geometry
trigonometry
ratio
circumcircle
geometry unsolved

Let $ (I)$ be the incircle of $ \triangle ABC$ which tangents to $ BC,CA,AB$ at $ A',B',C'$. Let $ O$ be any point in the plane of $ \triangle ABC$. $ A'O,B'O,C'O$ intersects $ B'C',C'A',A'B'$ at $ A_1,B_1,C_1$ and intersect $ (I)$ at $ A_2,B_2,C_2$. Prove that:
a. $ AA_1,BB_1,CC_1$ concurrent at $ O_1$.
b. $ AA_2,BB_2,CC_2$ concurrent at $ O_2$.
c. $ O,O_1,O_2$ are collinear.

[quote="Mashimaru"]Let $ (I)$ be the incircle of $ \triangle ABC$ which tangents to $ BC,CA,AB$ at $ A',B',C'$. Let $ O$ be any point in the plane of $ \triangle ABC$. $ AO,BO,CO$ intersects $ B'C',C'A',A'B'$ at $ A_1,B_1,C_1$ and intersect $ (I)$ at $ A_2,B_2,C_2$. Prove that:
a. $ AA_1,BB_1,CC_1$ concurrent at $ O_1$.
b. $ AA_2,BB_2,CC_2$ concurrent at $ O_2$.
c. $ O,O_1,O_2$ are collinear.[/quote]

I think it should be:
a. $ A'A_1,B'B_1,C'C_1$ concurrent at $ O_1$.
b. $ A'A_2,B'B_2,C'C_2$ concurrent at $ O_2$.

Look familiar... I'm trying to solve it.

No, it is exactly $ AA_1,BB_1,CC_1$ and $ AA_2,BB_2,CC_2$, because $ A'A_1,B'B_1,C'C_1$ trivially concur at $ O$.

a. can be solved by Trigonometry form of Ceva's theorem.
b. is just an special case of Ceva's nest theorem.

But if just simply applying those theorems, there is no way to solve c. so I think there could be something interesting.

I have solved c. by solving a. and b. again by calculating barycentric ratio of $ O,O_1,O_2$, with this you can see c. is trivial, but the calculation is terrible.

I hope for a synthetic idea to this problem.

Very nice problem! Have a look at http://www.mathlinks.ro/viewtopic.php?t=170047 . Enjoy!!!

I think there are some mistakes.Because $ A_2,A_1\in AO$ so in $ b,c$ they concur at $ O$  :?:

Sorry Training, I have edited the original problem posted above.

Thank you Mr.April, clearly the solution of PDatK40SP is fantastic :oops:

I have a generation for this problem:

Let $ A_1,B_1,C_1$ be 3 points on sides $ BC,CA,AB$ of $ \triangle ABC$ and $ O$ be any point in the plane. $ OA_1,OB_1,OC_1$ intersects the circumcircle of $ \triangle A_1B_1C_1$ at $ A_2,B_2,C_2$. Find the position of $ \triangle A_1B_1C_1$ such that $ AA_2,BB_2,CC_2$ concur.


geometry
geometric transformation
reflection
homothety
projective geometry
geometry proposed

Let $ ABC$ be a triangle let A-excircle $ (I_a)$ touch $ BC,CA,AB$ at $ D,E,F, EF\cap BC\equal{}P$ let $ PI_a$ intersect $ DE,DF$ at $ M,N$, let $ M',N'$ be midpoint of $ DE,DF$ prove that $ M'N,MN'$ and $ I_aD$ are concurrent.

The polar line of $ P$ wrt $ (I_a)$ clearly is $ AD$. Therefor, $ AD \perp PI_a$. This shows that the polar line of $ N$ is the line through $ B$, parallel to $ AD$. On the other hand, the polar line of $ M'$ is $ CI$. We will prove that their intersection, the pole of $ NM'$, is the reflection of $ C$ in $ I$.

Consider the homothecy centered at $ K$, with factor $ 2$. It maps the midpoint of $ BC$, which is $ L$, to $ D$. Furthermore, it maps $ I$ to a point $ J$, which lies on $ AD$. Indeed, $ J$ is the antipode of $ K$ in the incircle, hence the Monge-d'Alembert theorme on the circles $ (I), (I_a)$ and the degenerate $ BC$, show us that $ A,J,D$ are collinear. Therefor $ IL \parallel AD$. Consider a homothecy centered at $ C$ with factor $ 2$. It maps $ L$ to $ B$, and since $ IQ \parallel AD$, it maps $ I$ to the pole of $ NM'$. This shows that this pole is the reflection of $ C$ in $ I$. Similar proof holds for the pole of $ MN'$.

The line joining these two poles is the reflection of $ BC$ in $ I$, which is clearly parallel to $ BC$. This concludes the proof.

[quote="Jan"]
The line joining these two poles is the reflection of $ BC$ in $ I$, which is clearly parallel to $ BC$. This concludes the proof.[/quote]

More clear, polar line of then intersecton $ T$ of $ MN'$ and $ M'N$ is this line (the line joining these two poles is the reflection of $ BC$ in $ I$) so $ I_a,K,D$ are colinear. Nice proof.

Here is my proof, let $ PK$ is tangent of $ (I_a)$ (different $ PB$) this show that $ DEKF$ is harmonic quadrilateral so $ (DE,DF,DK,DP)$ is harmonic pencil let $ M'N'\cap EF\equal{}L$ note that $ I_aM'\perp DF,I_aN'\perp DE, I_aP\perp DK,I_aD\perp DP\Rightarrow (I_aM',I_aN',I_aP,I_aD)$ is harmonic pencil from this, let $ M'N'\cap EF\equal{} X,I_aD\cap M'N'\equal{}Y$ easily seen $ (M'N'XY)\equal{}(EFOX)\equal{}\minus{}1$ it is sufficient to show $ M'N,MN'$ and $ OY$ are concurrent.

I'm not quite sure I understood. Something went wrong with the names of your points I guess...
What is point $ O$? And I also think that $ M'N'$ and $ EF$ are parallel, hence $ X$ is a point at infinity?
Your proof looks very short and nice.

Anyway, it's clear up to $ (I_aM', I_aN', I_aP, I_aD)$ being a harmonic pencil. Note that this is also quickly seen from the fact that $ P$ is the pole of $ AD$.

Hi thanks, you are right sorry for some typos :blush: I think here is my complete solution

Here is my proof, let $ AD\cap (O)\equal{}K$ this show that $ DEKF$ is harmonic quadrilateral so $ (DE,DF,DK,DP)$ is harmonic pencil let $ M'N'\cap EF \equal{} L$ note that $ I_aM'\perp DF,I_aN'\perp DE, I_aP\perp DK,I_aD\perp DP\Rightarrow (I_aM',I_aN',I_aP,I_aD)$ is harmonic pencil from this, let $ MN\cap M'N' \equal{} X,I_aD\cap M'N' \equal{} Y$ from $ (I_aM',I_aN',I_aP,I_aD)\equal{}\minus{}1\Rightarrow (M'N'YX) \equal{} (NMI_aX) \equal{} \minus{} 1$ it is sufficient to show $ M'N,MN'$ and $ I_aY$ are concurrent.  :)

P/s I have edited  :P


We measure the angles of inclination of a tower, emerging on level ground, at distances of 50 m and 100 m from its base, respectively. The two measures sum up to 45°. Determine the height of the tower.

Let $a$ be the angle of inclination at distance of 50m, and $b$ be the angle of inclination at distance of 100m. Also, let $x$ be the height of the tower.

From the given information, we have $\tan a=\frac{x}{50}$ and $\tan b=\frac{x}{100}$.
Then, $\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}=\tan 45^\circ=1$

$\Longleftrightarrow \tan a+\tan b=1-\tan a\tan b$

$\Longleftrightarrow \frac{x}{50}+\frac{x}{100}=1-\left(\frac{x}{50}\right)\left(\frac{x}{100}\right)$

$\Longleftrightarrow \frac{3}{100}x=1-\frac{1}{5000}x^2$

$\Longleftrightarrow x^2+150x-5000=0$

Solving for $x$ (for $x>0$), we get $x=25\sqrt{17}-75\approx \boxed{28.08m}$


geometry
3D geometry
tetrahedron
geometry unsolved

The tetrahedron $ S.ABC$ has the faces $ SBC$ and $ ABC$ perpendicular. The three angles at $ S$ are all $ 60^{\circ}$ and $ SB \equal{} SC \equal{} 1$. Find the volume of the tetrahedron.

We can take the equilateral triangle BCS with area $ {\sqrt 3\over 2}$ as the tetrahedron base, and the height of the triangle BAC relative to base BC, $ \sqrt {3 \over 2}$, as the tetrahedron height; so the volume will be $ {\sqrt 2 \over 4}$.

humm. I remember, I met this statement somewhere and It's
[hide="here"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=376422&p=2411492#p2411492[/url][/hide]  :D
Apparently, that's your favorite saying.  :roll:

EDIT: I wanted to call attention to the fact that this user's post were all the same spam, but apparently the posts got deleted (for good reason), so please don't rate this down.


:D ,it has no nontrivial sol.
 observe its onto,let f(s)=0 and f(0)=c,show s=c=0.
 afterwards its standard.
 will post complete sol,if you dont get.

Please help, "who".
This is what I get:
For every $x$, $y=\frac{-f^{2}(x)}{2}$ $\to$ $f(x^{2}+y+f(y))=0$,
therefore, there's at least one zero $z$ with $f(z)=0$.
$x=0$, $y=z$ $\to$ $0=2z+f^{2}(0)$,
so there can be at most one zero at $z=\frac{-f^{2}(0)}{2}$.
$x=z$, $y=0$ $\to$ $f(\frac{-f^{2}(0)}{2}+f(0))=0$ $\to$ $(\frac{-f^{2}(0)}{2})^{2}+f(0)=z=\frac{-f^{2}(0)}{2}$.
It seems hard to solve!

Of course, one solution is $f(0)=0$, $z=0$.In this case,
$x=0$ $\to$ $f(y+f(y))=2y$.
$y=0$ $\to$ $f(x^{2})=f^{2}(x)$.
I think it will be easy to prove that $f(x)=x$.

[quote="xirti"]Find every function $f: \mathbb{R}\to \mathbb{R}$ such that : \[f(x^{2}+y+f(y))=2y+f^{2}(x)\] for all realx and y.[/quote]
I think this will help you.
[url]http://www.mathlinks.ro/Forum/resources.php?c=186&cid=41&year=2004&p=18289[/url]


topology
advanced fields
advanced fields unsolved

Let $ Y$ be a topological space and $ f: S^n\to Y$ continuous function. Prove that $ f$ is nullhomotopic iff $ f$ has a continuous extension $ g: \bar{B}^{n\plus{}1}\to Y$.

[hide="Hint"]The closed ball $ B$ is contractible (as a convex set), such that $ id_B\sim c$ implies $ f\sim f\circ c$. If $ \Phi: c \sim f$ is a homotopy, then define $ \Psi(x) \equal{} \Phi(\frac {x}{|x|},|x|)$, $ x\in B\setminus\{0\}$ and $ \Psi(0) \equal{} c$. Use the [url=http://en.wikipedia.org/wiki/Tube_lemma]tube lemma[/url] to show that $ \Psi$ is continuous at the origin.[/hide]

This is just http://www.mathlinks.ro/viewtopic.php?t=225434 ...


geometry
circumcircle
Euler
geometric transformation
reflection
rectangle
conics

Construct points $ A', B'$ on the rays $ \overrightarrow{E_aG}, \overrightarrow{E_bG},$ such that $ \overline {GA'} = 2\overline {E_aG}, \overline {GB'} = 2\overline {E_bG}.$ ($ A', B'$ are diametrically opposite to unknown $ A, B$ on unknown circumcircle $ (O).$) Let $ A'E_b$ cut $ B'E_a$ at $ C_0.$ The given $ G$ is also centroid of $ \triangle A'B'C_0.$ Perpendicular to $ A'B' \parallel E_bE_a$ through $ C_0$ is the C-altitude line $ h_c$ of unknown $ \triangle ABC.$ Let $ M_c'$ be midpoint of $ E_aE_b$ and let $ h_c$ cut $ E_aE_b$ at $ H_c'.$ Unknown $ \triangle ABC \sim \triangle E_aE_bE_c$ are centrally similar with center $ H$ (the unknown orthocenter) and coefficient 2. They have the same unknown Euler lines $ HG \equiv HG',$ where $ G' \not\equiv G$ is unknown centroid of unknown $ \triangle E_aE_bE_c.$ Construct centroid $ G_0$ of $ \triangle E_aE_bC_0.$ Perpendicular $ g_c$ to $ E_aE_b$ through $ G_0$ goes through unknown $ G'.$

Construct point $ P \in h_c,$ such that $ H_c'P = \sqrt {|E_aH_c' \cdot E_bH_c'|}.$ If $ H_c'$ is inside of the segment $ E_aE_b,$ then circle with diameter $ E_aE_b$ cuts $ h_c$ at $ P.$ If $ H_c'$ is outside of the segment $ E_aE_b$ (as in your sketch), then reflect $ E_b$ in $ h_c$ into $ F_b$ and circle with diameter $ E_aF_b$ cuts $ h_c$ at $ P.$ Construct equilateral $ \triangle H_c'PQ$ on the segment $ H_c'P$ and find its center $ R.$ Circle with center $ H_c'$ and radius $ H_c'R$ cuts $ h_c$ at $ S, T$ and $ H_c'S = H_c'T = H_c'R = \frac {H_c'P}{\sqrt {3}}.$ Parallels to $ E_aE_b$ through $ S, T$ cut the line $ g_c$ at $ V, U,$ respectively, forming a rectangle $ STUV$ with diagonal intersection $ W \in E_aE_b.$

The Euler line $ GH \equiv G'H$ is tangent to a conic $ \mathcal K$ with center $ W,$ its main axes identical with the rectangle midlines $ x \parallel SV,\ y \parallel ST$ through $ W.$ If $ H_c'$ is inside the segment $ E_aE_b,$ this conic is an ellipse with major axis length $ \max (ST, SV)$ and minor axis length $ \min (ST, SV).$ If $ H_c'$ is outside the segment $ E_aE_b,$ (as in your sketch), this conic is a hyperbola with major axis length $ SV$ and minor axis length $ ST.$ If you are curious why, ask.

To construct the conic tangents, find the conic foci $ F_1, F_2.$ That is: If $ H_c'$ is inside of $ E_aE_b$ (ellipse case) and $ ST > SV,$ then circle with diameter $ ST$ cuts $ y$ at the ellipse foci $ F_1, F_2$ and circle $ (W)$ with center $ W$ and equal diameter is the ellipse pedal circle. If $ H_c'$ is inside of $ E_aE_b$ (ellipse case again) and $ SV > ST,$ then circle with diameter $ SV$ cuts $ x$ at the ellipse foci $ F_1, F_2$ and circle $ (W)$ with center $ W$ and equal diameter is the ellipse pedal circle. On the other hand, if $ H_c'$ is outside of $ E_aE_b,$ then circle with diameters $ SU = TV$ cuts $ x$ at the hyperbola foci $ F_1, F_2$ and circle $ (W)$ with center $ W$ and diameter equal to $ SV$ is the hyperbola pedal circle.

Finally, construct the conic tangent from the given centroid $ G.$ That is: pick one focus, say $ F_1,$ and construct circle with diameter $ GF_1,$ intersecting the pedal circle $ (W)$ at $ X, X'.$ Then $ GX, GX'$ are the conic tangents from $ G,$ the Euler lines, intersecting $ h_c$ at the 2 possible orthocenters $ H, H'.$ Consider only one orthocenter $ H$ and construct $ E_c$ as orthocenter of $ \triangle E_aE_bH.$ Double the segments $ HE_a, HE_b, HE_c$ into $ A, B, C.$

[hide="Lemma about the conic."][b]Lemma:[/b] Let $ G$ be centroid and $ H$ orthocenter of a $ \triangle ABC.$ Let $ G_c, H_c \in AB$ be pedals of $ G, H.$ The triangle Euler line $ GH$ is tangent to a conic with center at the midpoint $ W$ of $ G_cH_c$ (that is, $ M_cW = \frac {_2}{^3} M_cH_c$), with the main axes parallel and perpendicular to $ AB,$ their lengths $ G_cH_c = \frac {_2}{^3} M_cH_c$ and $ \frac {_2}{^{\sqrt {3}}} \sqrt {|AH_c \cdot BH_c|},$ respectively.
_____________________________

$ CH$ cuts the circumcircle $ (O)$ again at $ K_c.$ By power of $ H_c$ to $ (O),$ $ H_cH = K_cH_c = \frac {AH_c \cdot H_cB}{H_cH}.$ Put coordinate origin at the midpoint $ M_c$ of $ AB$ and x-axis along $ AB.$ Let $ c = M_cB,$ $ d = M_cH_c,$ $ h = H_cC.$ Coordinates of $ C, G, H$ are $ C = [d, h],$ $ G = [\frac {_d}{^3}, \frac {_h}{^3}],$ $ H = [d, \frac {_{c^2 - d^2}}{^h}].$ Equation of $ GH$ is $ y - y_G = \frac {y_H - y_G}{x_H - x_G}\ (x - x_G).$ After substitution of the coordinates:

$ F(x, y, h) = 2dh(3y - h) + (h^2 - 3(c^2 - d^2)) (3x - d) = 0$

This is family of lines depending on an arbitrary parameter $ h.$ Equation of their envelope, if any, is obtained by eliminating $ h$ from the following 2 equations: $ F(x, y, h) = 0$ and $ \frac {\partial F}{\partial h} = 0.$

$ \frac {\partial F}{\partial h} = 6dy - 4dh + 2h(3x - d) = 6(dy + hx - hd) = 0\ \Longrightarrow\ h = \frac {dy}{d - x}.$

Plugging this to $ F(x, y, h) = 0$ and simplifying yields the envelope equation as

$ 3d^2y^2 (d - x) - 3(c^2 - d^2)(d - x)^2 (3x - d) = 0$

Reducing by $ 3(d - x) \neq 0$ (not really interested in isosceles triangles) and rearranging,

$ \frac {3}{c^2 - d^2}\ y^2 + \frac {9}{d^2} (x - \frac {_2}{^3}d)^2 = 1$

This is conic centered at $ W,$ with main axes parallel to coordinate axes, and with semimajor / semiminor axis lengths along x, y equal to $ \frac {_d}{^3} = \frac {_1}{^3} M_cH_c$ and $ \frac {_1}{^{\sqrt {3}}} \sqrt {|c^2 - d^2|} = \frac {_1}{^{\sqrt {3}}} \sqrt {|AH_c \cdot BH_c|}.$ The sign of $ c^2 - d^2,$ plus for $ H_c$ inside of $ AB$ and minus for $ H_c$ outside of $ AB,$ makes the conic either ellipse of hyperbola.[/hide]

Perpendicular to A'B'  through C0 is the C-altitude line hc of unknown triangle ABC ?

 :(

Do you want a proof or is a figure enough ?
[hide]It follows from the rectangle $ ABA'B'$ and from the common centroid $ G$ of $ \triangle ABC, \triangle A'B'C_0.$[/hide]

is close of correct but i think i don't see what you say about  x||SV and  y||ST

:(

Assume you got that rectangle STUV with diagonal intersection $ W \in E_aE_b.$ Lines parallel to its sides $ x \parallel SV, y \parallel SV$ through its diagonal intersection W are called rectangle (parallelogram) midlines and I have to label then, so that I can refer to them later. Midline $ x \parallel SV$ coincides with the line $ E_aE_b$ and midline $ y \parallel ST$ is perpendicular to $ E_aE_b$ at W. Now you have to decide, what kind of conic is the envelope of Euler lines, because construction of the foci depends on it. The 3 possibilities are:
1. $ H_c'$ inside of $ E_aE_b$ and ST > SV => ellipse with foci $ F_1, F_2 \in y.$
2. $ H_c'$ inside of $ E_aE_b$ and ST < SV => ellipse with foci $ F_1, F_2 \in x.$
3. $ H_c'$ outside of $ E_aE_b$ => hyperbola with foci $ F_1, F_2 \in x.$
I believe that your sketch calls for a hyperbola, but you have exact directions in each case, how to construct the foci and pedal circle. Anything else ?

P.S. See the attached figure depicting construction from the given points $ E_a, E_b, G,$ with approximately  the same positions as in your figure. The figure comes out large, because of unfortunate positions of the given points, but the construction works fine. Double click the figure and maximize the resulting window to see it in the actual size.

[hide="Another solution"]Easy to construct the line $ d\equiv h_c$ (one way has been showed in yetti's post). Let $ K$ be the midpoint of $ E_aE_b$. Then $ K$ is the midpoint of $ HM_c$ where $ H$ is the orthocenter and $ M_c$ is the midpoint of $ AB$. Let $ Q$ be the symmetric point of $ G$ wrt $ K$. $ GQ$ meets the line $ d$ at $ P$. See that $ KE_c\parallel M_cC\equiv M_cG \parallel QH$. Let point $ T\in PE_a$ such that $ TK\parallel E_aQ$.
Then, $ \frac{PE_a}{PT}\equal{}\frac{PQ}{PK}\equal{}\frac{PH}{PE_c}\Rightarrow E_aH\parallel TE_c \Rightarrow TE_c\perp E_bC$
Hence $ E_c$ is the just the intersection of the line $ d$ and the circle of diameter $ E_bT$.
Now, we can construct points $ H,A,B,C$ easily.[/hide]

Pedal circle of a conic is locus of pedals (feet of perpendiculars) from conic foci to conic tangents. It has the same center as the conic and radius equal to the conic semimajor axis. For a parabola, pedal circle degenerates to its vertex tangent. For proofs, see
[url]http://www.mathlinks.ro/viewtopic.php?t=210308[/url],
[url]http://www.mathlinks.ro/viewtopic.php?t=124926[/url].

[quote="N.N.Trung"]Your construction is very complicated, yetti.
Dear jrrbc and Mathlinkers, please post a short one.[/quote]
First, I posted another solution in the above post. Second, each solution always contains own beauties. Yett's solution used many nice geometric properties which can be useful in many problems. So if you are not contented with our solutions, you should find or post a nicer construction instead of giving criticisms and requests.
Best regards.

So sorry yetti for my words.  :fool: Thanks mr.danh for your advices  :yup: .I hope that there is no big problem :gathering:


function
number theory
least common multiple

Prove that for any multiplicative function f, $ f(m)f(n)\equal{}f(\gcd (m,n))f( \text{lcm} (m,n))$.

[u]Solution.[/u] Let $ m = \prod_{i = 1}^j{p_i}^{\alpha_i},\ n = \prod_{i = 1}^j{p_i}^{\beta_i}$.  Notice that the left-hand side of the equation is
\begin{eqnarray*}f(m)f(n) & = & f\left(\prod_{i = 1}^j{p_i}^{\alpha_i}\right)f\left(\prod_{i = 1}^j{p_i}^{\beta_i}\right) \\
& = & \prod_{i = 1}^j[f(p_i)]^{\alpha_i + \beta_i}.\end{eqnarray*}
Furthermore, the right-hand side of the equation is
\begin{eqnarray*}f(\gcd(m,n))f(\text{lcm}(m,n)) & = & f\left(\prod_{i = 1}^jp_i^{\min\{\alpha_i,\beta_i\}}\right)f\left(\prod_{i = 1}^jp_i^{\max\{\alpha_i,\beta_i\}}\right) \\
& = & \prod_{i = 1}^j[f(p_i)]^{\min\{\alpha_i,\beta_i\} + \max\{\alpha_i,\beta_i\}} \\
& = & f(m)f(n).\end{eqnarray*}
And the conclusion follows. $ \Box$


inequalities
real analysis
real analysis unsolved

Simple but not easy(at least, for me) problem on infinite series.

$ \Sigma_{n \equal{} 1}^{\infty} |a_n|^p < \infty , \Sigma_{n \equal{} 1}^{\infty} |a_n|^s < \infty$
$ 1 \leq p \leq s < \infty$

Show that

$ (\Sigma_{n \equal{} 1}^{\infty} |a_n|^s)^{\frac{1}{s}} \leq (\Sigma_{n \equal{} 1}^{\infty} |a_n|^p)^{\frac{1}{p}}$

This is a homogeneous inequality (of degree 1). That is, if we multiplied $ a$ by a positive constant, we would multiply both sides of the inequality by the same constant and the inequality would in effect remain the same.

The most basic trick for dealing with homogeneous inequalities is to set one important quantity equal to one.

So, suppose we have $ \|a\|_p \equal{} 1.$ What does that tell us about each $ |a_n|^p?$ What does that tell us about each $ |a_n|^s?$