus{} \cdot\cdot\cdot \plus{} (n \minus{} 1) \plus{} n}}\right)^{\frac {1}{n^2}} \equal{} \frac {1}{\sqrt [4]e}\ .$[/quote]

let
$ L\;\equal{}\;\lim_{n\to\infty}\ \left(\frac {1^1\cdot 2^2\cdot 3^3\cdot\cdot\cdot (n \minus{} 1)^{n \minus{} 1}\cdot n^n}{n^{1 \plus{} 2 \plus{} 3 \plus{} \cdot\cdot\cdot \plus{} (n \minus{} 1) \plus{} n}}\right)^{\frac {1}{n^2}}$

$ \equal{}\;\lim_{n\to\infty}\ \left\{ \left(\frac{1}{n}\right)^1 \cdot  \left(\frac{2}{n}\right)^2 \cdot  \left(\frac{3}{n}\right)^3 \cdot \;\;\;\ldots\;\;\; \cdot  \left(\frac{n\minus{}1}{n}\right)^{n\minus{}1} \cdot  \left(\frac{n}{n}\right)^n \right\}^{\frac {1}{n^2}}$

and so,
$ \ln\, L\;\equal{}\;\lim_{n\to\infty}\ \frac {1}{n^2}\;\left\{ 1\cdot \ln\,\left(\frac{1}{n}\right) \;\plus{}\; 2\cdot \ln\,\left(\frac{2}{n}\right) \;\plus{}\; 3\cdot \ln\,\left(\frac{3}{n}\right) \;\plus{}\; \;\;\;\ldots \;\;\; \plus{} (n\minus{}1)\cdot \ln\,\left(\frac{n\minus{}1}{n}\right) \;\plus{}\; n\cdot \ln\,\left(\frac{n}{n}\right) \right\}$

$ \equal{}\; \lim_{n\to\infty}\ \frac {1}{n^2}\; \sum_{k\equal{}1}^n \; k\cdot \ln\,\left(\frac{k}{n}\right)$

$ \equal{}\; \int_0^1 \; x\cdot\ln\,x\;\;\textbf dx \;\equal{}\; \frac{1}{4}\;x^2 \left(2\ln\,x \;\minus{}\; 1\right)\;\bigg|_0^1 \;\equal{}\; \minus{}\frac{1}{4} \;\;\;\quad \;\;\; \left[ \;\;\; \because \lim_{x\to 0} \;x\cdot\ln\,x \;\equal{}\; 0 \;\;\;\right]$

therefore,
$ L\;\equal{}\;e^{\minus{}\frac{1}{4}}$

Another method to solve this problem is to use the Glaisher Kinkelin constant technique as follows:

It is known that

$ \lim\limits_{n\rightarrow \infty}A_{n}\equal{}\lim\limits_{n\rightarrow \infty}\left(\frac{1^{1}\cdot 2^{2}\cdots n^{n}}{n^{n^{2}/2\plus{}n/2\plus{}1/12}e^{\minus{}n^{2}/4}}\right)\equal{}A$,

where $ A$ is the Glaisher Kinkelin constant.

Now, if $ x_{n}$ is your sequence, then a calculation shows that

$ x_{n}\equal{}A_{n}^{1/n^{2}}\cdot n^{1/(12n^{2})}e^{\minus{}1/4}$ which converges to  the desired limit.


inequalities
analytic geometry
geometry
symmetry
blogs
rearrangement inequality

Without using inequalities,coordinate geometry,find the type of triangle represented by the following equation

$a^2+b^2+c^2=ab+bc+ca$

(I know,really old question) :lol:

Well, since I don't see any $x$'s or $y$'s, may I safely assume that this is some sort of math pun?

O.k,I didn't mention $a,b,c$ to be sides of the triangle

Would it be
[hide]equilateral[/hide]
then?

I'm not sure how to prove it, but I suppose that the proof would involve symmetry or something (WLOG)...

There is a really easy method,really easy.I had my friends busting their head against the table when they tried so many cases and i finished it in a step  :D

[hide]You can tell that if a=b=c, by common sense and ingenuity, then the equality becomes true.

So it is an equilateral triangle.[/hide]

You can't say if $\ a=b=c$ the inequality becomes true, and thus it is proved.  For one, there could be other solutions.  I believe the intended solution is to multiply everything by $\ 2$ and then regroup all the terms on one side we will get $\ (a-b)^2 + (b - c)^2 + (c-a)^2 =0$ now since $\ a, b, c$ are real we find $\ a=b=c$ and thus it is equilateral.

Am I missing something?
[hide]If triangle abc is equilateral, then a=b=c. In this case, a single variable may be used to represent all the side lengths.
so... $a^2+a^2+a^2=aa+aa+aa$     or      $3a^2=3a^2$. [/hide]

mna851 is right.Mathematicus,read mna's post.

[quote="leo-pold"]I browsed around Getting Started today and realized that the problems are once again getting too difficult. Please refer to the posting guidelines to find out which forum your problems are appropriate for.

If you're not sure whether to post your problem in Getting Started of the Intermediate forum, post in the Intermediate forum.
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What's with the bulk advertising  :mad:

Yeah, someone posted a bunch of bulk medical advertising in my blog too :wallbash_red:

Is there a proof that a=b=c is the [b]only[/b] solution to the equality?

Bictor's method is probably the way to show that $\ a=b=c$ is the only solution.  We see that by AM-GM or rearrangement $\ a^2 + b^2 + c^2$ must always be greater than or equal to $\ ab + ac + bc$ with equality only holding when $\ a=b=c$

[quote="Bictor717"]It's the rearrangement inequality: $a^2+b^2+c^2\ge ab+bc+ca$, equality when $a=b=c$. mna851's solution offers a neat proof. (It generalizes to more terms of course.)[/quote]

Yeah, the solution is good, but didn't he say that you shouldn't use inequalities.  Though inequalities is a nice solution of doing things.


[quote="lucasm1"]Here is the problem:

3 travelers come to an inn. They order baked potatoes. The 1st traveler eats 1/3 of the potatoes, the 2nd traveler eats 1/3 of the REMAINING potatoes ,the 3rd traveler eats 1/3 of the REMAINING potatoes. There are 8 potatoes left. How many potatoes were there originally?[/quote]
Not really logic.
The first traveler gets $ \frac{1}{3}$ of the potatoes
The second gets $ \frac{1}{3}(1\minus{}\frac{1}{3})\equal{}\frac{2}{9}$ of the potatoes.
The third gets $ \frac{1}{3}(1\minus{}\frac{1}{3}\minus{}\frac{2}{9})\equal{}\frac{4}{27}$ of the potatoes.
In total, they ate $ \frac{1}{3}\plus{}\frac{2}{9}\plus{}\frac{4}{27}\equal{}\frac{19}{27}$ of the potatoes.
Therefore,
$ 8\equal{}(1\minus{}\frac{19}{27})x$
$ 8\equal{}\frac{8}{27}x$
$ x\equal{}27$

[hide]Work Backwards.

Before the third traveler eats, there are 12 potatoes.

Before the second traveler eats, there are 18 potatoes.

Before the first traveler eats, there are 27 potatoes.

Or note that eat time someone eats, the number of potatoes is multiplied by 2/3[/hide]

man its 27
as the keyword here is REMAINING

work backwards
1. 8=2/3
   12=3/3
2. 12=2/3
    18=3/3
3.  18=2/3
     27=3/3

hence, there was 27 potatoes
 :football: at the beginning :football:


function
induction
algebra proposed
algebra

Let $a\in\mathbb{R}-\{0\}$. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that $f(a+x) = f(x) - x$ for all $x\in\mathbb{R}$.

[i]Dan Schwartz[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31951]All the questions of the test[/url]

Partition reals onto disjoint arithmetic progressions with difference $a$. Then we may define $f$ in every progression $\{x_0+na\}$, by chosing $f(x_0)$ arbitrary. So, we have a lot of such functions. Maybe, I missed something? Or there are additional conditions, like continiuty?

Same idea as Fedor Petrov:
Let $g(x) = f(x) - \frac{x^2}{2a}+\frac{x}{2}$

Then $g(a+x)=g(x)$.

So it is enough to define arbitrarily $g(x)$ on $[0,a)$ and it will be periodic.
From $g(x)$ we construct $f(x)$.

There are no additional conditions. This is the problem.

The function $f(x)=-\frac{x(x-a)}{2a}$ satisfys the condition  :lol:
Is it the only one ??

I found something :
Putting $\alpha =f(0)$ we can show by induction that $f(n)=-\frac{n(n-a)}{2a} +\alpha$ for all $n \in \mathbb N$
So by using the method of Cauchy we can prove that $f(x)=-\frac{x(x-a)}{2a} +\alpha$ for all $x \in \mathbb R$

u can find it only for n=ak where k is a positive integer

No I don't think it's wrong. So what prowler said was wrong  :?

anyway, i found out what the error was.
the function was actually $g(x) = f(x) + \frac{x^2}{2a}-\frac{x}{2}$
Then $g(a+x)=g(x)$.

I think it is easier to observe $f(x) = g(x) - x[\frac{x}{a}]$. g is the same as above.


As shown below, find the radius of the smallest circle, if you are given that the radius of the largest one is 12.

[hide]I consider the image is a graph
Draw $ CH\perp Ox$. Let $ R$ be the radius of $ (C)$
By Pythagorean theorem:
In $ \triangle{ACH}$: $ AH^2\equal{}AC^2\minus{}CH^2\equal{}(12\minus{}R)^2\minus{}R^2\equal{}144\minus{}24R\plus{}R^2\minus{}R^2\equal{}144\minus{}24R$
in the trapezoid $ ABCH$: $ AH^2\equal{}(AB\plus{}HC)^2\minus{}(AB\minus{}HC)^2\equal{}(6\plus{}R)^2\minus{}(6\minus{}R)^2\equal{}24R$
So $ 24R\equal{}144\minus{}24R$ and therefore: $ R\equal{}3$
We done[/hide]

[quote="Unidranix"][hide]
In $ \triangle{ACH}$: $ AH^2 \equal{} AC^2 \minus{} CH^2 \equal{} (12 \minus{} R)^2 \minus{} R^2 \equal{} 144 \minus{} 24R \plus{} R^2 \minus{} R^2 \equal{} 144 \minus{} 24R$
[/hide][/quote]
 Can you prove that $ AC \equal{} 12 \minus{} R$? since it is not given that $ \overrightarrow{AC}$ meets the smallest circle at the point of the intersection!

I did much the same as unidranix ( nd also got $ r\equal{}3$), as it seems clear to me that the normal to the tangent at $ D$, say, where $ AC$ meets the circle is the same for the small circle as for the large one. So $ CD \equal{} r$ and $ OC \equal{} 12\minus{}r$.


Will the math jam tommorow (11/15/04) be like the other ones you had for the Putnam in that it is on a certain topic or is it more like randomly selected problems?

Nvm.  I just checked my mail and got a message about the math jam!  It says:

[quote]We will be focusing on "miscellaneous" problems, meaning problems which are not easily categorized as "analysis" or "algebra" (the subjects of the last two months' Putnam Math Jams).[/quote]

I should add that the problems in the Math Jam on Tuesday will not depend on calculus, linear algebra, or other such "college-level" mathematics, and thus may be appropriate for intermediate-to-advanced pre-college students.  (The only "advanced" concept I'm going to use is the idea of countability vs. uncountability.)


calculus
integration
inequalities
complex analysis

Let C be the arc of the circle $ |z| \equal{} 2$ that lies in the first quadrant. Show that
$ |\int_C {\frac {{\it dz}}{{z}^{2}\plus{}1}}| \le \frac{\pi}{3}$

This should be moved to Complex Analysis.

If $ \vert z \vert \equal{} 2$ then $ \vert z^{2} \plus{} 1\vert \equal{} \vert z^{2} \minus{} ( \minus{} 1) \vert \geq \vert z^{2} \vert \minus{} \vert ( \minus{} 1)\vert \equal{} \vert z \vert^{2} \minus{} 1 \equal{} 4 \minus{} 1 \equal{} 3.$ Therefore

$ \left \vert \int_{c} \frac {dz}{z^{2} \plus{} 1} \right \vert \leq
\left(\frac {1}{3}\right)\left(\frac {2\pi \cdot 2}{4}\right) \equal{} \frac {\pi}{3}.$

Regards,  :D

[quote="coquitao"]If $ \vert z \vert = 2$ then $ \vert z^{2} + 1\vert = \vert z^{2} - ( - 1) \vert \geq \vert z^{2} \vert - \vert ( - 1)\vert = \vert z \vert^{2} - 1 = 4 - 1 = 3.$ Therefore

$  \vert \int_{c} \frac {dz}{z^{2} + 1}  \vert \leq (\frac {1}{3})(\frac {2\pi \cdot 2}{4}) = \frac {\pi}{3}.$

Regards,  :D[/quote]

How did you get $ \frac {2\pi \cdot 2}{4}$ on the right side of the inequality?


geometry
integration
trigonometry
symmetry
calculus
calculus computations

Given two circles centered at $ A,\ B$ with radii $ a,\ b$ respectively such that $ AB \equal{} 1,\ a^2 \plus{} b^2\equal{}1$. Find the maximum area of the common part of the circles.

wlog we can have A be (0,0) and B be (1,0)

so we are looking at
$ x^2+y^2=a^2$, and $ (x-1)^2+y^2=b^2$

finding where they intersect:

$ x^2-(x-1)^2=2a^2 -1 \to x^2-x^2+2x-1=2a^2 u-1$

$ 2x=2a^2  \to x=a^2   \to y=\pm ab$

$ I_R(a)=2\int_{a^2 }^{a} \sqrt{a^2  -x^2} dx= 2\int_{\arcsin a}^{\frac{\pi}{2}} a^2  \cos^2 w dw$

$ =a^2 \int_{\arcsin a}^{\frac{\pi}{2}} 1+\cos 2w dw = \frac{a^2 }{2} \left( [\pi- 2\arcsin a ] -\sin (2\arcsin a ) \right)$

$ =\frac{a^2 }{2} \left( [\pi- 2\arcsin a ] -2ab \right)$

$ I_L(a)=2\int_{1-b}^{1-b^2} \sqrt{b^2-(x-1)^2} dx=2\int_{-\frac{\pi}{2}}^{-\arcsin b} b^2 \cos^2 w dw =$

$ =b^2 \int_{-\frac{\pi}{2}}^{-\arcsin b} 1+\cos 2w dw =\frac{b^2 }{2} \left( [\pi- 2\arcsin b ] -\sin (2\arcsin b) \right)$

$ =\frac{b^2  }{2} \left( [\pi- 2\arcsin b ] -2ab \right)$

$ I(a)=I_L(a)+I_R(a)$
$ = \frac{\pi}{2}- a^2\arcsin a -b^2 \arcsin b -ab$

$ b=\sqrt{1-a^2}, \frac{db}{da}=b'=\frac{-a}{\sqrt{1-a^2}}=\frac{-a}{b}$

$ \frac{dI}{da} = -2a\arcsin a -\frac{a^2}{\sqrt{1-a^2}} -2b b' \arcsin b -\frac{b^2 b'}{\sqrt{1-b^2}} -b-ab'$

$ =2a(\arcsin b -\arcsin a)=0$

when $ \arcsin a =\arcsin b$ and since $ 0\le a,b \le 1$

we know this means we must have $ a=b$

so $ a=b=\frac{1}{\sqrt 2}$ and the max value is

$ I(\frac{1}{\sqrt 2}) = \frac{\pi}{4} -\frac{1}{2}$

Your answer is correct.  :)

Can anyone solve the problem without using arc sine for Japanese high school students?

we can look at the definitions of I for the left and right

$ I_R(a)\equal{}2\int_{a^2 }^{a} \sqrt{a^2 \minus{}x^2} dx$

$ I_L(a)\equal{}2\int_{1\minus{}b}^{1\minus{}b^2} \sqrt{b^2\minus{}(x\minus{}1)^2} dx$

$ u\equal{}1\minus{}x$, $ du\equal{}\minus{}dx$

$ I_L(a)\equal{}\minus{}2\int_{b}^{b^2} \sqrt{b^2\minus{}u^2} du\equal{}I_R(b)$

$ \to I_L(b)\equal{}I_R(a)$

so since $ I(a)\equal{}I_R(a)\plus{}I_L(a)\equal{}I_R(a)\plus{}I_R(b)\equal{}I_L(b)\plus{}I_R(b)\equal{}I(b)$

then $ I(\cos \theta ) \equal{} I (\sin \theta), \forall \theta \in [0, \frac{\pi}{2}]$

this is symmetry about $ \theta\equal{}\frac{\pi}{4} \to (a,b) \equal{} (\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$

so extreme values will occur at $ (0,1), (\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}), (1,0)$
inspection shows us the middle one must be the max.

im not sure if you can integrate that without sin substitution (which to me is the same as arcsin?)


If you want to manipulate a single expression, is it bad style to just rewrite that expression over on new lines?

Ex:

$ x^2 + xy + y^2 + xy$
$ x^2 + 2xy + y^2$
$ (x + y)^2$

Instead of
\begin{align*}
x^2+xy+y^2+xy&=x^2+2xy+y^2\\
&=(x+y)^2
\end{align*}

I was reading about coding and proof writing, and it seemed like there were alot of rules and conventions I didn't now...

I think context is everything. If it's in a book explaining how algebra works it helps to separate the steps onto different lines. But if you are in the middle of something else and can assume that the reader is familiar with the algebra you can put it on one line and/or leave some of the steps out.

You might want to read some of thew following guides:
[url=http://www.maths.leeds.ac.uk/~khouston/httlam.pdf]Sample Chapter of How to Think Like a Mathematician[/url] which is a chapter on Writing Mathematics wrtten by Kevin Houston of Leeds University. The full book How to Think Like a Mathematician is available on his [url=http://www.maths.leeds.ac.uk/~khouston/httlam.html]website[/url] and I thoroughly recommend it to all mathematics undergrauates.

There's also a long list of books, articles and websites at [url=http://www.math.du.edu/~mkinyon/mathwrite.html]Writing in Mathematics[/url]. In particular, page 5 of [url=http://ems.calumet.purdue.edu/mcss/kevinlee/mathwriting/writingman.pdf]A Guide to Writing Mathematics[/url] by Kevin Lee of Purdue University addresses your question.


Let $P$ is any point in a triangle $ABC$.
Prove that $APsin\angle{BPC}+BPsin\angle{APC}+CPsin\angle{APB}\leq \frac{AB+BC+AC}{2}$.

Posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=6146 .

  darij


[quote="slowlearner"]3x+2y=16
2x-3y=-11

if solving for this problem do you try to solve for x,y or top row bottom row? Any help is greatly appreciated. thanks alot!![/quote]
[hide]You want both x's additive inverses of each other so you can cross it out. So the LCm of 2 and 3 is 6. So one has to be 6 and the other has to be -6. So multiply the first equation by 2 and the second one by -3. So now the first equation is 6x+4y=32. The second one is -6x+9y=33. So adding the two equations cancels out the x's and only leaves the y's so we can solve for that. So after adding both equaitons we get 13y=65. Solving for y is 5. So plugging that back in to any equations to find x you should get x=2.[/hide]

The answer to your question: it doesn't matter.

There are multiple ways to go about solving this problem. If you want to do substitution, pick an equation and pick a variable, it doesn't matter. If you want to add the equations together to eliminate a variable, you can again pick a variable to eliminate, though in this case it would be easier to eliminate $y$.

So just pick a method and pick a variable, whichever you feel is easiest.

A word of advice: there is nothing to "memorize" about solving equations. Especially of this simple form.

One is only taking 2 variables, 2 equations to 1 variable, 1 equation (that's all that solving does). How does that happen? A variable has to drop out. Everything done to solve a system is just trying to drop out variables.

That is one reason why substitution is nice; automatically, you have rid yourself of a variable through substitution.

Elimination is just setting up the equations so that coefficients cancel and the variable drops ($0x=0$)

3x+2y=16
2x-3y=-11

Let's cancel the $x$ for instance.
We have a 2 coeff and a 3 coeff. Just multiply the bottom by -3 to get a -6 coeff, and top by 2 to get a +6 coeff, then add. The essential idea is just to drop the variable.


Let $ \left \{ (A_i,B_i),1\leq i\leq M \right \}$ be a family of pairs of subsets of the set of integers such
that $ \left | A_i \right |\equal{}m$ for all $ i$ and $ \left | B_i \right |\equal{}n$ for all $ i$. Also, $ A_i$ and $ B_i$ are disjoint sets and $ (A_i\cap B_j)\cup (A_j\cap B_i)\neq\O$ for all $ i\neq j$. Prove that $ M\leq \frac{(m\plus{}n)^{m\plus{}n}}{m^mn^n}$.

you must be missing something here. "...a family of pairs of subsets of the set of integers...". there should be a restriction on these integers, else $ M$ is unbounded.


Support
geometry
geometric transformation
rotation

Send in your views about the problem below :-

A car(having 4 wheels) is moving on a horizontal circular path i.e, the road is not banked. If the car is accelerating on the path. Then tell whether front wheel (the 2 wheels near the centre of the path) or back wheel (the 2 wheels away from the centre of the path) will lift up earlier.

I feel the back wheel will lift earlier.

Support your answer with reasons.

The road is unbanked, so the centripetal force is completely provided by the friction of the road surface against the wheels of the car. The only forces acting are the weight of the car, through its COM, and the friction force, against the wheels, on the road surface.

I exclude the driving force propelling the car forward  :)

When drawing a sketch of the movement, let the car be on the left hand side of the sketch with the centre of rotation on the rhs.

The friction force acts on the bottom of the wheels and its line of action is to the right. This will give an ACW torque about the COM of the car. This ACW torque will tend to lift the tyres nearer to the centre of rotation, viz. the front wheels. If going fast enough, the car will topple over to the left, or outwards.

Of course, this is only true when $\mu$ is large enough such that the car would topple at all... with bad tires you'd just skid outwards (I have those kinds of tires... argh)

Torque is provided only by the centrifugal force, the weight and the normal reaction from the road.
At the time of just toppling, normal force will sfift towards the outer tyres and provide zero torque.
if the centrigufal torque is more than weight then car topples outward or else doesn't.
Also, if the centrifugal force is too large(in case of sharp turn at high speed) then friction cannot balance it and car will skid outwards.

Please correct me if I'm wrong, AJ :rotfl:


I'm currently reading a set of lecture notes on relativity and I'm struggling with Einstein notation. For example,

$ s^2\equal{}\eta_{\mu \nu}\Delta x^{\mu} \Delta x^{\nu}$

which is the spacetime interval or,

$ x^{\mu'}\equal{}\Lambda^{\mu'}_{\nu}x^{\nu}$

Can anyone explain this notation? Thanks.

It's just a shorter way to write a sum over one (or more) index (indices).
For example, x_i*y_i = x1y1 + x2*y2 + x3*y3 = Sum(x_i*y_i) for i=1,2,3
In relativity the index goes from 1 to 4 (there are four "dimensions").
So s^2= n11*dx1*dx1+n12*dx1*dx2+....

The general idea is that any index that shows up twice in the expression means that you sum over it.
In s^2 you have two repeating indexes so you sum for each one.


If we are given a point inside an equilateral triangle, and the distances a,b,c to each vertex, can we determine the sidelength? can we determine the angles between the lines connecting the point to each vertex?

It's rather trigonometry than geometry.
My solution is a bit long, because of using trigonometry.
But the answer is the following: yes, we can and the side of this triangle is a root of the next equation: $ x^4\minus{}x^2 \cdot(a^2\plus{}b^2\plus{}c^2)\plus{}a^4\plus{}b^4\plus{}c^4\equal{}a^2 b^2\plus{}b^2 c^2\plus{}a^2 c^2$.
Angles we can calculate with the help of Law of Cosines.

[asy]defaultpen(fontsize(8));
pair A=expi(pi/2), B=expi(7*pi/6), C=expi(11*pi/6), P=0.2*expi(5*pi/7), P1;
P1 = rotate(-60,A)*P;
draw(A--B--C--A);draw(A--P); draw(B--P); draw(C--P);
draw(P1--A);draw(P1--B);draw(P1--P);
label("$A$",A,(0,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));
label("$P$",P,(1,0));label("$P'$",P1,(-1,1));label("$a$",(A+P)/2,(1,-1));
label("$a$",(A+P1)/2,(-1,1));label("$a$",(P1+P)/2,(0,1));label("$b$",(B+P)/2,(1,-1));
label("$c$",(B+P1)/2,(-1,-1));label("$c$",(C+P)/2,(1,1));[/asy]
Rotate the triangle about point $ A$ by 60 degrees. Then you get $ \triangle AP'A$ is equilateral, so $ \triangle AP'B$ is congruent to $ \triangle APC$ (their corresponding sides are equal). You can calculate $ [BPP']$ by Heron's; $ [AP'P]\equal{}\frac{a^2\sqrt{3}}{4}$. Thus you now know $ [BPP']\plus{}[AP'P]\equal{}[AP'B]\plus{}[ABP]\equal{}[ABP]\plus{}[APC]$. In the same manner you can calculate $ [APC]\plus{}[BPC]$ and $ [BPC]\plus{}[ABP]$. Add them up and divide by 2 to get $ [APB]\plus{}[BPC]\plus{}[CPA]\equal{}[ABC]$. Now since $ \triangle ABC$ is equilateral, you can calculate according to $ [ABC]\equal{}\frac{s^2\sqrt{3}}{4}$, where $ s$ is the side length of $ \triangle ABC$.


function
complex analysis
complex analysis unsolved

Let f(z) be entire and let $ |f(z)| \geq 1$ on the whole complex plane. Prove that f is a constant.

I think this should be proved using Liouville's theorem which states that if f(z) is entire, and there is a constant M such that $ |f(z)| \leq M$ for all $ z \in \mathbb{C}$, then f is constant. How would I finish the proof exactly?

Apply Liouville to $ g(z)\equal{}1/f(z)$.

[quote="Third Edition"]Apply Liouville to $ g(z) \equal{} 1/f(z)$.[/quote]

Than you, that is the way to get it done. I have an answer in my book that says:
1/f is entire and $ |\frac{1}{f(z)}| \leq 1$ on  $ \mathbb{C}$. Therefore 1/f is constant by Liouville's theorem. Hence f is constant.

Any more detail on this? What does g(z) mean?


I am hereby announcing that the official Mathlinks chat has been created. This thread was posted in the Questions, Suggestions, & Announcements Forum, but nobody goes there so I will also start the topic here.
MODS DO NOT MOVE THIS PLEASE

So, here are the instructions:
[quote="ZetaX"]

First to all who want to chat:

So I proclamate now (later on we can still choose another one) EuIRC:
- irc.euirc.net::6665-6669
- channel: #mathlinks

For those new to that there is some html-client:
- go to http://html.chat.euirc.net/
- enter some nickname, leave the password and Username unchanged
- choose some random channel at the bottom (lets not bother with them Wink ), for example "#euirc - General information channel"
- click on "go"
- in the empty line you see type: "/join #mathlinks" and press enter.
- voila, there you are
[/end quote]

Also:
One who doesn't want to chat is not forced to come there.
But everyone is invited to try chatting.

Please do not spam too much, but this forum is for chatting on Math-related subjects.
Please join and enjoy!

Have fun!

It's an announcement: the mods did not want to make a global announcement about this so me and ZetaX decided to announce this in various highly-populated forums. :)

Yes.  SC also has it's own chatroom, called "scarmlstuff" on AIM.  We use that, since AIM is much better than IRC.


A car travels from $ A$ to $ B$ at an average speed of $ 50$ miles per hour. What average speed is needed on the return journey to make the average speed there and back $ 60$ miles per hour?

I make it $ 75$ miles per hour:

If the distance from $ A$ to $ B$ is $ 300$ miles, then the outward journey takes $ \frac{300}{50} \equal{} 6$ hours. If the return journey takes $ x$ hours, then $ \frac{600}{6 \plus{} x} \equal{} 60$. Therefore $ x\equal{}4$ and so the average return speed is $ \frac{300}{4} \equal{} 75$.

OMG what a stupid mistake I made... Don't look at my post :oops:

It's the harmonic mean: $ \frac{2\cdot50x}{50\plus{}x}\equal{}60$, so $ x\equal{}75$
When there are 2 equal distances, it's harmonic mean, when there are 2 equal times, it's arithmetic mean...


floor function
combinatorics unsolved
combinatorics

For a given integer number n greater than one, prove that there exists a proper subset A of {1,2,3...n} such that
(a) |A|$ \le 2[\sqrt{n}] \plus{} 1$
(b) {|x-y|: x,y in A}, x distincts y }={1,2,3...n-1}

Take $ \displaystyle \ell \equal{} \Big \lfloor \frac {n} {\lfloor \sqrt{n} \rfloor \plus{} 1} \Big \rfloor$. Since $ \left ( \lfloor \sqrt{n} \rfloor \plus{} 1 \right )^2 > n$, it follows $ \ell \leq \lfloor \sqrt{n} \rfloor$. Now take $ A \equal{} \{1,2,\ldots,\lfloor \sqrt{n} \rfloor\} \cup \{k\big (\lfloor \sqrt{n} \rfloor \plus{} 1\big ) \ : \ k \equal{} 1,2,\ldots,\ell \} \cup \{n\}$.
Clearly $ |A| \leq 2\lfloor \sqrt{n} \rfloor \plus{} 1$, and $ A$ also has property (b)

Determine now a non-trivial lower bound: let $ A$ be such a set, then $ |A| \geq \lfloor \sqrt{2n} \rfloor$.

[hide]Since any $ 1 \leq k \leq n\minus{}1$ can be written as $ |x\minus{}y|$, associate it with (one of) the doubleton set $ \{x,y\}$. Clearly different values are associated with different doubletons, hence $ \genfrac {(} {)} {0pt} {} {|A|} {2} \geq n\minus{}1$, i.e. $ |A|^2 \minus{} |A| \minus{} 2(n\minus{}1) \geq 0$, therefore $ |A| \geq \frac {1 \plus{} \sqrt{1 \plus{} 8(n\minus{}1)}}{2} \geq \sqrt{2n} \geq \lfloor \sqrt{2n} \rfloor$.[/hide]


topology
geometry
3D geometry
sphere
function
abstract algebra
algebra

Prove that the set of solutions of the system of equations: $ x_1^2\plus{}...\plus{}x_n^2\equal{}1$ and $ x_1\plus{}...\plus{}x_n\equal{}0$ is an n-2 dimensional submanifold of $ \mathbf{R}^n$

Geometrically it's  the intersection of the sphere and hyperplane, Indeed it's n-2 dimensional submanifold  ($ \equal{}\mathbb{S}^{n\minus{}2}$)

I suspect that the real purpose of the problem is to provide a simple exercise on using the (implicit function theorem-based) definition of a submanifold.  That is, set up the intersection as the zero set of a function $ F$, and then show that at each point of the intersection, the kernel of $ DF$ has $ n\minus{}2$ dimensions.


Elmo makes $ N$ sandwiches for a fundraiser.  For each sandwich he uses $ B$ globs of peanut butter at 4 cents per glob and $ J$ blobs of jam at 5 cents per glob.  The cost of the peanut butter and jam to make all the sandwiches is $ \$$2.53.  Assume that $ B, J,$ and $ N$ are all positive integers with $ N > 1$.  What is the cost of the jam Elmo uses to make the sandwiches?

$ \textbf{(A) } \$1.05 \qquad \textbf{(B) } \$1.25 \qquad \textbf{(C) } \$1.45 \qquad \textbf{(D) } \$1.65 \qquad \textbf{(E) } \$1.85$

[hide]We see that 253=n(4b+5j).  253 factorizes as 11*23, so n is either 11 or 23.  If n is 23 we have 4b+5j=11, which has no positive integer solutions.  If n is 11 we have 4b+5j=23, which has the solution j=3, b=2.  The cost of the jam Elmo uses is 11(5*3)=11*15=165.  D.[/hide]


calculus
integration
trigonometry
limit
function
real analysis
real analysis unsolved

Find $\int_{0}^{\infty}\frac{(\sin{x})^{n}}{x^{n}}dx,n\in \mathbb{N}$ (this integral is motivated by another problem mentioned here ).
Also evaluate; $a_{n}: =\int_{0}^{\infty}\frac{(\sin(x) )^{n}}{x^{n}}dx,$
$\lim_{n\rightarrow \infty}a_{n}=?$

Your integral can be calculated exactly.

Let consider the case when $n=2p$. Then I obtained, assuming that I made no computational mistakes,  that:

$\lim\int_{0}^{\infty}\frac{(\sin(x))^{2p}}{(x)^{2p}}dx=\lim_{p\rightarrow \infty}\int_{0}^{1}\frac{2p}{(1+4y^{2})(1+42^{2}y^{2})\cdots (1+4p^{2}y^{2})}dy$.

Why this limit is 0?. Any ideas?

[quote="didilica"]Why this limit is 0?. Any ideas?[/quote]
For $n\ge 2,\ \left|\frac{\sin^{n}x}{x^{n}}\right|\le \frac{\sin^{2}x}{x^{2}},$ and the latter is an absolutely integrable function on $[0,\infty).$

For $x\ne0,\lim_{n\to\infty}\frac{\sin^{n}x}{x^{n}}=0,$ pointwise.

Hence by the Lebesgue Dominated Convergence Theorem, we can interchange integral and limit to conclude that

$\lim_{n\to\infty}\int_{0}^{\infty}\frac{\sin^{n}x}{x^{n}}\,dx=0.$


The U.S. postage rates on newspapers and periodicals are to become, starting January 1, 1875,
  (1) 2 cents per pound if published once per week, or more frequently, or
  (2) 3 cents per pound if published less frequently than once per week.
the postage must be paid in advance, by the publisher.  But, rather than putting a stamp on each item, the publisher will submit one form for a large group of items, and put the stamps on the form.  The form has room for five stamps only, and only the special newspaper and periodical stamps can be used.

the design problem for the Post Office is to choose the number of different stamps and the denominations for the stamps that will allow any shipment from one to 2,400 pounds at either rate to be paid with no more than five stamps.

How much are the special newspaper stamps worth? :huh:


limit function Asymptote logarithms calculus derivative algebra

This was my high school final exam ... i dont know if it is suitable for this forum... sorry if you have to move it...

I) Calculate the following limits:

$\displaystyle\lim_{x\to\+1}\frac{1}{x-1}-\frac{1}{Lnx}$

$\displaystyle\lim_{x\to\--2}\frac{2+cos(x+2)}{(x+2)^3.sin(x+2)}-\frac{3}{(x+2)^4}$

II) Study and represent the following function (without f'')

$f(x)=(e^x-7)^\frac{2}{3}-(e^{2x}-49)^\frac{2}{3}$

III) Clasiffy the folowing series in absolutely convergent, conditionally convergent or divergent

$\displaystyle\sum_{i=1}^{\infty}\(Ln(1+1/n)^{1+1/n}+Ln(1-1/n)^{1-1/n})$

$\displaystyle\sum_{i=1}^{\infty}\(n(n-cotan(1/n)-1/3)$

Discuss depending on x real:

$\displaystyle\sum_{i=1}^{\infty}\frac{Ln(n)}{n}.x^n$

$\displaystyle\sum_{i=1}^{\infty}\((1+1/n)^n(x-1))^n$

IV) Probe that for each real x in $[1/2,1]$
$atanx-Ln(1+x^2)\geq pi/4-Ln2$

V)  Study and represent the following function in $[-2pi, 2pi]$
$f(x)=\sqrt{2}.L|tan(x/4)|-x$

Wow I like your high school.
What do you mean by study and represent the function?
Notice also that one of your problems upset the latex.

[hide="a few at a glance"] #1 is 0 by l'hospital, ...[/hide]
I'll work on this later, but good problems!

Well, you can use Taylor or L'Hopital if you want (we studied that in class so you are able to use it)...
A complete study and representation of a function includes (for my teacher of course, maybe it is not complete):

Existence and limits where it doesn't exist
Zeros and sign of the function
Infinite branches and asymptotes
First derivate (zeros (max and mins) sign, existence (where it doesnt exist study semitangents)
Second derivate (flexion points and concavity)

We often dont have to do a complete study because it takes a lot of time and usually the second derivate is quite a complicate expression.

sorry if i didnt write the maths concepts properly (I noticed that english teachers are not likely to teach you this kind of useful english)

The second limit is quite easy if you can use taylor expansions.

For II, the function is $f(x) = \left( {e^{x}}-7 \right) ^{2/3} \left( 1- \left( {e^{x}}+7 \right) ^{ 2/3} \right)$ defined for $x \ge \ln 7$.

$f'(x) = -2/3\,{\frac{{e^{x}} \left( -\sqrt [3]{{e^{x}}+7}+2\,{e^{x}} \right) }{\sqrt [3]{{e^{x}}-7}\sqrt [3]{{e^{x}}+7}}}$.

It is quite easy to see that the only zero for the numerator is $x = 0$, for which the function is not defined. And the derivative is negative for all $x \ge \ln 7$ and $f$ obviously tends to negative infinity as $x\to\infty$. This analysis should be enough, right? I assume you don't have to sort out concavity/convexity seeing as you're not supposed to use $f''$. It looks like the derivative tends to negative infinity as $x\to\ln 7^+$, but that's probably not requied.

edit: there are obviously no asympotes.

V does not look hard. Just do seperate studies in the intervals $[0,2\pi)$ and $(-2\pi,0)$ to avoid the absolute value sign.

i think $(e^x-7)^{2/3}$ can be defined for all reals: $(e^x-7)^{2^{1/3}}=(e^x-7)^{\frac{1}{3}^2}$ I dont see why you dont define it for zero , for instance.

Well, the reason I took it as undefined is that it's really part of complex analysis to discuss non-integer power functions of negative numbers. You COULD define $x^{1/n}$ for negative x for odd integer n I suppose, but then how do you make sure that $x^{1/3}$ is really the same thing as $x^{2/6}$? I think it's wrong to have $(-7)^{1/3}$ defined, but of course it's possible that your school sees it as defined.

Edit: To clarify, the usual "easy way" (not complex analysis) to define power functions is $x^\alpha = e^{\alpha\ln x}$ and that $0^\alpha = 0$ for alla positive $\alpha$. This definition does not allow for negative x as you can see (but for positive integer n we can identify the power function with its polynomial instead). There are just so many problems in defining the power functions for all real x without using rather advanced mathematics.

well, that's quite interesting stuff... at school we do not define $b^{a_n}$ for b negative, but if we are talking about $(a_n)^b$ it doesnt matter if $a_n$ is negative or not as long as you can find an only real that satisfy... but of course, what you say makes a lot of sense, because $(-7)^{1/3}$ should be equal to $(-7)^{2/6}$ and it is not. Anyway, in my exam, I studied the function for all reals and it was coherent.  :D  :D  :D

Well, your high school certainly seems to do a lot of mathematics. I will not have the methods to solve III and IV until in a couple of weeks, and I've been studying mathematics at university for more than half a semester now (we will define the Riemann integral on monday, which I suppose these problems are related to).

The problems are not related to Rienmann or maybe they are and I dont know  :blush:  because I havent studied that yet... I learnt at school basic rules and easy methods to integrate, but they were out of the national plan so they werent asked at the exam... The reason why I studied that in high school is that when you are 16 you choose between 3 packages of subjects (I chose the one with Math A, Math B, Math C, Physics, Chemistry, Phylosophy  :huh:  and English, but just a few people choose that, the most prefer biology and other stuff or literature, history, etc)...

Dang choices in high school that is revolutionary!  I like your high school so much and I go to arguably the most difficult high school in the US.  It is only arguable since I haven't gone to every high school in the US, but it's definately top 3 in topics that matter.

My highschool = TJHSST = Thomas Jefferson High School for Science and Technology.


Find $x;y;z \in Z^{+}$ such that:
$yx^{z}= x+y^{z}$

1. $y=0\to x=0$, z - any.
2. $y=1$ - no solutions.
3. $y\ge 2\to y|x$ - no solutions.


greatest common divisor
number theory
least common multiple

THE greatset common divisior of 3 positive numbers is 1. but the greatset common divisors of the 3 possible pairings of the numbers are 7, 17,and 5. Thier LCM is 1190. What is the least possible sum of the 3 numbers?

[hide]Let the three numbers be a,b, and c.

WLOG,

(a,b)=7, (b,c)=17, (c,a)=5 and (a,b,c)=1.

thus we can write a,b,c as:

a=35m, b=119n, c=85p for natural numbers m, n, and p.

Since [a,b,c]=1190=595[m,n,p], [m,n,p]=2.

Hence, the least possible sum would come out if m=2, and n=p=1.

in that case, the sum would be 274. :) [/hide]


Prove that the points in the complex plane representing the complex numbers  u, v and w cannot all lie on the same side of the real axis if u+v+w =uvw

[hide="Er..."] It's late at night, so I'm not sure if this is right, but:

$ u, v, w$ can be replaced with their negatives, so suppose WLOG by contradiction that $ u, v, w$ all lie in the upper half plane.  Then the argument of $ u \plus{} v \plus{} w$ is a weighted average of the arguments of $ u, v, w$ (and in particular is between the smallest and largest of the three arguments) but the argument of $ uvw$ is the sum of the arguments of $ u, v, w$ (and in particular is greater than the largest of the three arguments).  Contradiction. [/hide]


ratio
number theory
relatively prime
HOLI 2021

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

[hide="Answer"]$A$'s success rate was highest on Day 2, so we set $B$'s points on Day 1 as low as possible. He had a lower success rate than $A$, but had a positive integer score, so he had 1 point out of 2, since $A$'s success rate was over $\frac{1}{2}$. On Day 2, he had 498 possible points. $A$'s success rate on Day 2 was $\frac{7}{10}$. $\frac{7}{10}\cdot 498=348.6$, so $B$ scored $348$ points on Day 2, and his total ratio was $\frac{349}{500}$, so $m+n=\boxed{849}$.[/hide]

p14: Say Beta attempted $p$ points on the first day and got $q$ of those, and attempted $500-p$ on the second day, getting $r$ of those. Then $\frac{q}{p} < \frac{8}{15}$ and $\frac{r}{500-p} < \frac{7}{10}$. Cross multiplying and simplifying gives $15q < 8p$ and $10r < 3500-7p$, and we want to maximize $\frac{q+r}{500}$. Rewriting our two inequalities gives $q < \frac{8}{15}p$ and $r < \frac{3500-7p}{10}$. It's also given that $p < 300$. Plugging in our inequalities to the desired expression gives $q+r < 350-\frac{p}{6}$. Hence we set $p = 2$, so $q = 1, r = 348$. Beta's success ratio is $\frac{349}{500} \implies \boxed{849}$.