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How to express an angle between two angle bisectors in interior angles of a convex quadrilateral?

$\angle LEC = \angle BEA = 180^{\circ} - \frac{\angle A}{2} - \angle B$ $\alpha + 180^{\circ} - \frac{\angle A}{2} - \angle B + 180^{\circ} - \frac{\angle C}{2} = 180^{\circ}$ (in $\triangle LEC$) $\...
Lion Heart's user avatar
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Pentagon $(ABCDE)$ is inscribed in a circle of radius $1$. If $\angle DEA=\angle EAB = \angle ABC$ and $m\angle CAD=60^{\circ}$ and $BC=2AB$.

Let's define: $$ \angle BOA=\angle EOD=\alpha, \quad\text{and consequently:}\quad \angle COB=\angle AOE=120°-\alpha. $$ From $BC=2AB$ we obtain the equation: $$ \sin{120°-\alpha\over2}=2\sin{\alpha\...
Intelligenti pauca's user avatar
2 votes

Find the segment "DC" in the obtuse triangle below

Let $\angle BAD = \angle CAD = \alpha$ then $\angle BAD =\angle BCA = 2\alpha$ Draw $DE$ such that $AE=QE$ ($E$ is the midpoint of $AQ$) then $AE=QE=DE=5$ and $\angle ADE = \angle DAE = \alpha$ then $\...
Lion Heart's user avatar
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2 votes

Find the segment "DC" in the obtuse triangle below

Let $R$ be the midpoint of $AQ$, so $\lvert AR\rvert = \lvert RQ\rvert = 5$. Since $\triangle ADQ$ is right-angled, then $AQ$ is the diameter of its circumcircle, so $R$ is its center, which means ...
John Omielan's user avatar
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1 vote

Find the segment "DC" in the obtuse triangle below

Using your result of $\measuredangle CQD = 90^{\circ} + \theta$ and the Law of sines with $\triangle CDQ$ gives $$\begin{equation}\begin{aligned} \frac{\lvert DC\rvert}{\sin(90^{\circ} + \theta)} &...
John Omielan's user avatar
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Find the angle $x$ in the triangle $APQ $ below

Another geometric approach: In picture CE is the bisector of angle ACB. S is mid point of arc BC so $MS\bot BC$. We have: $\overset {\huge\frown}{BS}=4\alpha $ $\Rightarrow\angle MBC=\angle MCB$ $\...
sirous's user avatar
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2 votes
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Find the minimum value of the sum of the squares of the distances from M to the lines AB, AC and BC

As you said, the sum of the three squared distances, by Pythagorean theorem are $ f = 3 (1)^2 + (M'X)^2 + (M'Y)^2 + (M'Z)^2 $ Now we know that these altitudes satisfy $ \frac{1}{2} ( 2 \sqrt{3} ) ( M'...
Quadrics's user avatar
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If ABCD is a 2*2 square , E is midpoint of AB ,F is the midpoint of BC , AF and DE intersect at I , BD and AF intersect at H , Find area of BEIH.

Tha area ot $ABF$ is $1$. Triangle $BHF$ is similar to $AHD$ with ratio $\displaystyle{1\over2}$, hence the height of $HF$ is $\displaystyle{2\over3}$ and its area $\displaystyle{1\over3}$. Triangle $...
Intelligenti pauca's user avatar
1 vote

If ABCD is a 2*2 square , E is midpoint of AB ,F is the midpoint of BC , AF and DE intersect at I , BD and AF intersect at H , Find area of BEIH.

I don’t want to show you the answer, just show some hints: Extend DE and CB to point G. We can get that triangle BEG is actually congruent to triangle AED by proving using AAS or ASA. So $GB=DA$. We ...
user1358538's user avatar
7 votes
Accepted

The maximum area of a pentagon inside a circle

Consider any two consecutive sides $AB$, $BC$ of a convex polygon inscribed in a given circle. Suppose to keep all vertices fixed apart $B$: the area of the polygon is maximum when $B$ is the midpoint ...
Intelligenti pauca's user avatar
3 votes

The maximum area of a pentagon inside a circle

Let's look at general (convex) $n$-sided polygons inscribed in a unit circle. Join the centre of the circle to each vertex of the polygon to form $n$ isosceles triangles. Say their apex angles are $\...
Chris Lewis's user avatar
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4 votes
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Find the angle $x$ in the triangle $APQ $ below

As in the figure below, let $R$ be the third vertex of the equilateral triangle $AQR$ (with $R$ on the same side as $B$, with respect to $AC$). $ARB$ is isosceles by construction and the hypotheses, ...
dfnu's user avatar
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1 vote
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Calculate the value of the expression $\sin^2 HAB+ \sin^2 HAD+ \sin^2HAA'$.

Let $H$ be ANY point inside the parallelepiped (but it could even be outside) and let $H_1$, $H_2$, $H_3$ be its projections onto lines $AB$, $AD$, $AA'$ respectively. Then we have: $$ \cos\angle HAB={...
Intelligenti pauca's user avatar
3 votes

Find the angle $x$ in the triangle $APQ $ below

You've made a good start. I assume the angles in your diagram were provided with the problem. Thus, as you've determined, $\measuredangle ABP = 120^{\circ} - 2\alpha$ and $\measuredangle BPA = 60^{\...
John Omielan's user avatar
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2 votes

Show that: $(\frac{MN}{MA})^2+ (\frac{MP}{MB})^2+ (\frac{MQ}{MC})^2+ (\frac{MR}{MD})^2 \geq \frac{4 }{9}$

Affine transformation Euclid has already mentioned that you can make affine transformations to move the points $A,B,C,D$ to arbitrary positions. Let me explain what that means since you're not ...
Einar Rødland's user avatar
0 votes

Prove that each diagonal of a quadrilateral lies either entirely in its interior or entirely in its exterior.

I tried to prove this with the axiom of internal angle region and external angle region: First i wanted to show that for 4 distinct non-collinear points on a plane you can draw (4*3)/(2!) = 6 line ...
Gustavo Carvalho's user avatar
1 vote

Seeking "900 Geometry Problems" Book – Any Leads on Its Whereabouts?

I think you heard it a little wrong. The title is "110 Geometry Problems for the International Mathematical Olympiad" authored by Titu Andreescu & Cosmin Pohoata Check out : https://www....
Prem's user avatar
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1 vote

Applying an $SO(3)$ geodesic onto a unit vector results in a circle on a sphere

For the converse, there is a quick geometric construction. Consider the plane containing the circle and any vector $\omega=[\omega_x,\omega_y,\omega_z]^T$ normal to the plane. Pick any $b_0$ on the ...
pwensing's user avatar
0 votes

Calculate the value of the expression $\sin^2 HAB+ \sin^2 HAD+ \sin^2HAA'$.

Hint: As you can see in picture triangle A'BD is equilateral and together with vertex A makes a pyramid AH is the altitude of the pyramid and also it is the altitude of three equal triangles ABY,AGD ...
sirous's user avatar
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0 votes

From a point $P=(3,4)$, perpendiculars PQ and PR are drawn to line $3x+4y-7=0$ and a variable line $y-1=m(x-7)$ respectively

I have read both answers, and see that both of them haven't done the question thoroughly. Both the websites listed above have done the wrong procedure. Here's the method I used to solve the question ...
Sid N's user avatar
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0 votes

Show that the intersection of planes $(VMC),(VNA),(VPB)$ is line if and only if the pyramid has $2$ lateral sides congruent.

You already proved that if the pyramid has $2$ lateral sides congruent, then the intersection of planes $(VMC),(VNA),(VPB)$ is the line. In the following, let us prove that if the intersection of ...
mathlove's user avatar
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0 votes

What, if anything, is this metric on $\mathbb{R}^2$ named? And, what do the open balls in this metric space geometrically look like?

In general, you have three cases. If $r < ||c||$ then $B_d(c,r)$ just consist of a small segment of the circle around $0$ with radius $||c||$ centered on $c$. If $||c|| < r < 2\cdot ||c||$ ...
quarague's user avatar
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0 votes

In a geodesic triangle, is the longest side opposite to the largest angle?

Here is an argument that works in the constant curvature case. Consider the circle inscribed in the triangle $ABC$. The points of contact of the inscribed circle with the sides of the triangle ...
Mikhail Katz's user avatar
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1 vote
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Proof a curve is a geodesic on a sphere S

I'll still post a solution because it could be handy in terms of the surfaces more complicated than a sphere. The basic idea was to somehow define the tangent plane, or, rather, separate it from the ...
Egor Larionov's user avatar
0 votes

Find the point $P$ on an ellipse such that $\overline{AP} + \overline{BP}$ is minimum for given points $A$ and $B$

Given a point $P = E(t)$ for some $t$, it lies on a unique ellipsoid (of revolution) whose foci are $A$ and $B$. It was shown here that the equation of this ellipsoid is $ (p - C)^T (a^2 I - {UU}^T ) ...
Quadrics's user avatar
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Euclid's fourth postulate

I don’t think Euclid ever moves anything. He postulates (1) that you can draw a straight line between any two points, (2) that you can extend any line, and (3) that you can make a circle using one end ...
Edward Porcella's user avatar
1 vote
Accepted

Calculate the angle $x$ in the quadrilateral $ABCD$

In the figure circles s and d have common chord CD so OF is perpendicular bisector of CD. We have: $\overset {\huge\frown}{DH}=\overset {\huge\frown}{CN}$ $\Rightarrow \overset {\huge\frown}{NH}=\...
sirous's user avatar
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3 votes

Show that: $(\frac{MN}{MA})^2+ (\frac{MP}{MB})^2+ (\frac{MQ}{MC})^2+ (\frac{MR}{MD})^2 \geq \frac{4 }{9}$

Let $\displaystyle \overrightarrow{BM}=\overrightarrow{BE}+\overrightarrow{BF}+\overrightarrow{BG}$, where $\overrightarrow{BE}=x\overrightarrow{BA}$, $\overrightarrow{BF}=y\overrightarrow{BC}$, $\...
grj040803's user avatar
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2 votes

Why is the intersection point of the diagonals of a regular polygon always the center?

Assume the center of the circumcircle is the origin. Note that a diagonal through vertex $x$ passes through the origin iff $-x$ is also a vertex iff the polygon has an even number of sides.
Ted Shifrin's user avatar
1 vote

Confusion about inverse kinematics of simple double-jointed arm

False alarm. It was just as I feared. The silly mistake was made on Desmos itself, and I couldn't see it no matter how much I stared at my math. I'd written $(L_1 + L_2)y\cos b$ rather than the ...
HydroPage's user avatar
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1 vote
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solution-verification | compare to angles in a rectangular parallelipiped

Because $YD=2a\sqrt{2}, DN=3a\sqrt{2}, DN=a\sqrt{38}$ by the thorem of the cosinus the $cos \angle YDN = \frac{11}{24}$ I think $DN=a\sqrt{38}$ is not correct. It should be $\color{red}{YN}=a\sqrt{38}...
mathlove's user avatar
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2 votes
Accepted

Calculate the angle $x^o $ in the triangle below

The answer is $\boxed{\angle AFE = 25^\circ}$. Two outer angle bisectors and one inner angle bisector are concurrent. In this case, for the triangle $\triangle ABE$, outer angle bisectors of vertex $B$...
Euclid's user avatar
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Finf the segment EC in the triangle ABC below

You can also reflect C over PE and show that triangles ABG and GBC' are congruent, so $BC'=6\rightarrow CC'=3\Rightarrow CE=\frac {CC'}2=1.5$
sirous's user avatar
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2 votes

The First Singular Homology Group $H_1(X)$ and the Fundamental Group

I don't find this proof the most conceptually enlightening but it is efficient. I will discuss that and also, for me, a nicer proof. As commented, $\theta$ is a quotient map. If you think carefully ...
FShrike's user avatar
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0 votes

Find the coordinates of the closest point on the surface of the ellipsoid to the line in space

Your ellipsoid is given by $ r^T Q r = 1 \tag{1} $ where $r = [x,y,z]^T $ and $Q = \begin{bmatrix} a_1^{-2} && 0 && 0 \\ 0 && a_2^{-2} && 0 \\ 0 && 0 &&...
Quadrics's user avatar
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0 votes

How to determine the elevation at the edges of a tilted rectangle?

If the tile were expanded to a $120\mathrm cm$ x $120\mathrm cm$ square under the same conditions, the elevation at $B$ would be$$\frac{6}{5}\cdot 1.2=1.44\mathrm cm$$and for the tile to drain from $B$...
Edward Porcella's user avatar
0 votes
Accepted

Unit ball with dual set

Suppose $B(0,r)\subset X$, then suppose that $\exists y\in X^*,\;\lvert\lvert y\rvert\rvert>\frac{2}{r}$. Then, $x=\frac{r}{2}\frac{y}{\lvert\lvert y\rvert\rvert}$ belongs to $X$, but $\langle x,y\...
Maxime's user avatar
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4 votes
Accepted

Finf the segment EC in the triangle ABC below

Reflect $A$ over $BP$ to $A' \in BC$. Notice two things: $|PA'|=|PA|=|PC|$, so $\triangle PA'C$ is isosceles with $|PA'|=|PC|$, thus $|A'E|=|EC|$. $|BA'| = |BA| = 6$ To conclude, the remaining $|A'C|...
Euclid's user avatar
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3 votes

G.P. geometry relation

Progressions with a common ratio govern some nested polygon structures. Below is such a structure involving triangles. Start with $\triangle ABC$. Along ray $\overset{\rightarrow}{AB}$ identify point ...
Oscar Lanzi's user avatar
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3 votes

Largest Area Triangle in the Vesica Piscis

Here is a solution in the "17th century spirit" where extremal solutions were found based on the computation of infinitesimal quantities. I assume that we look for an optimal solution under ...
Jean Marie's user avatar
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2 votes

How to prove opposite angle bisector theorem for convex quadrilaterals?

Regardless of where the bisectors of $\widehat B$ and $\widehat D$ intersect, let them meet $AC$ at $E$ and $F$ respectively. The angle you are looking for is $|\widehat{LEA} - \widehat{LFA}|$. We ...
Saeed's user avatar
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1 vote

Find the measure of $\measuredangle BAC$ in the triangle below given two sides and an angle

Do you realise your angle might not be unique? I've drawn a line segment with a length of $10$, an angle of $34.5°$ on it, and then I drew a circle with a radius of $6$, and this is what I came up ...
Dominique's user avatar
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Find the measure of $\measuredangle BAC$ in the triangle below given two sides and an angle

$\mathsf{BF \perp AC (F \in AC)\\ \triangle BFC: BC^2 = 9l^2+13l^2 = 250l^2 \therefore BC = 5\sqrt{10}l \therefore 5\sqrt{10}l = 10 \implies l = \frac{\sqrt{10}}{5}\\ BF = 9l = \frac{9\sqrt{10}}{5}\\ \...
peta arantes's user avatar
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Proof of Miquel's six circle theorem

There is a nice correspondence between Miquel's problem in 2D and configurations of planar slices of a sphere. See Miquel's Six Circles in 3d and also Problem 11 and its solution on this page. The ...
brainjam's user avatar
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1 vote

Find the measure of $\measuredangle BAC$ in the triangle below given two sides and an angle

I would use the "sine rule: $\frac{sin(34.5)}{6}= \frac{sin(A)}{10}= \frac{sin(B)}{b}$. $sin(A)= (10/6) sin(34.5)= (5/3)(0.5664)= 0.994$ $A= arcsin(O.994)= 70.7$ degrees. 70.7=
George Ivey's user avatar
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1 vote

Show that $S(M) \leq 5a^2$

We have $$BM^2 + MC'^2 \le BM^2 + MC'^2 + 2BM.MC'$$ $$BM^2 + MC'^2 \le (BM + MC')^2$$ Note that $BM + MC' = BC' = \sqrt{2}a$. Also note that $MC \le a$ (why?). You did most of the job.
Saeed's user avatar
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0 votes

Find the angle $ECA$

The question can be put in more general terms. Are there triangles which can be dissected in four subtriangles, connected in point E on the altitude. The given triangle is just one example. And ...
Paul vdVeen's user avatar
2 votes
Accepted

How to prove opposite angle bisector theorem for convex quadrilaterals?

Let point $E$ on $AD$ intersection of $AD$ and extension of $BL$ . $\angle BED = \angle A + \frac{\angle B}{2}$(exterior $\angle$ of $\triangle$) $\alpha + \angle A + \frac{\angle B}{2} + \frac{\...
Lion Heart's user avatar
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1 vote

How to prove opposite angle bisector theorem for convex quadrilaterals?

In your first figure: $$ \begin{align} \alpha &=\pi-\angle BLD\\ &=\pi-(2\pi-\angle B/2-\angle D/2-\angle C)\\ &={\angle B+\angle D\over2}+\angle C-\pi\\ &={2\pi-\angle A-\angle C\...
Intelligenti pauca's user avatar
2 votes
Accepted

Show that $S(M) \leq 5a^2$

Let's use a coordinate system with origin at $B$, the $x$-axis along $BC$ and $y$ along $BB'$. Then the coordinates of point $M$ will be $(x,x)$ (since it is moving on the $BC'$ diagonal), with $x\in[...
Andrei's user avatar
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