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Show that triangle $ABC$ is isosceles.

We can construct two isosceles triangles: $\triangle AEC$ by drawing a segment $AE$ at a $10^\circ$ angle to $AM$; and given the intersection of $AE$ and ray $BM$ at point $F$, $\triangle AFB$. $EE_\...
Ted Black's user avatar
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-1 votes

How to find the legs of a right triangle if its hypotenuse is numerically equal to its area?

[ TO FIND THE LEGS OF THE RIGHT TRIANGLE HAVING ITS HYPOTENUSE AS IT'S AREA Consider ABC a triangle such that AC is the hypotenuse while AB (b) and BC (a) will be the sides (legs). Now , $ AC = \frac{...
Aditya's user avatar
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1 vote

Solve the system $ - \frac{x^2}{2} + y^2 + z^2 = 2a^2 \quad \quad x^2 + - \frac{y^2}{2}+ z^2 = 2b^2 \quad \quad x^2 + y^2 - \frac{z^2}{2}= 2c^2$

If $a$, $b$ and $c$ are sides of a triangle then $a<b+c$, so $$x^2=\frac49\left(-a^2+2b^2+2c^2\right)>\frac49\left(b^2-2bc+c^2\right)>0$$so $x\in\mathbb{R}$. Likewise, by symmetry, $y,z\in\...
Sai Mehta's user avatar
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How to find the legs of a right triangle if its hypotenuse is numerically equal to its area?

The right angle of the triangle lies on a circle whose diameter is the hypotenuse. In order for the area to equal the hypotenuse, the altitude on the hypotenuse is $2$. These two facts exactly ...
David K's user avatar
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1 vote

Show that triangle $ABC$ is isosceles.

This solution was given by user vineet on AoPS. Construct an equilateral triangle $\triangle AMD$ with base $AM$, on the opposite side of $B$. Let $E = DB\cap CM$. Claim 1: $BM$ is the perpendicular ...
D S's user avatar
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2 votes

Integer Right Triangle with Repdigit Area

For a right triangle with sides $a,b,c$, $c$ being the hypotenuse, the area is $S = ab/2$ thus $ab = 2S$, therefore $a$ can only be a divisor of $2S$. Suppose we already know about divisors of $2S$, ...
ioveri's user avatar
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2 votes

How to find the legs of a right triangle if its hypotenuse is numerically equal to its area?

In the triangle $ABC$ right angled at $C$, call $F$ the foot of the perpendicular from the vertex $C$ on the hypotenuse. Denote the lengths of the mentionned segments $c,a,b,h$ respectivelly. The use ...
user376343's user avatar
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2 votes

How to find the legs of a right triangle if its hypotenuse is numerically equal to its area?

You have two equations: $ c = \dfrac{1}{2} a b $ and $ c^2 = a^2 + b^2 $ So $ \dfrac{1}{4} a^2 b^2 = a^2 + b^2 $ Let $x = a^2, y = b^2 $ then $ x y = 4 (x + y) $ where $ x \gt 0 , y \gt 0 $ Hence, $ x(...
Hosam Hajeer's user avatar
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2 votes

How to find the legs of a right triangle if its hypotenuse is numerically equal to its area?

Let $ACB$ be a triangle right angled at $C$ let $AB=c, AC= b, BC = a$. Then by Pythagoras theorem $a^2+b^2=c^2$ but given $c=\frac{ab}{2}$. Plugging in and simplifying you get $\frac{1}{a^2}+\frac{1}{...
Amrit Awasthi's user avatar
3 votes

Show that triangle $ABC$ is isosceles.

Here is an elementary proof of this statement that does not involve trigonometry. By plane separation axiom, line $AC$ separates a plane into two half-planes. Construct an equilateral triangle $CAK$ ...
Rusurano's user avatar
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2 votes
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Integer Right Triangle with Repdigit Area

I will check up almost all numbers up to 100 digits (except some numbers) that there is no other repdigit which is the area of a Pythagorean Triangle except $666666.$ I will use the CAS Maxima. Its ...
miracle173's user avatar
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4 votes

Show that triangle $ABC$ is isosceles.

Here is a more general hint towards solving these kinds of problems. When you see a problem where degrees are given in $10x$ increments (this is sometimes disguised a bit), the problem is likely ...
RobinSparrow's user avatar
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1 vote

Show that triangle $ABC$ is isosceles.

HINT.-Let $a,b,c$ the sides. The angles to determine are $\alpha=\angle{MCB}$ and $\beta=\angle{MBC}$. These are solution of the system $$\alpha+\beta=80^{\circ}\\ac\sin 50^{\circ}=bc\sin (20+\beta)=...
Piquito's user avatar
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5 votes
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Show that triangle $ABC$ is isosceles.

By law of sines, $$\frac{AM}{\sin 30°}=\frac{AC}{\sin 110°} \tag{1}\label{1} $$ $$\frac{AB}{\sin 150°}=\frac{AM}{\sin 20°} \tag{2}\label{2}$$ We can combine ($\ref{1}$) and ($\ref{2}$) to get $$\frac{...
Gwen's user avatar
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0 votes

Number of triangles in a graph based on number of edges

Here's a simple proof with, in fact, an optimal bound. For each vertex $v_i \in V$, let $V_i, E_i$ be the set of vertices adjacent to $v_i$ and the set of edges from $E$ that join those vertices. The ...
Sgg8's user avatar
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0 votes

Given the base and angles of an isosceles triangle, how to find length of the two sides?

You have been given the base: a You have been given the base angle: 𝜃 Then the length of the side (b) is given by the formula: b = a / (2 * Cos(𝜃))
Robert Livingston's user avatar
1 vote

Integer Right Triangle with Repdigit Area

In a Pythagorean triple, side-$A\,$ can be can any number greater than $\,2\,$ and side-$B\,$ can be any multiple of $\,4.\,$ Using brute force, testing all side lengths from $\,3\,$ to $\,2,000,000\,$...
poetasis's user avatar
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1 vote

Looking for a simple proof of this statement about angle bisectors

This is a bit quicker. Use the same relation of $2\alpha, 2\beta, a$ on the triangles with angles $2\alpha, 2\phi, \beta$ and $2\beta, 2\delta, \alpha$ to get $$\alpha+\phi = 90^\circ - \frac{\beta}{2}...
JMP's user avatar
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1 vote

Looking for a simple proof of this statement about angle bisectors

Considering the figure we have: $\widehat{BDC}=90^o+\frac{\widehat{BAC}=a}2$ In quadrilateral FDJI we have: $\widehat{FIJ}=180-(\widehat{FDJ}=90+\frac a2)=90-\frac a2$ I is on perpendicular bisector ...
sirous's user avatar
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-2 votes

Angle in a triangle

The rule for tangent is $\tan={\frac{opposite}{adjacent}}$, So, the opposite side of $∠𝐴$ in triangle ABC is BC, and the adjacent side of $∠𝐴$ is AB, so for triangle ABC, it is $\frac{BC}{AB}$ For ...
Washington Fig's user avatar
2 votes

If in $\triangle ABC$, $r=1,a=3$,then find least possible area of $\triangle ABC$

HINT.- Since area of $\triangle {ABC}$ is equal to $\dfrac{\overline{BC}\times h}{2}$ we can find the smallest possible $h$ and classic construction of the tangents shows that this occurs when $\...
Piquito's user avatar
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1 vote

If in $\triangle ABC$, $r=1,a=3$,then find least possible area of $\triangle ABC$

Here is one way using Lagrangian. You have found out that $\Delta =s$. Using your notations, the Heron formula implies $(s-3)(s-b)(s-c)-s=0$. We want to minimize $s$. The Lagrangian is $$L(b,c,\lambda)...
toronto hrb's user avatar
2 votes
Accepted

If in $\triangle ABC$, $r=1,a=3$,then find least possible area of $\triangle ABC$

Let the center of the circle $O = (0,0)$, $B = (-b,-r)$, $C = (c,-r)$, and $\beta = \angle CBO$, $\gamma = \angle BCO$​. Let the height of triangle (distance between $A$ and $BC$) be $h$. Minimizing ...
Y.D.X.'s user avatar
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2 votes

If in $\triangle ABC$, $r=1,a=3$,then find least possible area of $\triangle ABC$

Let put the triangle into the coordinate axes such that $B=(0,0)$ and $C=(3,0)$, then the inscribed circle touches $BC$ at point $D=(a,0)$ (where $a\in[0,3]$ can vary), see the picture. Let $b=\angle ...
van der Wolf's user avatar
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1 vote
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Knowing that the area of a lateral face is equal to the area of the base, find the measure of the angle formed by the planes $(MBN)$ and $(ABC)$.

Hint (almost solution). First solution. One way to solve this problem without thinking too much, just doing calculations, is to apply the orthogonal projection theorem. When orthogonally projected ...
greyls's user avatar
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1 vote

Does $ \frac{\sin(\theta-\alpha)}{\sin\alpha}=\frac{\cos(\theta+\gamma-\alpha)}{\cos(\gamma-\alpha)}$ have an analytical solution for $\alpha$?

You can rewrite your Equation (1) as $$\sin\!\left(A-B\right)\cos\!\left(A+B\right)=-\sin\!\left(A-C\right)\cos\!\left(A+C\right)$$ with $$A=\frac12\left(\theta+\gamma-2\alpha\right)\,,\quad B=\frac12\...
Toffomat's user avatar
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4 votes
Accepted

Does $ \frac{\sin(\theta-\alpha)}{\sin\alpha}=\frac{\cos(\theta+\gamma-\alpha)}{\cos(\gamma-\alpha)}$ have an analytical solution for $\alpha$?

Short answer: $$\alpha=\arctan\left(\frac1{\cot\theta-\tan\gamma+\sqrt{\tan^2\gamma+\cot^2\theta+1}}\right)$$ is the positive $\alpha.$ Directly from the geometric question, if we write in coordinate ...
Thomas Andrews's user avatar
0 votes

Draw tangents at 3 random points on a circle to form a triangle. Show that the probability that a random side is shorter than the diameter is $1/2$.

Let $T_1,T_2,T_3$ be the tangent points so that $A,T_1,B$ colinear. $$P(AT_1\leq r)=\frac12$$ $$P(BT_1\leq r)=\frac12$$ Since, the events $(1)$ and $(2)$ are independent, we have $$P(AT_1+BT_2=AB\leq ...
Bob Dobbs's user avatar
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0 votes

Prove |ab − a′b′| ≤ |b||a − a′| + |a||b − b′| + |a − a′||b − b′|

Using $|u+v|\le |u|+|v|$ and $|uv|=|u|\,|v|$ we have \begin{align} |ab-a'b'|&=|ab-a'b+a'b-a'b'|=|(a-a')b+a'(b-b')|\\ & \le|b|\,|a-a'|+|a'(b-b')| =|b|\,|a-a'|+|a'(b-b') +a(b-b')-a(b-b')| \\ &...
van der Wolf's user avatar
  • 2,360
2 votes

Prove that the segment $BE$ intersects the common point of two circles

Let's mark the point of intersection of two circles as $P$. We know that $DC = DP$, as radiuses of $C_{2}$. Now let $E'$ be the point of intersection of $BP$ and $AD$. We want to prove that $E' = E$. $...
Meison's user avatar
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2 votes
Accepted

Prove that the segment $BE$ intersects the common point of two circles

Let $C \neq X = C_1 \cap C_2$. Note that $$\begin{align} \angle BXC &= 180^\circ -\angle BAC \\ &= 180^\circ -\angle ACB \\ &= 180^\circ - \angle ADB \\ &= \angle BDE \end{align}$$ ...
D S's user avatar
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0 votes
Accepted

Draw tangents at 3 random points on a circle to form a triangle. Show that the probability that a random side is shorter than the diameter is $1/2$.

Assume the radius of the circle is $1$. Call the three random points $A,B,C$. Point $A$ is the lowest point on the circle. The arc from $A$ to $B$ measured anticlockwise has length $2x$, and the arc ...
Dan's user avatar
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1 vote
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Find all positive integers $k$ for which, for any $a,b,c$ the inequality $k(ab+bc+ac) \geq 5(a^2+b^2+c^2)$ holds, $a,b,c$ are the sides of a triangle

I don’t quite understand your logic. Your reasoning shows that, rather, $k\le 10$. But you didn’t show “$2(ab + bc + ac) \geq a ^ 2 + b ^ 2 + c ^ 2\implies$ there is a triangle with side lengths $a, b$...
Aig's user avatar
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1 vote
Accepted

The volume of a tetrahedron $ABCD$ is $\frac{1}{6}$. Determine $CD$ knowing that $\angle ACB=45$ and that $AD+BC+\frac{AC \sqrt{2}}{2}=3$.

Let's define: $x=BC$, $y=AC/\sqrt2$, $z=AD$. We obtain then, from the given relations: $$ x+y+z=3 $$ and $$ xyh=1, $$ where $h$ is the distance of $D$ from plane $ABC$. The tetrahedron can exist only ...
Intelligenti pauca's user avatar
4 votes

How to find an acute angle of a right triangle inscribed in a square?

I'm afraid you're likely to kick yourself…  (This doesn't need any trigonometry, just simple angle sums.) First, consider the angles in the bottom triangle.  Its left-hand angle is marked as $\alpha$, ...
gidds's user avatar
  • 179
6 votes
Accepted

How to find an acute angle of a right triangle inscribed in a square?

$?+90+(90-\alpha)=180$ so $?=\alpha$.
marty cohen's user avatar
2 votes

Find the area of the region enclosed by $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ and the $x$-axis.

I really liked your intuition about the $\dfrac{\pi^2}{8}$ value of the area. I tried to achieve $y=f(x)$ but couldn't and so I decided to approximate the area as follows: I determined the points $A, ...
Piquito's user avatar
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3 votes

Find the area of the region enclosed by $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ and the $x$-axis.

$${\sin x\over\sin y}=\frac{\sin x+\sin y}{\sin x \cos y+\sin y \cos x}\implies \sin^2 x\cos y+\sin x \cos x \sin y =\sin y \sin x+\sin^2y$$ If we substitute $g = \sin y$ and $v = \sin x$, we get: $$ ...
Masd's user avatar
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1 vote

Draw tangents at 3 random points on a circle to form a triangle. Show that the probability that a random side is shorter than the diameter is $1/2$.

Here is an extended comment: There is a geometric equivalence to an earlier question of yours, also a geometric probability question wherein the answer of $\frac 12$ resists explanation. Claim. Let $...
Carl Schildkraut's user avatar
0 votes

Three randomly placed points inside a unit circle. Probability that the formed triangle contains the centre of the circle inside?

Here is an approach using random variables and straightforward geometrical reasoning. Let points A, B, and C be located inside a circle of center M, with the following qualifications: point A is on ...
user2585330's user avatar
2 votes

Conjecture: Two different random triangles (both based on random points on a circle) have the same distribution of side length ratios.

The angles in $T1$ are half the corresponding angles at the centre of the circle, so they are unuformly distributed on the triangle $\alpha+\beta+\gamma=\pi$. The angles in $T2$ are either $\pi-2\...
Empy2's user avatar
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0 votes

Can any right triangle be inscribed in a circle?

Any right angle triangle can be inscribed in a circle, granted that the hypotenuse is the diameter The following proof uses the "Triangle Proportionality Theorem": If a line parallel to one ...
Reuben's user avatar
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0 votes
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Connection between trigonometric ratios and similar triangles.

The two triangles are similar (same angles) but not necessarily congruent. For the generic triangle on the left, the hypotenuse has length $c$, but for the right triangle embedded inside the unit ...
Sammy Black's user avatar
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1 vote

How must I find the third vertex of an equilateral / right isosceles triangle given the coordinates of 2 vertices?

This is really trivial. For the isosceles right triangle with given hypotenuse endpoints $A(x_1,y_1)$ and $B(x_2, y_2)$, you just need to find the midpoint of $AB$: $M = \dfrac{1}{2} (A + B) = \dfrac{...
Hosam Hajeer's user avatar
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1 vote

How must I find the third vertex of an equilateral / right isosceles triangle given the coordinates of 2 vertices?

I'm trying to write the most simple way I know . You actually can get 2 isoceles right triangles if you're given only two points as vertices. Let the other vertex have coords $C(h,k)$. $A(x_1,y_1)$ ...
Gwen's user avatar
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1 vote
Accepted

How must I find the third vertex of an equilateral / right isosceles triangle given the coordinates of 2 vertices?

You can show your isosceles triangle equality $$(x-x_1)^2+(y-y_1)^2=(y-y_2)^2+(x-x_2)^2$$ leads to $$(2x-x_1-x_2)(x_2-x_1) = (2y-y_1-y_2)(y_1-y_2)$$ and so to $$\frac{x-\frac{x_1+x_2}2}{y_1-y_2} = \...
Henry's user avatar
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5 votes

Draw tangents at 3 random points on a circle to form a triangle. Show that the probability that a random side is shorter than the diameter is $1/2$.

Here is my non-intuitive answer. Assume the circle is $x^2+y^2=1$, and the points are: $A\space(\cos(-\alpha), \sin(-\alpha))$ where $0<\alpha<2\pi$ $B\space(\cos \beta, \sin \beta)$ where $0<...
Dan's user avatar
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3 votes
Accepted

Conjecture: The line joining the incenter and the circumcenter always subtends an obtuse angle at centroid

Denote the incenter by $I$, circumcenter by $O$, the centroid by $G$. Assume that the triangle is not equilateral, so that the points $I, G, O$ are distinct. In order to prove that $\angle IGO>90^\...
timon92's user avatar
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0 votes

Using Euclidean geometry, how to find $x$?

As already shown, $BD=AB$, also since $\widehat{DAC}=\widehat{ACB}=x$ we infer $BC\parallel AD$, hence $\widehat{DBC}=\widehat{ADB}=3x$. Take $E$ reflection of $D$ about $BC$, it belongs to $AB$ and $...
Stan FULGER's user avatar
0 votes

Which theorem should be used to solve this question?

Take $E$ - reflection of $C$ about $AD$, it belongs to $AB$ and $\widehat{BED}=91^\circ$, i.e. $O$, the circumcenter of $\triangle BDE$ belongs to $BC$ and $BO=OD=DE=CD$, hence $\widehat{BCD}=\widehat{...
Stan FULGER's user avatar

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