New answers tagged euclidean-geometry
5
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A question about a pair of triangles in a triangle
Here's an outline of a barycentric demonstration of the concurrence of the long diagonals. Verifying calculations is left as an exercise to the reader.
For some $\alpha$, $\beta$, $\gamma$, we can ...
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2
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How to prove that four points are concyclic under these given conditions?
Since $ABCD$ is a parallelogram and $E, M$ are the midpoints of $BC, AD$ respectively,
$EM \parallel AB,\; BE \parallel AD,\; ED \parallel BM$.
Given $BK \perp AB \Rightarrow BK \perp EM$.
Similarly, $...
4
votes
How to prove that four points are concyclic under these given conditions?
Let $\Gamma$ be the circle through $K,E,D$. Define points $X = BK \cap CD$ and $Y = BK \cap EM$.
From equal heights of parallelograms $BEMA$ and $ECDM$, we have $BY=YX$. Hence, by $SAS$ congruence we ...
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How to prove that four points are concyclic under these given conditions?
HINT.-In your five figures angle $\angle{KED}$ is right so in your last figure, points $K,E,D$ (whose coordinates are known) determine a circle whose diameter is $\overline{KD}$. One way to finish is ...
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1
vote
Accepted
Is this result already a known theorem in geometry?
This is a community-wiki answer recording the discussion from the comments, so that this question may be marked as answered.
Yes, it is well known that the area of any quadrilateral is $\frac12 pq \...
1
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Existence of this trick-of-the-eye figure in 3-dimensional space
Perhaps I have misunderstood, but I believe this configuration satisfies the conditions of the problem--i.e., four planar quadrilaterals that form a Möbius strip. It's not pretty or symmetric, so I ...
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2
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Expected area of an inscribed triangle in a sphere
WLOG, let $A$ be at the north pole. The distribution of the angle to $B$ is proportional to the perimeter of the meridian that contains $B$, hence to $\sin\phi$ (colatitude). The normalization factor ...
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2
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Expected area of an inscribed triangle in a sphere
Here is an approach on similar lines, the arguments that make it work are collected and displayed in the right order to lead to a quick solution.
(This is a very old question, i want to see if there ...
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2
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Accepted
Concyclicity of Fermat points in a cyclic quadrilateral
Note that $ABF_DF_C$ is cyclic. Indeed, the (oriented) angles $AF_CB$ and $AF_DB$ are equal to $\dfrac{2\pi}{3}$. Similarly, $BCF_AF_D$, $CDF_BF_A$ and $DAF_CF_B$ are cyclic.
Now,
\begin{align*}
\...
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0
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What auxiliary construction helps solve this angle chasing problem?
First of all, the given angles immediately give $AD=AC=CB$.
Let $E$ be the reflection of $D$ in $AB$. Then, since $AE=AD$ and $\angle EAD = 2\angle BAD = 60^\circ$, the triangle $ADE$ is equilateral.
...
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2
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Elementary geometry: showing that four points are coplanar
The tetrahedrons $ABCD$ and $A'B'C'D'$ are congruents, hence they have the same volume. Since the points $A, B, C, D$ are coplanar, the volume is zero.
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4
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Elementary geometry: showing that four points are coplanar
$\triangle ABC$ and $\triangle A'B'C'$ are congruent, as are $\triangle BCD$ and $\triangle B'C'D'$. $ABCD$ is coplanar, so $|AD|$ takes the maximum (resp. minimum) distance possible given the ...
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0
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What is the name of this result about isosceles triangles?
If we think that B and C are on the circle whose center is A and radius is p. Then we can easily get $q^2-p^2=-uv$. This is the power of a point theorem:From a piont P to draw a line intersecting on A ...
0
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Non-concyclicity of the circumcenters of complementary triangles in a quadrilateral
Now, by experimenting with Geogebra, I've come up with a relationship around the angles, which I think will bring us a lot closer to the solution.
$\angle{DAB}-\angle{BCD}=\angle{O_DO_AO_B}-\angle{...
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1
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Non-concyclicity of the circumcenters of complementary triangles in a quadrilateral
The statement is in fact equivalent to the cyclicity of $ABCD$, with a degenerate case.
Since the property is invariant under similarities, we may assume
$$
A=(a,am),\quad B=(b,-bm),\quad C=(-c,-cm),\...
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0
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How to construct the following isosceles trapezoid with compass and a straightedge?
In the picture $MN=3u$ where u is the unit.The circle c is at center M with radius u.The key point is to find the position of point A. $OA\approx MG$ where G is the intersection of the perpendicular ...
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What auxiliary construction helps solve this angle chasing problem?
In the picture circle d is on points b. c and D.DH is drawn perpendicular to CB.We have:
$\angle CDA=\frac12 \overset{\large\frown}{AI}=66^o$
$\angle ACD=\frac12 (\overset{\large\frown}{GI}+\overset{\...
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3
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Calculating the radius of a semicircle inscribed in a triangle
A simple rough solution: The center of the semi circle is on a perpendicular to BC at P. We have:
$PK\bot AB$
$IP\bot PK\Rightarrow IP||AB$
Calculations:
$$\frac{99}{sin 75^o}=\frac {AB}{\sin 45^o}=\...
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0
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How can you demonstrate that $x = a + b + c$ without using trigonometry?
A remark with an additional question.
The algebraic relation $\boxed{c^2=a^2-b^2+bc}$ can be deduced using purely geometric methods, as can $x=\dfrac{ab}{a-b}=\dfrac{a^2}{\sqrt{4a^2-3c^2}}$.
It ...
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3
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How to construct the following isosceles trapezoid with compass and a straightedge?
Given
$ABCD$ is an isosceles trapezoid with $BC \parallel AD$,
$MN$ is a segment such that $M\in AB$, $N\in CD$, $BC\parallel MN\parallel AD$,
$AM : MB = DN : NC = 1 : 2$, and
$MN = AB = CD$,
the ...
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4
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How to construct the following isosceles trapezoid with compass and a straightedge?
Hint:
(This space intentionally left blank.)
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2
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Accepted
How can you demonstrate that $x = a + b + c$ without using trigonometry?
Let $ABC$ be a triangle with $AC=a$ and $\angle A=\angle C=80^\circ$. Let $CD$ be the bisector of $\angle ACB$. Extend $CA$ beyond $A$ and take points $E,Z$ such that $AE=b$, $EZ=c$. Let $ZD\cap BC=H$....
4
votes
Accepted
How to construct the following isosceles trapezoid with compass and a straightedge?
Draw a line $a$ for major base $AD$ and fix point $A$ on it. Choose the unit length at will (this will set the scale of the figure).
Draw point $H$ on $a$ such that $AH=4$. Draw line $HO$ ...
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7
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Calculating the radius of a semicircle inscribed in a triangle
Here's a more-explicit, geometric approach than in my previous answer ...
Let $O$ be the point of tangency, and let $R$ be the center of the circle; we know $\overline{OR}\perp\overline{BC}$. Define $...
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16
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How can you demonstrate that $x = a + b + c$ without using trigonometry?
You have the right idea about drawing additional lines, but with the following showing how it can be simply done by adding just $2$ lines inside of $\triangle ABC$, as indicated in an updated version ...
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6
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Calculating the radius of a semicircle inscribed in a triangle
Working in some generality, I took point
$$P = \frac12(1-\lambda)B+\frac12(1+\lambda) C$$
(In OP, $\lambda=-1/3$.) Writing $U$ and $V$ for the endpoints of the diameter, and $M$ for the center, I ...
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3
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Calculating the radius of a semicircle inscribed in a triangle
A synthetic solution to Petersen’s problem may be of interest here.
Given triangle $ABC$ and point $D$ on $BC$, draw $DE\perp AB$, $DF\perp DE$, join $EF$, and make $DG\perp BC$.
Since $DF\parallel ...
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4
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Determining a quadrilateral up to similarity with four angles
In this answer we prove that the quadrilaterals found by @Intelligenti pauca have the desired properties.
Consider a convex quadrilateral $ABCD$ with $\angle DBA = 30^\circ$, $\angle CBD=30^\circ$, $\...
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6
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Calculating the radius of a semicircle inscribed in a triangle
The unit vector along $BA$ is
$u_1 = ( \cos 60^\circ, \sin 60^\circ ) = (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} )$
The unit vector along $CA$ is
$u_2 = (- \cos 45^\circ, \sin 45^\circ) = (\dfrac{-1}{\sqrt{...
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9
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Calculating the radius of a semicircle inscribed in a triangle
Here's one more idea which proves to be slightly simpler. Using the definitions of points $D, E$ in my other answer (without scaling), let $PE$ be the line $y = m(x - 33)$ with variable slope $m$; $DP$...
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8
votes
Accepted
Calculating the radius of a semicircle inscribed in a triangle
I would try an algebraic approach. Let $O$ be the center of the semicircle, of radius $R$ (the coordinates are $(0,R)$), and the intersections of the dotted line with $AB$ and $AC$ be $M$ and $N$. The ...
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9
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Calculating the radius of a semicircle inscribed in a triangle
Scale everything down by a factor of $33$: for example, $BP = 1, BC = 3$.
Let the point where the circle intersects $AB$ be $D$ and let the point where it intersects $AC$ be $E$. We also set up ...
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1
vote
Accepted
On Line-Plane Perpendicularity in Euclidean Geometry
This is Euclid, Elements, Book XI, Proposition 4. The proof is long and involves one triangle congruence after another (that too in different planes), but the diagram is quite clear, and the proof is ...
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0
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Coaxality of three circles in a complete cyclic quadrilateral
Comment:
As can be seen in the picture the orthocenter of triangle PQR is coincident to point O which is the center of circumcircle of the quadrilateral ABCD. Also N is the center of the nine point ...
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11
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Determining a quadrilateral up to similarity with four angles
There can be more than one possible quadrilateral, given your four angles. I don't have a proof (yet), but found a counterexample with GeoGebra, see figures below.
EDIT.
Here's a way to find all the ...
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3
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Determining a quadrilateral up to similarity with four angles
Edit:
the answer here means there was an error in my original argument. I've rewritten it to connect it to that answer.
The discussion below treats this structure as a linkage with two fixed vertices ...
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5
votes
Accepted
Reflections of a point with respect to the sides of a triangle lie on a conic passing through the triangle’s vertices
With a considerable coordinate bash, and a great deal of simplification, I find this equation for the curve in $u:v:w$ barycentric coordinates:
$$
\frac{1}{\ddot{u}\,\ddot{v}\,\ddot{w}}\left(\frac{a^2}...
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1
vote
Accepted
Three Tangents Lemma Using Directed Angles
$$\begin{align}
\measuredangle FEM &= \measuredangle FEB + \measuredangle BEM \\
&= \measuredangle FCB + \measuredangle MBE \\
&= \measuredangle HCB + \measuredangle CBH \\
&= \...
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4
votes
Accepted
Find the other two angles of a triangle when one angle and the ratio of two medians are given.
Use cosine law. Let the original triangle lengths be $2a$, $2b$, and $2c$. Then in $\triangle ACF$ we have $$AF^2=a^2+4b^2-4ab\cos 73^\circ$$
In $\triangle ECB$ you get $$BE^2=4a^2+b^2-4ab\cos 73^\...
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2
votes
Seeking beautiful proofs of plane geometry problems by lifting them to three dimensions
One I had come across $3$ or $4$ months ago:
Finding if $4$ points in the $x-y$ plane lie in a circle (there is some edge case for a line also, beware.)
Consider the points $(x_1,y_1),(x_2,y_2),(x_3,...
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Three Tangents Lemma Using Directed Angles
I did not get directed angles and did not use the hint. This is my approach:
In the picture circle $c$ is on points $A$, $E$ and $F$. Circle $d$ is on points $E$, $F$, $B$ and $C$. $GE$ is tangent to ...
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0
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How to find the value of this angle? A problem from an olympiad.
Hint: In the picture we have:
$\angle ATC=\angle ALI=90^o\Rightarrow BC||LI\Rightarrow \overset{\Large \frown}{BL }=\overset{\Large \frown}{CI }$
Which results in:
$\angle BAL=\angle CAI$
You need to ...
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1
vote
What auxiliary construction helps solve this angle chasing problem?
I apologize in advance, because this feels like a betrayal of the spirit of the problem. To find $x$, I did try to use geometric methods, but I decided to force a trigonometry shortcut to determine ...
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3
votes
Prove that a line is closed in a chain of orthogonal projections onto the circumcircle of a triangle.
@richrow has given a nice answer.
Let me give a cousin explanation here.
I borrow to this answer the notations $\ell_a$, $\ell_b$, $\ell_c$ and $s_a, s_b, s_c$ (see Fig. $1$).
I propose to establish ...
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0
votes
Side of the small pentagon formed by diagonals of a regular pentagon
The diagonal $\overline{BD}$ is parallel to side $\overline{AE}$ and $\overline{EC}$ is parallel to side $\overline{AB}$, so $\square{ABQE}$ is a parallelogram and ftom the congruence of opposite ...
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Finding $BP:PA$ in $\triangle ABC$ with right angle at $A$, $D$ the midpoint of $BC$, and $P$ on $AB$ such that $\angle APC=\angle BPD$
Construct two infinite sequences of points $P, P', P'' \cdots$ and $D, D', D''$ on $AB$ and $CB$ respectively, such that $\angle CPD = \angle PDP' = \angle DP' D = \angle P' D' P'' \cdots$.
Without ...
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2
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Finding $BP:PA$ in $\triangle ABC$ with right angle at $A$, $D$ the midpoint of $BC$, and $P$ on $AB$ such that $\angle APC=\angle BPD$
Reflect $C$ through $AB$ to get $C'$. Then $DC'$ and $AB$ are both medians, so $P$ is the centroid of $\triangle ACC'$, so the ratio is $2:1$.
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Finding $BP:PA$ in $\triangle ABC$ with right angle at $A$, $D$ the midpoint of $BC$, and $P$ on $AB$ such that $\angle APC=\angle BPD$
I can do it using trigonometry.
\begin{align*}
\triangle CPD &= (PC)(PD)\sin 2\theta = 2(PC)(PD)\sin \theta\cos\theta\\
\triangle BPD &= (BP)(PD)\sin \theta = \triangle CPD\\
2(PC)(PD)\sin \...
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3
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Finding $BP:PA$ in $\triangle ABC$ with right angle at $A$, $D$ the midpoint of $BC$, and $P$ on $AB$ such that $\angle APC=\angle BPD$
HINT) Let $E$ be the projection of $D$ on AB. Then the triangles $CAP$ and $DEP$ are similar. Also $DE$ is half of $AC$.
Can you take it from here?
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