Hot answers tagged

6

A hint: $3^\circ=45^\circ+30^\circ-72^\circ$. For $72^\circ$ however you have to know how to construct a regular pentagon or some golden ratio triangle.


5

An alternative geometric approach is presented. Consider vectors $\mathbf{x}=(\sqrt{a_1},\sqrt{b_1},\sqrt{c_1})\in \mathbb{R}^3$ and $\mathbf{y}=(\sqrt{a_2},\sqrt{b_2},\sqrt{c_2})\in \mathbb{R}^3$. Then, considering the vector dot product, we have $$\mathbf{x}\cdot\mathbf{y}=\lvert\mathbf{x}\lvert\lvert\mathbf{y}\lvert\cos\theta\\\Rightarrow\sqrt{a_1a_2}+...


5

Why? As $\sin A,\sin B,\sin C>0,$ $$\dfrac{\sin A+\sin B}2\ge\sqrt{\sin A\sin B}$$ $$\implies(\sin A+\sin B)(\sin B+\sin C)(\sin C+\sin A)\ge8\sin A\sin B\sin C$$ which is definitely $>\sin A\sin B\sin C$


5

As long as this triangle is Euclidean, each of the sin values is positive (or maybe 0 but that case is easy). When you multiply these factors out, you will get $\sin A \sin B \sin C+$..(other positive stuff) which is definitely greater than $\sin A \sin B \sin C$


4

The length is $2AD+DE$. Obviously $DE=2$, so we have to find $AD$. Consider the triangle $ADF$. $DF=1,\angle AFD=60^o$, so $AF=\sqrt3DF=\sqrt3$. Hence $AB=2+2\sqrt3$.


3

I asked this from our teacher and he give this answer may helps also this answer don't uses trigonometry: $BNC=90$degrees and $BAN=45$ then we have $AN=BN$ also because $BNC$ is right angle $MN=BM$ also because $NBM=60$,the triangle $BNM$ is equilateral.Also $MNA=30$ then because $MN=AN$ the angle $NMA=75$ then we have $AMC=45$.


3

It's easy. Assume that, $a_i$'s, $b_i$'s and $c_i$'s are corresponding sides. Now squaring both sides, and cancelling common terms both side, you get the identity something like this,$$\sum\limits_{\text{cyclic}}(\sqrt{a_1b_2}-\sqrt{b_1a_2})^2=0.$$ Which means, each square term is individually $0$. So, $$\sqrt{a_1b_2}-\sqrt{b_1a_2}=0\\\implies\frac{a_1}{...


3

No constructions needed. ADY and ABX have half the base and the same altitude (to different sides) so their areas are 1/4 of ABCD. XYC has half the base and half the altitude so its area is 1/8 of ABCD. Together they make 1/4+1/4+1/8=5/8, so what remains is 3/8.


3

If $a,b,c$ are sides of triangle and we assume that $\displaystyle a,b,c,\frac{a+b+c}{2}$ is an arithmetic sequence, then we can write \begin{align}a+c&=2b\tag{1}\\b+\frac{a+b+c}{2}&=2c\tag{2}\end{align}


3

X must be a right angle because it faces a $180°$ arc. Therefore, $z=180-90-25=65°$. What you have is correct. You know that y must be $90°$ because it also faces an $180°$ arc, and because triangle ACD is isosceles, W must be $\frac{180-90}{2}$ or $45°$.


3

First, note that a given triangle and a given point of perspectivity determine three rays making some set of angles; we'll say that the "outer" rays form respective angles $\alpha$ and $\beta$ with the "middle" ray. (Having taken opposite rays if necessary, we may assume $\alpha + \beta \leq 180^\circ$.) Also, if a target triangle has a "similar copy" that's ...


3

Following E.Girgin suggestion plus Richmond's method: Steps: Construct a regular hexagon $ABCDEF$ with centre $O$; Construct a square in such a way that $\widehat{AOG}=75^\circ$; let $B'=BC\cap OG$ and take $A'\in FA$ such that $OA'\perp OB'$; Let $\Gamma$ be the circle with centre $O$ through $A'$ and $M$ be the midpoint of $OA'$; let $N$ be the ...


3

Since the cotangent function is odd, we may assume $A,B,C\geq 0$ without loss of generality. So $A,B,C$ are the angles of a triangle. Let we deal first with the case of an acute triangle. By the Cauchy-Schwarz inequality: $$\cot(A)^2+\cot(B)^2+\cot(C)^2 \geq 3\left(\frac{\cot A+\cot B+\cot C}{3}\right)^2 = \frac{\cot(\omega)^2}{3}$$ where $\omega$ is the ...


3

For another way, since $\cot^2x$ is convex when $x \in [0,\pi]$ and we need to worry only of positive $x$, so by Jensen's inequality: $$ \sum_{cyc} \cot^2 A \geqslant 3 \cot^2\frac{A+B+C}3=1$$ As equality is possible when $A=B=C=\frac{\pi}3$, we have the minimum.


3

Nope, $AC$ need not equal to other $5$ edges and the tetrahedron need not be regular. As long as the remaining $5$ edges are equal in length, we have $BC = CD = DB \implies \triangle BCD$ is equilateral $\implies K$ coincides with centroid of $\triangle BCD$. $AB = BD = AD \implies \triangle ABD$ is equilateral $\implies M$ coincides with centroid of $\...


3

$$(a-c)(a+c)^2+bc(a+c)=ab^2$$ $$\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$ $$\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$ $$\implies (a-b+c)(a(a+b)-c^2)=0$$ Dividing the both sides by $a-b+c\gt 0$ gives $$c^2=a^2+ab$$ from which we have $c\gt a$. Case 1 : $...


3

As mentioned, the limiting case is when the small circle touches the larger circle internally. Then, we have the figure below:- Applying Pythagoras theorem to the red triangle, we have $OM^2 = 1 – (o.5L)^2$. Applying Pythagoras theorem to the green triangle, we have $(R + d)^2 = (1 – R)^2 – OM^2$ Eliminating $OM^2$ from the two equations, we get $d$ in ...


3

We have: $$ \frac{r_a}{bc}=\frac{\Delta}{(s-a)bc}=\frac{\frac{1}{2}\sin(A)}{s-a}=\frac{a}{4R(s-a)}\tag{1}$$ hence we have to prove that: $$ 2+\sum_{cyc} \frac{a}{s-a} = \frac{4R}{r} \tag{2}$$ that is equivalent to: $$ 2r+\sum_{cyc}\frac{a}{\cot\left(\frac{A}{2}\right)} = 2R\tag{3}$$ or to: $$ 2r+\sum_{cyc}2R(1-\cos(A)) = 2R\tag{4}$$ that follows from Carnot'...


2

This will be an answer that uses vector calculus. Let $a,b,c$ be the vector-coordinates from the origin $O$ to $A,B,C$ respectively. As you want to rotate around $B$, it is useful to translate $B$ to the origin, and to translate back when you have found the coordinates. As $A$ and $C$ are on the line $BA$ that you want to rotate, it is indifferent if you ...


2

Draw the diagonal $AC$. We see that $\triangle ACY$ has the same area as $\triangle ADY$, by choosing $CY$ and $DY$ to be the bases, since they have the same height. In the same way, $\triangle ACX$ has the same area as $\triangle ABX$. Therefore, the quadrilateral $AYCX$ takes up half of the parallelogram. It remains to show that $\triangle CXY$ has area ...


2

Use $$S_{ABC}=\frac12 a\cdot h_a$$ and $$S_{ABCD}= a\cdot h_a$$ Let $Ar(ABCD)=S$, then $Ar(\triangle ABX)=\frac14S$, $Ar(\triangle ADY)=\frac14S$, $Ar(\triangle CXY)=\frac18S$ Then $Ar(\triangle AXY)=S-\frac14S-\frac14S-\frac18S=\frac38S$


2

By Stewart's theorem $$ 4m_b^2 = 2a^2+2c^2-b^2,\qquad 4m_c^2 = 2a^2+2b^2-c^2, $$ hence: $$ 4(m_c^2 - m_b^2) = 3(b^2-c^2)$$ and $c>b$ is equivalent to $m_c < m_b$.


2

$\qquad\qquad\qquad$ Let $E$ be a point inside $\triangle{OBC}$ such that $\triangle{EBC}$ is an isosceles triangle with $\angle{EBC}=\angle{ECB}=15^\circ$. Then, since $\triangle{OEB}$ is an isosceles triangle with $OB=BE$ and $\angle{OBE}=60^\circ$, we know that $\triangle{OEB}$ is an equilateral triangle, and so $\triangle{OEC}$ is an isosceles triangle ...


2

The sum of two opposite angles $ = 90^0 + 90^0$. Hence there is a circle (red) passing through OAXB where X is the point of intersection of the tangents.


2

The answer is $\color{red}{45^\circ}$. You may embed the construction in the complex plane by assuming $C=1, B=-1$. Then $A$ has to fulfill: $$A-C = \lambda e^{5\pi i/6},\qquad A-B = \mu e^{\pi i/12},\qquad (\lambda,\mu\in\mathbb{R}) $$ but that leads to $A=\eta(1+i)$, and the claim follows. An alternative, elementary way. By assuming $BM=MC=1$ we have: $$...


2

This is immediate from the Cauchy-Schwarz inequality. Assuming you are not familiar with that we proceed from first principles: $(x\sqrt{a_1}+\sqrt{a_2})^2+(x\sqrt{b_1}+\sqrt{b_2})^2+(x\sqrt{c_1}+\sqrt{c_2})^2\ge0$ so $(a_1+b_1+c_1)x^2+2(\sqrt{a_1a_2}+\sqrt{b_1b_2}+\sqrt{c_1c_2})x+(a_2+b_2+c_2)\ge0$ This quadratic in $x$ cannot have two real roots we must ...


2

By Ceva, $AP,BQ$ and $CR$ are concurrent if and only if $\frac{|BP|}{|CP|}\cdot\frac{|CQ|}{|AQ|}\cdot\frac{|AR|}{|BR|}=1$. Similarly $AP',BQ'$ and $CR'$ are concurrent iff $\frac{|BP'|}{|CP'|}\cdot\frac{|CQ'|}{|AQ'|}\cdot\frac{|AR'|}{|BR'|}=1$. So it suffices to show that $\frac{|BP|}{|CP|}\cdot\frac{|CQ|}{|AQ|}\cdot\frac{|AR|}{|BR|}=\frac{|CP'|}{|BP'|}\...


2

Put $x=\cot A,y=\cot B$. Using the standard formulae we have $z=\cot C=-\cot(A+B)=\frac{1-xy}{x+y}$. So we want to minimize $$f(x,y)=\frac{(x^2+y^2)(x+y)^2+(xy-1)^2}{(x+y)^2}$$ where $x,y$ can take any real values. If we want to minimize $(x^2+y^2)(x+y)^2+x^2y^2-2xy+1$ subject to $x+y=k$, then using Lagrange multipliers we find $4x^3-2y+6x^2y+6xy^2+2y^3=\...


2

Let the sides of the triangle be $a,b,c.$ Let $R$ be the circumradius of the triangle We have $$0< 2 R=(\sin A)/a=(\sin B)/b=(\sin C)/c.$$ So the inequality is equivalent to $$(a+b)(b+c)(c+a)>a b c.$$ In a triangle we have $a+b>c>0$ and $b+c>a>0$ and $c+a>b>0 .$ So $(a+b)(b+c)(c+a)>c a b.$ Alternate proof: In a triangle the ...


2

Let $F$ be the intersection point of $x$ with $z$. Then, draw a circle whose center is $F$ with radius $EF$. The intersection points of the circle with $x$ are $B,C$. Now extend the sides of the right triangle $\triangle{BCE}$ to have $A$.



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