given three distinct lines $g,h,l$ meeting in one point $P$. I want to construct a triangle with vertices on $g,h,l$ such that those lines $g,h,l$ become its angle bisectors.  In general, if we consider a triangle $\Delta(ABC)$ with angles $\alpha,\beta,\gamma$ whose angle bisectors meet at a point $P$, then I recognized the following relations between the angles:  $$\angle (APB)=\frac{1}{2}(\pi+\gamma), \angle (BPC)=\frac{1}{2}(\pi+\alpha), \angle (CPA)=\frac{1}{2}(\pi+\beta).$$  Now suppose we choose points $C,C'\in g$ on different sides of $g$ with respect to $P$ and let $Q\in l,R\in h$ such that $C'$ lies in the interior of the angle $\angle (QPR)$. I started the construction as follows:   Construct the angle $\angle (QPR)-\pi/2$ on one side of $g$ at the point C. Then we get a intersection point $A$ with the ray $\overrightarrow{PQ}$.  Construct the angle $\angle (QPR)-\pi/2$ on the other side of $g$ at the point C. Then we get a intersection point $B$ with the ray $\overrightarrow{PR}$.  Now connect the points $A,B$   But now I'm worried about the angles at $A,B$. I do not know if the angles have the right size and the lines $l,h$ through $A,B$ are indeed angle bisectors? Maybe my construction is wrong. Do you know how to do it right?  Best wishes
122903
given three lines $\ell_1,\ell_2, \ell_3 $ which intersect in one point $P$. How can one construct a triangle such that the given lines become its angle bisectors?  So far I tried to find conditions on how the three given lines have to intersect such that the desired triangle exists. There are altogether six rays $\omega_1^1, \omega_1^2; \omega_2^1,\omega_2^2;\omega_3^1,\omega_3^2$ - any line $\ell_j$ determines the two rays $\omega_j^1,\omega_j^2$- emanating from the common point $P$.  Any of the three vertices $V_1,V_2,V_3$ of the desired triangle has to lie on exactly one line, i.e.  $V_j\in \ell_j=\omega_j^1\cup \omega_j^2$. Hence we have either $V_j\in \omega_j^1$ or $V_j\in \omega_j^2$.  Therefore the triangle should exist if and only if it is possible to choose three rays $\tilde{\omega}_1\in \lbrace \omega_1^1, \omega_1^2\rbrace $, $\tilde{\omega}_2\in \lbrace \omega_1^2, \omega_1^2\rbrace$, $\tilde{\omega}_3\in \lbrace \omega_1^3, \omega_1^3\rbrace$ such that any three of the rays $\tilde{\omega_1}, \tilde{\omega_2}, \tilde{\omega_3}$ form an angle less than $\pi$.  But I do not know how to construct the triangle?!   I started to draw a circle around $P$ with any radius. Now I can choose a Point $V_1$ on the ray $\tilde{\omega_1}$ outside the circle. And now I can draw the tangents through $V_1$ to the circle. I think those tangents will meet the other rays at some points $V_2,V_3$ (Why?). Now we can draw the line $V_2,V_3$. But this line needs to be tangent to the circle and I'm not sure about that. Will this construction work, or is it done in a different way?  Best regards
24652
Let $H$ be the orthocentre of an acute triangle $ABC$. Consider the circumcenters of triangle $ABH,BCH,CAH$ .Prove that they are the vertices of a triangle that is congruent to $ABC$.   My Work:   Observe that $AH$ is perpendicular to $BC$ &amp; $O_BO_C$. (Let $O_B$ be the circumcentre of $AHC$ and the others are defined same).  Clearly,$AH\perp BC$ &amp; $AH\perp O_BO_C$. Therefore $BC||O_BO_C$  Similarly $AC||O_AO_C$ &amp; $AB||O_AO_B$  But, I failed to show $O_BO_C=BC$  You may also show that  $AO_C=AO_B$
76111
Suppose $a,b,c$ lie in the unit circle of the complex plane and satisfy $a+b+c = 0$. Then, $a,b,c$ form the vertices of an equilateral triangle. Try So, I want to show that $|b-a|=|c-a|=|b-c|$. We are given $|a| = |b| = |c| = 1$ and $a+b+c=0$. Since $|z| = z \overline{z} $, we have $$ (b-a)( \overline{b}-\overline{a} ) = (c-a)( \overline{c} - \overline{a} )$$ which simplifies to $$ |b| - b \overline{a} - a \overline{b} - |a| = |c| - c \overline{a} - a \overline{c} - |a| $$ Hence, $$ c \overline{a} + a \overline{c} - b \overline{a} - a \overline{b} = 0$$ $$ a( \overline{c} - \overline{b} ) + \overline{a} ( c-b) = 0 $$ Here I am stuck. Perhaps this is the wrong approach?
29160
Let $PQRS$ be a trapezoid. The bisectors of legs $PS$ and $QR$ intersect the opposite legs at $M$ and $N$ so that triangles $PMS$ and $QNR$ are formed.  Prove that $\measuredangle PMS= \measuredangle QNR$.  The question may seem easy but I fail to move on, playing with angles didn't help. I'd appreciate some hint from somebody with an eye for geometry and triangles.
10093
Self-studying “Basic Mathematics” by Serge Lang, got stuck doing final problem of chapter 5.    I took a sample triangle and labeled two lines and points.    As L1 is the perpendicular bisector of segment PQ, I know KP = KQ. Points K, Q and M form a right triangle.  That’s the information I gathered, do not know how to proceed from there. Any help would be appreciated.
121932
We have an acute triangle with angles $\alpha, \beta, \gamma$ and we need to tell whether the following inequality is true for all such triangles:  $$(\tan(\frac{\alpha}{2}))^2+(\tan(\frac{\beta}{2}))^2+(\tan(\frac{\gamma}{2}))^2\geq1$$  We are also asked if $1%$ is the greatest number, for which it is true.  I'm having huge problems when trying to solve such optimization calculus problems. I think we must somehow find triangle for which we can prove, that the upper sum is minimum, than evaluate it and prove that it is equal to $1$. As with most optimization problems, it turns out that if $\alpha=\beta=\gamma=\frac{\pi}{3}$, then the sum evaluates to $1$. But how to prove that this is infimum of this sum?
121821
I found this exercise in Bollobas: Modern Graph Theory "Construct a triangle-free graph of chromatic number 1526"  It is added not to use results from the chapter about Ramsey Theory.   Now my qestions:  (1) Is there a nice way to construct such a graph?  (2) If yes then what is so special about this specific graph?  My ideas:  Grötzsch's theorem states that every triangle-free planar graph may be 3-colored, so we know that we can exclude planar grpahs and focus on non-planar graphs.  I read something about the Mycielskian on   I think this is the key to the solution but I do not see how to follow a concrete construction.
56200
I've read that if the complex numbers $a_1$, $a_2$ and $a_3$ are the vertices of a triangle in the complex plane such that $$ a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_1a_3 $$ then the vertices are actually those of an equilateral triangle.  I tried to see why this by first shifting the vertex $a_1$ to the origin, and considering the triangle with vertices $0$, $a_2-a_1$ and $a_3-a_1$. The given equation holds if and only if $$ (a_2-a_1)^2+(a_3-a_1)^2=(a_2-a_1)(a_3-a_1) $$ which is equivalent to saying $$ \frac{a_2-a_1}{a_3-a_1}+\frac{a_3-a_1}{a_2-a_1}=1. $$ Letting $\frac{a_2-a_1}{a_3-a_1}=x$, this gives the quadratic $x+\frac{1}{x}=1$, and so I solve for $x$ to be  $$ \frac{a_2-a_1}{a_3-a_1}=x=\frac{1\pm\sqrt{-3}}{2}=e^{\pm i\pi/3}.$$  I'm not sure where I'm going now. How can I conclude that such a relation implies the vertices form an equilateral triangle? Thank you.
99893
Prove,by vector method,that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides.    My Attempt: Let the trapezium be $OABC$ and that the O is a origin and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}$.Then the equation of $OB$ diagonal is $\vec{r}=\vec{0}+\lambda \vec{b}................(1)$ And equation of $AC$ diagonal is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a}).......(2)$ And the equation of the line joining the mid points of $OA$ i.e.$\frac{\vec{a}}{2}$ and $BC$ i.e. $\frac{\vec{b}+\vec{c}}{2}$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2}).........(3)$. Here $\lambda,\mu,t$ are scalars. I do not know how to solve $(1)$ and $(2)$ and put into $(3)$ to prove the desired result. Please help me.Thanks.
83715
I want to calculate the area of the largest square which can be inscribed in a triangle of sides $a, b, c$ . The "square" which I will refer to, from now on, has all its four vertices on the sides of the triangle, and of course is completely inscribed within the triangle, and by "largest square", with the maximum area of such squares.   I did some research. I found that in the case of right triangles, there seemed to be only two kinds of squares being formed. One with a side of $\frac{ab}{a+b}$ and the other of $\frac{abc}{c^2 + ab}$, $c$ being the hypotenuse.   After some time, I realized that there was a common formula for all triangles. And that is:  $$side = \frac{ch}{c+h}$$       or $$\frac{1}{side} = \frac{1}{c} + \frac{1}{h}$$ where, $h$ is height on base $c$, or $$h = \frac{2\sqrt{(s(s-a)(s-b)(s-c))}}{c}$$    I achieved the above results using similar triangles, and have added a figure, to make the point clear. But that is all, I can prove. I tested for many triangles, and found that the smaller side always yields the larger square, but I am not able to prove it. For example, I can say that the largest square of a $4,7,10$ triangle, has area of approximately $\fraq{16}{3}$. I also want to make another conjecture (with an intuitive proof) that any triangle will give at most three squares.  So I want to ask you, are my calculations, proofs ... correct? Have I got them right? And those predictions? Are they right, and if so, proofs? Conclusively, do we have a sure-fire formula to get the area of the largest square?
117503
Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is  a. a circle if $ T $ is a circle. b. a circle minus (0, 0, 1) if $ T $ is a line.  Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$  To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:  $$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$  We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:  $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.   I really am at a  loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?
62773
$ABC$ is an acute triangle. The angle bisector $AD$, the median $BE$ and the altitude $CF$ are concurrent. Prove that angle $A$ is more than $45$ degrees. Here $D,E,F$ are  points on $BC,CA,AB$ respectively.  This question was asked in 4th All-Soviet Union Mathematics Competition  in 1970 and the wording of the problem is same.
56913
How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:   $$\frac{2 \sqrt{bcs(s-a)}}{b+c}$$  I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.
62765
I want to draw several different triangles ABC with the angles marked and the angle value written in it. Thanks to answers on on another question, I've managed to get this code:  \documentclass{article} \usepackage{tikz} \usepackage{tkz-euclide} \usetkzobj{all} %% om allerhande objecten te gebruiken zoals gradenboog... \usetikzlibrary{calc,intersections,through,backgrounds,snakes} \usepackage{pgfplots} \pgfplotsset{compat=1.8} \usepgfplotslibrary{statistics}   \begin{document}  \makeatletter \newcommand\ddToddmmss[1]{% \pgfmathtruncatemacro\deg@tmp{#1}% \pgfmathsetmacro\min@tmp{(#1-\deg@tmp)*60}% \pgfmathtruncatemacro\Min@tmp{\min@tmp} \pgfmathtruncatemacro\sec@tmp{(\min@tmp-floor(\min@tmp))*60}% \ensuremath{\deg@tmp^\circ \Min@tmp ' \sec@tmp ''}} \makeatother   \begin{tikzpicture} \tkzInit[ymin=0,ymax=9,xmin=0,xmax=15] \tkzClip \tkzGrid \tkzDefPoints{1/3/A, 7/8/B, 12/1/C} \tkzDrawPolygon(A,B,C) \tkzDrawPoints(A,B,C) \tkzLabelPoints(A,B,C)  \tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](C,A,B); \tkzFindAngle(C,A,B) \tkzGetAngle{angleCAB}; %\FPround\angleBAC\angleCAB{2} \tkzLabelAngle[pos = 2.3](C,A,B){\ddToddmmss{\angleCAB}};    \tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](A,B,C); \tkzFindAngle(A,B,C) \tkzGetAngle{angleABC}; %\FPround\angleABC\angleABC{2} \tkzLabelAngle[pos = 2.3](A,B,C){\ddToddmmss{\angleABC}};  \tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](B,C,A); \tkzFindAngle(B,C,A) \tkzGetAngle{angleBCA}; %\FPround\angleBCA\angleBCA{2} \tkzLabelAngle[pos = 2.3](B,C,A){\ddToddmmss{\angleBCA}}; \end{tikzpicture}   \begin{tikzpicture} \tkzInit[ymin=0,ymax=9,xmin=0,xmax=15] \tkzClip \tkzGrid \tkzDefPoints{1/3/A, 7/8/B, 12/5/C} \tkzDrawPolygon(A,B,C) \tkzDrawPoints(A,B,C) \tkzLabelPoints(A,B,C)  \tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](C,A,B); \tkzFindAngle(C,A,B) \tkzGetAngle{angleCAB}; %\FPround\angleBAC\angleCAB{2} \tkzLabelAngle[pos = 2.3](C,A,B){\ddToddmmss{\angleCAB}};    \tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](A,B,C); \tkzFindAngle(A,B,C) \tkzGetAngle{angleABC}; %\FPround\angleABC\angleABC{2} \tkzLabelAngle[pos = 2.3](A,B,C){\ddToddmmss{\angleABC}};  \tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](B,C,A); \tkzFindAngle(B,C,A) \tkzGetAngle{angleBCA}; %\FPround\angleBCA\angleBCA{2} \tkzLabelAngle[pos = 2.3](B,C,A){\ddToddmmss{\angleBCA}}; \end{tikzpicture}  \end{document}     Problem is, in version 2, where point C is higher than point A, he takes the wrong angle.  Any suggestions?
87966
Let $x_1, \ldots ,x_n$ be distinct points in $\Bbb{R}^d$. Prove that there is a permutation $\sigma$ of $[n]$ such that the $\angle x_{\sigma(i−1)}x_{\sigma(i)}x_{\sigma(i+1)} &lt; \pi/2$ for any $i = 2,\ldots, n−1$.  So I have got this question, and I am not able to understand what &quot;a permutation $\sigma$ of $[n]$&quot; mean? Can someone explain to me this question and provide some hints to approach it? P.S. I have edited out some parts as it was a bit irrelevant.
23251
The formula for finding the angular bisectors of two lines $ax+by+c=0$ and $px+qy+r=0$ is $$\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm\frac{px+qy+r}{\sqrt{p^2+q^2}}$$ I understand the proof of this formula but I do not understand how to determine which sign is for acute bisector and which one for obtuse. I can find the angle between a bisector and a line, and if it comes less than $45^\circ$ then it is acute bisector. But that is a lengthy method and involves calculation. My book says, if $ap+bq$ is positive then the negative sign in the formula is for acute bisector. I want a proof of this method. Edit: Using the method for finding the position of two points with respect to a line is okay for the proof.
107441
I had some troubles with this problem :     Let $ABCD$ be a convex quadrilateral. $M$ and $N$ are the midpoints of   the diagonals $AC$ and $BD$. The sides $AB$ and $CD$ are extended   until they intersect. The intersection point is $E$. The sides $AD$   and $BC$ are extended until they intersect. The intersection point is   $F$. Let $P$ be the midpoint of the segment $[EF]$. Prove that $M$, $N$, $P$ are    collinear.     First, I found that a quadrilateral in which the opposite sides interesect is also known as a complete quadrilateral. Then, the line $M-N-P$ is known as Newton-Gauss line and the problem above as Newton's Problem.  I've taught about solving it using areas. I've used the property that median divides the triangle in two echivalent triangles (with the same area). Many properties can be derived from it.  I haven't figure out, but I'm interested in a proof using areas. I would appreciate some suggestions.  Thanks!