New answers tagged

0

It's true $-$ Euler was the first to show that if the incenter lies on the Euler line that the triangle is isosceles. Euler's 1763 paper, Solutio facilis problematum quorundam geometricorum difficillimorum, is nicely discussed in Ed Sandifer's How Euler Did It: The Euler Line and Sandifer briefly discusses Euler's handling of the case where the orthocenter, ...


2

ere's a short coordinate proof: Suppose the triangle has side length $2$, one vertex lies at the origin, and the right half of the $x$-axis bisects the triangle. The circle has radius $\sqrt{2}$, so its Cartesian equation is $x^{2} + y^{2} = 2$. The yellow/blue chord lies on $x = \frac{1}{2}\sqrt{3}$, which intersects the circle at $\frac{1}{2}(\sqrt{3}, ...


1

The conjecture already fails at $3$-dimension. For $3$-dimension, consider the square anti-prism with vertices at $$\left( \pm \sqrt{3}, \pm \sqrt{3}, \sqrt{2} \right),\quad ( \pm \sqrt{6}, 0, -\sqrt{2} )\quad\text{ and }\quad( 0, \pm \sqrt{6}, -\sqrt{2})$$ It has volume $V = 16(\sqrt{2}+1)$ and diameter $d = \sqrt{20 + 6\sqrt{2}}$. Its "compactness" ...


1

I would recommend using a tag for each coordinate for easier labeling rather than for the entire point as mentioned ... for center C, point A, point B respectively as $$ (xc,yc); \, (xa, ya); \, (xb,yb ) ; $$ so that we have a combination of x-, y- coordinates: $$ (xb,yb ) = (xc,yc) + r * ( \sin \frac{t^0 \pi}{180 }, \cos \frac{t^0 \pi}{180 }). $$ Or, ...


1

All points on that given circle can be described by the vector equation $$ (x, y) = (cx, cy) + r (\sin \alpha, \cos\alpha) $$ where the second summand is the vector starting from the center point $(cx, cy)$ with endpoint on the circle of radius $r$, having an angle $\alpha$ to $P_a = (cx, cy + r)$. In other words: one rotates $P_a$ by angle $\alpha$ around ...


1

$$ \lambda=\dfrac{4\cdot\text{trapezia}}{8\cdot\text{triangles}}=\dfrac{4\cdot\left(\dfrac{a+b}{2}\right)a}{8\cdot bc}=\dfrac{a+b}{4\left(\sqrt{2}-1\right)b} \tag{01} $$ since $$ c=\left(\sqrt{2}-1\right)a \tag{02} $$ Now $$ 4\cdot\text{trapezia}+4\cdot\text{triangles}= 1\cdot \text{square} $$ that is $$ ...


0

The statement is known as the Van-Aubel theorem. Here is a simple proof: Apply Menelaus theorem for triangles $\triangle ACD$ and $\triangle ABD$. You get: $$\frac {AP}{PD} * \frac {DB}{BC} * \frac {CE} {EA} = 1$$ $$\frac {AF}{FB} * \frac {BC}{CD} * \frac {DP} {AP} = 1$$ Now $$\frac {AE} {CE} = \frac {AP}{PD} * \frac {DB}{BC} $$ and $$\frac {AF}{FB} = ...


0

Hint. Use a unit square, since the question asks for the ratio of areas, and $1$ is simpler than $11$. Now let $x$ be the distance between any intersection point and the nearest corner of either square. By symmetry, all sides of the octagonal area of intersection are equal in length to $x\sqrt{2}$, so the side of the square is $(2+\sqrt{2})x$. Since this ...


0

$3.664779=4.000000-\tan(37.064450°/2)$. See picture below.


0

Given two points $P$ and $Q$ you can parametrize the line between them by $r(t) = P + t(Q-P)$ for $t\in[0,1]$, this has the orientation going from $P$ to $Q$. In your example you have three points, $P=(0,0),Q=(1,0)$ and $R= (0,2)$. You should draw the triangle that these points define and observe that the anti-clockwise direction starting at $P$ goes to $R$ ...


1

$$3 = \dfrac{2R \sin B \sin C}{R(\sin B \sin C -\cos B \cos C)}\tag1$$ and then simplified to: $$3 = \dfrac{R \sin B \sin C}{-R \cos B \cos c}$$ (and it is there that I think I made a mistake). Yes, it is wrong. Dividing both the numerator and the denominator of RHS in $(1)$ by $\cos B\cos C$ gives $$3=\frac{2R\tan B\tan C}{R(\tan B\tan C-1)}$$ ...


0

You could model your operations as $$ \DeclareMathOperator{vsplit}{vsplit} \DeclareMathOperator{hsplit}{hsplit} \vsplit((x, y, w, h)) = ((x-w/4, y, w/2, h), (x + w/4, y, w/2, h)) \\ \hsplit((x, y, w, h)) = ((x, y-h/4, w, h/2), (x, y+h/4, w, h/2)) $$ These functions take a rectangle, described as four tupel $(x,y,w,h)$ where $(x,y)$ are the coordinates of its ...


0

In modern mathematics we have several ways of formalizing infinity. The one that is most relevant to your question was provided by Abraham Robinson; see here. Following his framework, there are both infinitesimals and infinite numbers. Thus if $H$ denotes an infinite number, one can indeed divide a unit interval into $H$ parts. Each part would have ...


0

The question you ask (which is a variation on the theme of Gordan's lemma) appears exactly (and with almost exactly the same notations) as the "conversely" part of proposition 1.1 on page 3 in Tadao Oda's Convex Bodies and Algebraic Geometry (An Introduction to the Theory of Toric Varieties) (Springer 1985). The key point is that if $v$ belongs to the ...


0

First draw two line segments through the point, one parallel with each side of the angle, and bounded by your original lines. See that each of the lines touches one of the sides of the angle. Now rotate one of the lines away from the other, around your fixed point, until it sits on top of the other line. You will see that through the rotation one end began ...


1

Here's a proof. The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have $$ l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$ and $$h_a = \frac{2 T}{a}$$ and cyclic shifts of those. With these formulae, the required inequality gets $$ ...


1

By the "chord-chord" aspect of the Power of a Point Theorem, we have $$|\overline{AX}||\overline{XB}| = |\overline{CX}||\overline{XD}|\quad\to\quad p\cdot p = (p+q)\cdot q \quad\to\quad\frac{p}{q} = \frac{p+q}{p} = \frac{|\overline{CX}|}{|\overline{XE}|}$$ The relation between $p$ and $q$ says exactly that $\frac{p}{q}$ is the golden ratio, $1.618\dots$. ...


0

Observe that $DXY$ is similar to $YOZ$ Implying $DX/XY=YO/OX$ i.e $(a+b)/a=a/b$ $OY$ is constructed similar to $OX$


17

Suppose we have five positive integers $a_1\le a_2\le a_3\le a_4\le a_5$ such that any three are the sides of an obtuse triangle. Then by repeated substitution of $a_n^2+a_{n+1}^2<a_{n+2}^2$, $$a_5^2>a_4^2+a_3^2>2a_3^2+a_2^2>3a_2^2+2a_1^2$$ But $a_1+a_2>a_5$, so $$(a_1+a_2)^2>a_5^2>3a_2^2+2a_1^2=(a_2+a_1)^2+a_2^2+(a_2-a_1)^2$$which is ...


4

The sentence says something like this: If $a$ and $b$ are any two lines in the plane $\alpha$ with cutting point $C$ with respect to the original geometry, then, in general, there will exist lines in any of the 4 angle spaces around $C$, that will neither intersect $a$ nor $b$; if on the other hand there are no such lines in 2 opposite angle spaces, then the ...


1

So applying cosine rule I got $$\small PI^2=4R^2+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} -16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)$$ I think that you have a typo (the red part) : $$\begin{align}&\small PI^2=4R^2\color{red}{\cos^2A}+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} ...


0

Let X be the mid point of BC. Join PX and QX. Now,P,Q,X are the midpoints of the sides of the triangle PQX. Therefore $ar(\triangle APQ)$=$ar(\triangle PXQ)$=$ar(\triangle PBX)$=$ar(\triangle QXC)$.(by mid-point theorem). Let,area of each triangle mentioned above =$x$ sq. units. So,total area of triangle=$x+x+x+x=4x$ sq.units Also,note that ...


0

Start with similarity: $\triangle ABC \sim \triangle APQ$ with dimension ratio $2$, and therefore area ratio $2^2 = 4$. Join $P$ to $R$ and observe $\triangle PSR = \triangle SQR$ (same base, same height "SBSH") If you let $\triangle SQR = x$, then $\triangle PRQ = 2x$. Now $\triangle APQ = \triangle PRQ$ (SBSH, since the base $PQ$ is common and the height ...


1

Here's another way to think about it, and it may be more revealing. Geometrically, an ellipse is defined as the locus of all points that have a fixed sum of distances from two fixed foci. The length of your one fixed side gives you the distance between the foci, and the foci occur at either end of this segment. The third point can occur anywhere along the ...


1

Let (area of) $\triangle ECF = x$. Then $\triangle FCD = \triangle ECF = x$ (same base, same height - I'll just call this SBSH for brevity). So $\triangle ECD = 2x$ and $\triangle EBC = \triangle ECD = 2x$ (SBSH) Hence $\triangle EBD = 4x$ and $\triangle AED = \triangle EBD = 4x$ (SBSH) Therefore $\triangle ABD = 8x = 8\triangle CEF \ \ (QED)$


0

Hint: Here are the eight triangles: (Large Version)


0

Area BED = 1/2 Area ABD as BE = 1/2 BA and BD = BD. (so they have the same base but one has only half the height. Area CED = 1/2 Area BED as CD = 1/2 BD and E = E. (One has half the base but they both have the same height.) Area CFD = 1/2 CED as CD = CD and DF = 1/2 DE. (So they both have same base but one has half the height.)


-1

ED is the median of △(ABD) and therefore △(AED)=△(BED)=1/2 x △(ABD). EC is the median of △(BED) and therefore △(EBC)=△(ECD)=1/2 x △(BED). CF is the median of △(ECD) and therefore △(CEF)=△(CDF)=1/2 x △(ECD). From the three equations above △(CEF)=1/8 x △(ABD) or 8 x △(CEF)=△(ABD)


0

Well $\triangle$AED=$\frac{1}{2}\triangle ABD$. So the problem reduces to showing that $$\triangle CEF = \frac{1}{4} \triangle BED$$ Since F is the midpoint of ED, we have by similarity that: $$|EB| = 2|FB|$$ in other words: $$\triangle CFD = \frac{1}{4} \triangle BED$$ Likewise, also by similarity, $$\triangle BEC = \frac{1}{4} \triangle ABD = ...


1

It is not possible. A triangle in the plane can be uniquely solved only in these cases: You know Three sides Two sides and the included angle Two sides and an angle not included between them, if the side length adjacent to the angle is shorter than the other side length. A side and the two angles adjacent to it A side, the angle opposite to it and an ...


8

The general rule that relates the sides of a triangle is called the law of cosines. It regards the following situation, where you know an angle of the triangle and want to relate the side lengths: It is stated as follows: $$C^2=A^2+B^2-2AB\cos(\alpha).$$ Note that this has a very similar form to the Pythagorean theorem - just with an extra adjustment for ...


5

Sorry for the short answer. Cosine Rule, whose special case is the Pythagoras theorem. $$ c^2= a^2 + b^2 - 2 a b \cos C$$ when $C = 90^0. $


0

The points $P(1,0,0),\;Q(0,1,0),\;R(0,0,1)$, forming an equilateral triangle, each lie on both the sphere and the plane given. This is sufficient to determine that the centre of the circle is $C(\frac 13,\frac 13,\frac 13)$ and that the radius of the circle is $$|PC|=\frac {\sqrt{6}}{3}.$$ The normal to the plane of the circle is ...


0

I assume it's an isosceles trapezoid: Draw another height with length $6$. The bottom will be $2x+3$ which is $8$. While $x=5/2$ it is not difficult to see you have $5k-12k-13k$ triangle. So the hypotenuse will be $13/2$. Now you can calculate the perimeter.


1

Suppose by contradiction there exists a line $a$ passing through a single point $A$. By axiom 3 there exist two other points $B$ and $C$ and by axiom 1 they belong to a line $r$ parallel to $a$. By axiom 1 there exists a line $s$ passing through $A$ and $B$ and by axiom 2 there exists a line $t$ passing through $C$ and parallel to $s$. Line $t$ does not ...


0

The radius equals the height of the equilateral triangles of side $s$. By Pythagoras, $$h^2+\left(\frac s2\right)^2=s^2$$ so that $$h=\frac{\sqrt 3}2s.$$


0

Label the center of the circle. Draw six lines from the the center to the circle to the vertices of the hexagon. (These lines will be longer than the radius.) This will divide the circle into six triangles. Question for you: Tell me every thing you can about these triangles. In particular, what are the lengths of the lines from the center? Now draw six ...


0

Draw the six isosceles triangles. Divide each of these triangles into two right angled triangles. Then you have $s = 2x = 2 (r \sin \theta)$ where $r$ is the radius of the circle, $\theta$ is the top angle in the right angled triangles and there are in total $12$ of these triangles so its easy to figure out $\theta$. $x$ is the short side in these right ...


1

When you say edges,you mean sides of triangle? I'm going to write the important stuff and leave computation to you. Okay, denote $h$ as height of triangle , $a=BC,b=AB=AC$,D altitude of height h from vertex A, $S1$ center of bigger circle,$S2$ center of smaller circle, M feet of altitude from $S1$ to $AC$, $N$ feet of altitude from $S2$ to $AC$. Triangles ...


1

It's a good beginning. If you call the (yet unknown) height of the triangle $h$ and draw the perpendiculars from the two circle centres to $AC$, you get some similar right triangles that you can use to derive relations between $r$, $R$, and $h$.


1

$$\triangle APQ=🔺 ABD\iff\triangle DOQ=🔺 POB$$ Adding $\triangle OBQ,$ $$\triangle PBQ=🔺 DBQ$$ $$\implies DP \parallel BQ\iff DP \parallel BR$$ $\implies\square DPBR$ is a parallelogram $\implies DR=PB$ But $DC=DR+RC, AB=AP+PB$ as $\square ABCD$ is a parallelogram


0

You can construct the parallel line. Draw a line from the given point cutting the given line. Construct a line through the given point making the alternate angles, $a$ in the diagram match. Now if the lines meet to the left, the sum of the angles in the triangle exceed $\pi$. Similarly if they meet to the right the sum exceeds $\pi$. Hence the lines do ...


3

Hint: These are the two ellipses: Solution: The red ellipse seems to be a rotated ellipse, centered around the origin. The minimal distance then is the length of the major semi-axis. So we try to find the rotation: We can write the second the second equation as $u^t Q u = 1$ with $u = (x, y)^t$ and $$ Q = \begin{pmatrix} 2 & 3 \\ 3 & 10 ...


-1

Assume that the sum of the angles of any triangle is $180$ degrees. Let $L_1$, $L_2$ be two straight lines that intersect in the angle $\gamma > 0$. Now, let $\bar{L_3}$ be a line segment that intersects both $L_1$ and $L_2$. Assume that $L_1$ and $L_2$ meet on the side of $\bar{L_3}$ where the two interior angles:$\alpha, \beta$ on that same side formed ...


5

To show that $\pi$ is constant we must show that given two circles of diameters $d_1$ and $d_2$ and circumferences $c_1$ and $c_2$, respectively, that $\frac{c_1}{d_1}=\frac{c_2}{d_2}$. If $d_1=d_2$ then the two circles are congruent because one can be placed upon the other and they will line up. Without loss of generality we can assume $d_1\lt d_2$. Draw ...


8

This is not a very rigorous proof, but it is how I was taught the fact that the circumference of a circle is proportional to its radius. Consider two concentric circles as in the diagram above. The radius of the smaller one is $r$, while that of the larger one, $R$; their circumferences are $c$ and $C$ respectively. We draw two lines through the center ...


2

There is some confusion on your induction. $n=1$ does not make sense, since polygons have at least 3 sides (a triangle). This step, although not really wrong, is unnecessary. You are simply renaming $n$ as $k$. This is ok, might you should be careful when renaming $n=k$ in step 2 (I'm being somewhat pedantic here). Induction works as follows: take a ...


1

Assume the $n$-gon is centered at the origin. Pick any mouse and let $\phi$ be the constant angle between the mouse's instantaneous direction of travel (i.e., the tangent line to the spiral) and the radial line from the origin. Then the equation in polar form of a mouse's trajectory is $$ r = a\exp(-b\theta) $$ where $\tan \phi = \frac 1b$ and $a$ is the ...


0

For the second case. Draw $\overline{BE}$ and let the measures of arcs $CB$ and $ED$ be $\alpha$ and $\beta$. By the inscribed angle theorem, the measure of an angle inscribed in a circle is half the measure of its intercepted arc. Therefore $\angle{E} = \frac{\alpha}{2}$ and $\angle{EBD} = \frac{\beta}{2}$ By the exterior angle theorem $m ...


3

The given points: p1 = {0, 5}; p2 = {5, 0}; p3 = {6, 5}; A vector perpendicular to p2 - p1: normal = RotationTransform[90 °][p2 - p1]; The reflection transform of a mirror placed at p1 having normal as its normal -- i.e., the mirror that is perpendicular to the line through p1 and p2: rF = ReflectionTransform[normal, p1]; The image of p3 as given by ...



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