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12

Vectors: The area of a parallelogram embedded in a three-dimensional Euclidean space can be calculated using vectors. Let vectors $AB$ and $AC$ point respectively from $A$ to $B$ and from $A$ to $C$. The area of parallelogram ABDC is then $$\left|AB \times AC\right|$$ so that the area of a triangle is half of this, giving $$A_{\text{triangle}} = \frac{1}{2} ...


12

Let the legs measure $a$ and $ka$ with $k>1$. Then the approximation given is: $\frac{7ka}{8}+\frac{4a}{8}=\frac{(7k+4)a}{8}$. The real measure is $\sqrt{k^2a^2+a^2}=\sqrt{k^2+1}a$. So how does $\frac{7k+4}{8\sqrt{k^2+1}}$ behave? Not that bad. Here is a graph: So as the graph shows, in the interval $(1,\infty)$ the best approximation occurs at ...


6

Let $M,A,B,C$ be the geometric images of the complex numbers $z,z_{1},z_{2},z_{3}$, let $f(x)=1$, then use Lagrange interpolation formula applied to the polynomial $f(x)$, we have ...


5

Since the sine theorem implies: $$\sum_\text{cyc}a\sin A = \frac{1}{2R}\sum_\text{cyc}a^2 \tag{1}$$ we just need to prove: $$ \sum_\text{cyc} a \cos A = \frac{abc}{2R^2}=\frac{2\Delta}{R}\tag 2$$ that is trivial since twice the (signed) area of the triangle made by $B,C$ and the circumcenter $O$ is exactly $aR\cos A$:


5

Firstly, in figure 1, Using Napier's Analogy in $\Delta ABC$, we have, $$\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cdot \cot \left( \frac{A}{2}\right)$$ Now, in figure 2, Using Sine Law in $\Delta ABM$, we have, $$\dfrac{x}{\sin \left(\frac{A}{2} + \alpha \right)}=\dfrac{c}{ \sin(\angle BMA)}$$ Similarly, in $\Delta ACM$, ...


5

The method provides a not-so-great approximation when the two catheti are roughly of the same length. In this case, $$\frac{7}{8}a+\frac{1}{2}b\approx\frac{7}{8}a+\frac{1}{2}a=\frac{11}{8}a=1\mathrm.375 a\approx(1\mathrm.414...)a=\sqrt{2}{a}\approx\sqrt{a^2+b^2}.$$


4

Numerical optimization reveals how truly astonishing the approximation is: Taking a hint from @dREaM's answer and normalizing the length of the shortest side, for side lengths $1$ and $k$ we are interested in the ratio $$\frac{ak+b}{\sqrt{k^2+1}}$$ as a function of the coefficients $a$ and $b$ (in the Tamil approximation, $a=7/8$ and $b=1/2$). We want the ...


4

$\triangle ABC \sim \triangle DEF$.


3

$s=pr$ where $p=\frac{a+b+c}{2}$ and $r$ is the radius of the inscribed circle. $s=\sqrt{r\cdot r_a\cdot r_b\cdot r_c}$ where $r_a,r_b,r_c$ are the exradii of excircles.


3

Using the law of cosines $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$ and the law of sines $$\sin B=\frac{b}{2R},\ \ \ \sin C=\frac{c}{2R}$$ where $R$ is the radius of the circumscribed circle of $\triangle{ABC}$, we have $$\frac{b^2+c^2-a^2}{2bc}=\frac{\frac{b}{2R}}{2\cdot\frac{c}{2R}}\iff \frac{b^2+c^2-a^2}{2bc}=\frac{b}{2c}$$$$\iff b^2+c^2-a^2=b^2\iff ...


3

As I said in my comment above, one can define: $\alpha=1/BC$, $\beta=1/AC$, $\gamma=1/AB$ so that we are required to find the point $P$ such that $\alpha PA + \beta PB + \gamma PC$ attains its minimum value. Following the method outlined here let's denote by $\vec{i}$, $\vec{j}$ and $\vec{k}$ the unit vectors along $\vec{PA}$, $\vec{PB}$ and $\vec{PC}$ ...


3

Let the triangle be $\triangle ABC$. Let $\angle A=60^{\circ}$. Let the incircle touch the triangle on the points $A_1,B_1,C_1$ (opposite to $A,B,C$, respectively). Let $A_1B=a,\,A_1C=b$. Then $C_1B=a,\, B_1C=b$. Let $AC_1=AB_1=x$. Then by the Law of cosines: $$(a+b)^2=(x+a)^2+(x+b)^2-2(x+a)(x+b)\cos 60^{\circ}$$ Solve this quadratic equation. The area is ...


2

Hint: $\sin(\theta)=\frac{5}{8}$ because


2

Expressing the side lengths a,b & c in term of the radii a',b' & c' of the mutually tangent circles centered on the triangle vertices (which define the Soddy circles) $$a=b'+c'\\b=a'+c'\\c=a'+b'$$give the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$


2

This is not an answer but it is too long for a comment. Interested by pew's answer, I focused on $$I=\int_1^3 \left(1- \frac{ak+b}{\sqrt{k^2+1}}\right)^2 \>dk$$ Using a CAS, $$I=\frac{1}{4} a^2 \left(8+\pi -4 \tan ^{-1}(3)\right)+a \left(b \log (5)-2 \sqrt{2} \left(\sqrt{5}-1\right)\right)+b^2 \left(\tan ^{-1}(3)-\frac{\pi }{4}\right)+2 b ...


2

Geometric Proof: In this proof, $M$ is not assumed to be inside $ABC$. Let $P$ and $Q$ be the points such that $APBC$ and $MQBC$ are parallelograms. Note that $APQM$ is also a parallelogram. Hence, $AP=BC=MQ$, $BP=CA$, $BQ=MC$, and $PQ=MA$. Consider the quadrilateral $APQB$. We have by Ptolemy's Inequality that $$MC\cdot BC+MA\cdot AB=AP\cdot BQ+PQ\cdot ...


1

By what you mean as area of an triangle, that is the area of an triangle in terms of squares of unit length. But balls arranged in equilateral triangle do not pack in square cells; rather, they are more closely related to hexagonal tiling. There are two directions: either arrange triangle numbers in the form of right-angled isosceles triangles, or use ...


1

The symbol for denoting similar triangles is ($\color{blue}{\sim}$) Notice, suppose $\triangle ABC$ & $\triangle PQR$ are similar then in LaTex it is written as $\text{"\triangle ABC \sim \triangle PQR"}$ surrounded in-between by 2 or 4 dollar signs which appears as follows $$\color{blue}{\triangle ABC \sim \triangle PQR}$$


1

No. Consider $x_1-x_2=4$, $y_1-y_2=4$ then $(x_1-x_2)^2+(y_1-y_2)^2=32$ and $|x_1-x_2|+|y_1-y_2|=8$. But $x_1-x_2=6$, $y_1-y_2=1$ then $(x_1-x_2)^2+(y_1-y_2)^2=37$ and $|x_1-x_2|+|y_1-y_2|=7$. So this would give the wrong comparison.


1

These diagrams should make it clear. In each diagram, the brown thick segments are the relevant triangle, with shaded interior. The black segments are altitudes, and the dashed segments are extensions of the sides or the altitudes. The points and line segments are exactly the same in both diagrams: only the interpretation changes, along with the triangle ...


1

Hints: $2e = 10$ $\frac{d}{e} = \frac{5}{12}$ $d^2 + e^2 = c^2$ $\tan a = \frac{d}{e}$ $a + b = 90^\circ$


1

An approach to Q2 ... Divide each angle into sixths, with $A = 6\alpha$, $B = 6 \beta$, $C = 6 \gamma$, so that $\alpha + \beta + \gamma = 30^\circ$. Position $\triangle ABC$ on the coordinate plane with $A$ at the origin and the positive $x$-axis bisecting $\angle A$, as shown. Then, with $r$ the circumradius of $\triangle ABC$, we have $$\begin{align} A ...


1

          @StevenTaschuk's construction, with right triangle $T$ base $1$ and altitude $2$. Centroid of $T$ is the purple dot.


1

For Q1: Let $M$ be the midpoint of $BC$; then $AB\sin\angle BAM = AC\sin\angle CAM$. So if we take a triangle with $\angle A=90^\circ$ and, say, $AB=1$ and $AC=10000$, we'll have $\angle BAM$ much larger than $\angle CAM$, and so the median $AM$ won't lie in the middle third of $\angle A$.


1

The proof is correct. $\angle B = \angle C$ is very vague. He should have written $\angle ABC = \angle ACB$. But we can see that $\angle ACD = \angle ACB = \angle ABC = \angle ABD$.


1

you can use http://triancal.esy.es (online triangle calculator) This problem


1

If we start with the following pythagorean triplet $$ m^2 + n^2 , m^2 - n^2 , 2mn$$ Then your method produces, $$ m^2+n^2=\frac{7}{8} \times(m^2-n^2)+mn$$ $$\frac{m^2}{8}+\frac{15}{8}\times n^2 - mn=0$$ $$m^2-8mn+15n^2=0$$ $$(m-3n)(m-5n)=0$$ So, the method above will work perfectly for above pythagorean triplets with m=3n or m=5n.


1

I have a little more:) $$ S = 4R^2(\sin A + \sin B + \sin C)\sin\frac A2\sin\frac B2\sin\frac C2\\ S = \frac{r^2}{4}\frac{\sin A + \sin B + \sin C}{\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}\\ S = r^2\left(\cot\frac A2+\cot\frac B2 + \cot\frac C2\right)\\ S = r^2\cot\frac A2 \cot\frac B2 \cot\frac C2\\ S = 2p^2 \frac{\sin A\sin B\sin C}{(\sin A+\sin B+\sin ...


1

If $W$ is the "hypotenuse-face" of a right-corner tetrahedron, and $X$, $Y$, $Z$ are the (right-triangular) "leg-faces", then $$W^2 = X^2 + Y^2 + Z^2$$ where, yes, we are squaring areas. (This fact is actually equivalent to Heron's formula for non-obtuse triangles. You can extend it to include obtuse triangles by allowing the tetrahedron to have ...


1

You may refer to the following figure: The three red lines are the altitudes of the triangle. $$\cos A=\frac{\sin B}{2\sin C} \\ \implies \frac{AD}{AB}=\frac{1}{2}\frac{AE/AB}{AE/AC} =\frac{1}{2}\frac{AC}{AB} \\ \implies AD=\frac{AC}{2}$$ Then $BD$ cuts $AC$ in half. As $\triangle ADB \cong\triangle CDB$ $(SAS)$ and $BA=BC$, $\triangle ABC$ is ...



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