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12

All you need is the lengths of each side of the triangle. By Heron's Formula, we know for a triangle with sides $a,b,c$, we have $$A=\sqrt{s(s-a)(s-b)(s-c)}\text{ ,where }s=\frac{a+b+c}2$$ Reference: https://en.wikipedia.org/wiki/Heron%27s_formula EDIT: In response to suggestion by @Hurkyl , I now add the case of ASA and SsA. (ASA) With a known ...


6

The area of a triangle can also be calculated as $\frac12 rp$, where $r$ is the radius of the incircle and $p$ is the perimeter.


6

Your daughter might like to think about two ways of tackling this ... Approach A Draw a triangle of side 2, and fit four unit triangles into it. Extend that picture to a triangle of side 3: how many new unit triangles of side one can you fit into the newly added strip (the trapezoid 2 units along the top, 3 along the bottom)? So how many triangle fit into ...


5

Let $a,b,c$ be the lengths of the sides and $s=(a+b+c)/2$. By Heron's formula, We have $$\sqrt{s(s-a)(s-b)(s-c)}=2s,$$ i.e. $$(-a+b+c)(a-b+c)(a+b-c)=16(a+b+c)\tag1$$ Here, let $x=-a+b+c,y=a-b+c,z=a+b-c$, then we have $$a=\frac{y+z}{2},\quad b=\frac{z+x}{2},\quad c=\frac{x+y}{2}\tag2$$ So, $(1)$ can be written as $$xyz=16(x+y+z)\tag3$$ Here, note that ...


4

If the side lengths are $l,\frac{2}{3}l,\frac{4}{9}l$ then the circumradius is given by: $$ R=\frac{abc}{4\Delta}=\frac{\frac{8}{27}l^3}{l^2\sqrt{\left(1+\frac{2}{3}+\frac{4}{9}\right)\left(1-\frac{2}{3}+\frac{4}{9}\right)\left(1+\frac{2}{3}-\frac{4}{9}\right)\left(-1+\frac{2}{3}+\frac{4}{9}\right)}}$$ by Heron's formula, hence: $$ R = \frac{24}{\sqrt{1463}} ...


3

This is basically a long comment about OP's linked solution to the $2$-ricochet case: You're right, it's a "mess". :) Don't feel too bad, however. It took me a couple of tries to find a clean approach. A first attempt involved an irreducible cubic; the second, a cubic with an extraneous linear factor. However, a little perseverance and geometric insight ...


3

Using Euler's theorem in geometry, $$r^2=R(R-2r)\iff r^2+2Rr-R^2=0$$ $$\implies r=\dfrac{-2R\pm\sqrt{4R^2+4R^2}}2=R(\sqrt2-1)$$ as $R,r>0$ Now using this, $$\dfrac rR=\cos A+\cos B+\cos C-1$$


3

Let $S$ denote the area of the triangle, so: $$ S = \frac{ab\sin{(\gamma)}}{2} = \frac{cb\sin{(\alpha)}}{2} = \frac{ac\sin{(\beta)}}{2} $$ And as for $\theta \in[0,\pi]$: $0\leq\sin{\theta}\leq1$, we get as you said: $$ S\leq\frac{ab}{2}\\ S\leq\frac{ac}{2}\\ S\leq\frac{bc}{2}\\ $$ So: $$ 4S \leq 2ab\\ 4S\leq 2ac\\ 4S\leq 2bc\\ $$ And for $a,b,c\in ...


3

There are a lot of answers to this question--all* of them will involve knowing 3 parameters about the triangle; the formula you use depends on which 3 you know. So far we've seen Heron's formula if you know three sides and the law of cosines approach if you know two sides and an angle (note that also subsumes the formula for a right triangle). I'll add the ...


3

There is another general formula for the area of a triangle: $A=\frac{1}{2}ab\sin\theta$ where $\theta$ is the angle enclosed by the sides of length $a$ and $b$. This is simply a generalisation of $A=\frac{1}{2}bh$ when the concept of height doesn't correspond to a side of the triangle.


3

Following ASCII advocate's idea, the problem boils down to proving that: $$ 4\Delta^2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq (a+b+c)^2 \tag{1}$$ or, using Heron's formula, $$ (a+b-c)(a-b+c)(-a+b+c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq (a+b+c). \tag{2}$$ Through Ravi substitution that turns out to be equivalent to: $$ ...


3

Let me try. One has $$\frac{B'A}{B'C} = \frac{BA}{BC} \Rightarrow \frac{B'A}{AC} = \frac{BA}{BA+BC},$$ $$\frac{C'A}{C'B} = \frac{CA}{CB} \Rightarrow \frac{C'A}{AB} = \frac{CA}{CA+CB}$$. So, one has $$\frac{S_{AB'C'}}{S_{ABC}} = \frac{AB'.AC'}{AC.AC} = \frac{BA.CA}{(BA+BC)(CA+CB)}.$$ Similarly, one has $$\frac{S_{A'BC'}}{S_{ABC}} = \frac{BA'.BC'}{BA.BC} = ...


3

The following works only for the particular case when the circle PHNQ passes through A too. $\alpha + \gamma = \beta + \gamma$ (angles in the same segment) $= \delta$ (exterior angles cyclic quad) $= 90^0$! figure-1 For the general case, I have to give up after a long period of trial. However, I would like to share some of the interesting findings that ...


2

HINT I would say, if $\tan \alpha = \frac{3}{4}$ then $\sin \alpha = \frac{3}{5}$ It simply find from a right triangle with sides 3, 4, 5.


2

Using 2(Area) = $r(a+b+c) = ah_a = bh_b = ch_c$ the inequality can be written in the more appealing form $$ \sum \frac {1}{a^2} \leq \frac{1}{r^2} $$ which is the question of maximizing $\sum a^{-2}$ for fixed $r$. Maybe that has a geometric solution? Of course there is equality for an equilateral triangle. Descartes formula seems potentially useful: ...


2

Let me try. Let $d$ be a distance between the circumcentre and incentre. From Euler's theorem, you have $$d^2= R(R-2r).$$ When the circumcentre is on the incircle, we have $d=r$. Then, $$r^2 = R^2 - 2Rr.$$ So, you get $$\left(\frac{r}{R}\right)^2 + 2\frac{r}{R} - 1 = 0.$$ Then, you get $$\frac{r}{R} = -1 + \sqrt{2}$$ (Do you see why not ...


2

The main observation is that R is the center of an excircle of $\triangle APQ$. Also note that because $\angle BAR = \angle CAR$ we have BR=CR s.t. R can also be defined as the intersection of the bisector of $\angle A$ and the bisector of BC. Let R' be the intersection of the angle bisectors of $\angle QPB$ and $\angle PQC$. Since R' is the center of an ...


2

Applying sine rule in $\triangle ABD$ as follows $$\frac{\sin\angle ABD}{AD}=\frac{\sin\angle ADB}{AB} $$ $$\implies \color{red}{\sin \angle ABD=\frac{AD}{AB}\sin\angle ADB}$$ Now, the circumscribed radius $R_b$ of $\triangle ABD$ is given as $$R_b=\frac{\text{side of triangle }}{2\times\sin \text{(opposite angle)}}$$$$=\frac{AD}{2\sin \angle ...


2

with $a=y+z,b=x+z,c=x+y$ we get $$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}$$ by AM-GM we have $$(x+y)(x+z)(y+z)\geq 8xyz$$ thus our term above is $$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}\geq \sqrt{8}\sqrt{2}=4$$


2

Let $a,b,c$ be the side lengths. Then, $b,c$ can be written as $$b=\frac 23a,\quad c=\left(\frac 23\right)^2a$$ where $a$ is the longest side. So, we have, by the law of cosines, $$a^2=\left(\frac 23a\right)^2+\left(\left(\frac 23\right)^2a\right)^2-2\cdot\frac 23a\cdot\left(\frac 23\right)^2a\cos A$$ $$\Rightarrow \cos A=-\frac{29}{48}.$$ Hence, $$a=2R\sin ...


2

Assume that: $$M=\alpha A+\beta B+\gamma C,\quad \alpha,\beta,\gamma\geq 0,\;\alpha+\beta+\gamma=1, $$ i.e. let $[\alpha,\beta,\gamma]$ be the barycentric coordinates of $M$. The line through $M$ and $A$ has equation $\gamma y-\beta z=0$ while the line through $A$ and $B$ has equation $z=0$. By corollary $18$ of Volonec, $$\cot\widehat{MAB} = \cot A+(\cot ...


2

The side of the triangle in G.P. with common ratio $\frac{2}{3}$ can be taken as $$a=\frac{3}{2}x, b=x, c=\frac{2}{3}x$$ Where, $x$ is a positive real number. Hence, using cosine formula we get $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ $$=\frac{\frac{9}{4}x^2+x^2-\frac{4}{9}x^2}{2\cdot\frac{3}{2}x \cdot x }=\frac{101}{108}$$ $$\implies \sin C=\sqrt{1-\cos ^2 ...


2

Using trilinear coordinates, we have $I=[1;1;1]$ and $G=\left[\frac{1}{a},\frac{1}{b},\frac{1}{c}\right]=[h_a;h_b;h_c]$, so: $$ P=[0;1,1],\quad Q=[1;0;1] $$ and $G$ lies on $PQ$ iff $\det[G,P,Q]=0$. Since: $$\det\begin{pmatrix}h_a & h_b & h_c \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}=h_a+h_b-h_c$$ the claim trivially follows.


2

You have $\log A=\log a + \log b - \log 4, \frac {dA}A=\frac {da}a+\frac {db}b, \frac {dA}A=4\%-3\%=1\%$


2

The simplest way to remember how to calculate is by taking $\frac{1}{2}$ the value of the determinant of the matrix $$ \begin{bmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{bmatrix} $$


2

Yes, it is the right track. You may also directly prove that $PQNBM$ is a cyclic pentagon. Assuming $D=(0,0),C=(1,0),N=(1,x_0)$, the equation of the $MD$-line is given by: $$ y = \frac{1+x_0}{1-x_0}\,x $$ by the tangent addition formula, hence $M=\left(\frac{1-x_0}{1+x_0},1\right)$ and $R=\left(\frac{1}{1+x_0},\frac{1+x_0}{2}\right)$. The equation of the ...


2

Here is the diagram: The two smaller triangles are similar and the ratio of their legs is the ratio of their bases since the two $12$ length sides are also the radius of the circle tangent to the larger triangle.


2

Let the tangent length shown in sketch be T. The power of circle $$ T^2 = (15-R) (15+R) \tag{1}$$ From similar triangles, (radius/hypotenuse) of right side right angled triangle: $$ \frac{T}{R}= \frac{15}{20}= \frac{3}{4} \tag{2}$$ Solving $$ R= 12, \; T = 9 \tag{3} $$ Arc Length is quarter circle $$ =\pi \, R/ 2 = \frac{ \pi \cdot 12}{2} = 6 \pi ...


1

Using $$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$ $$\Rightarrow \frac{a+b+c}{2}r\cdot\frac{1}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac 14$$ and $$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{-a+b+c}{2}r_a=\frac{a-b+c}{2}r_b=\frac{a+b-c}{2}r_c$$$$\Rightarrow rr_a=\frac{a+b+c}{-a+b+c}r^2,\quad ...


1

Brute force method: Firstly, Heron's formula can be simplified in this form: The area $\Delta$ of a triangle with sides $a,b,c$ is given by $16\Delta^2 $ $= (a + b + c)(b + c - a) (c + a - b)(a + b - c) \\= 2(a^2 b^2 + b^2 c^2 + c^2 a^2) - (a^4 + b^4 + c^4)$ Therefore area of triangle with sides as $\sqrt a, \sqrt b, \sqrt c$ is $\frac 1 4 \sqrt{2 (ab + bc ...



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