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7

The Pythagorean theorem applies only to triangles with a right angle.


5

In this diagram, the large triangle has sides of $2$ while the four smaller ones have sides of $1$. So place five pandas into those four small triangles. The maximum distance between any two points in any small triangle is clearly just $1$. Use the pigeonhole principle and finish from here.


4

Update (Full solution) You may assume that one vertex of the given lattice triangle $T$ is $(0,0)$. Apply the linear transformation $A$ with matrix $$[A]=\left[\matrix{1&-1\cr 1&1\cr}\right]$$ to $T$. This amounts to a counterclockwise rotation by $45^\circ$ and a linear scaling by $\sqrt{2}$. The resulting triangle $T'$ is again a lattice ...


4

By using the Ravi transformation let $a=x+y$, $b=y+z$ and $c=z+x$, then \begin{align} \frac a{c+a-b}+\frac b{a+b-c}+\frac c{b+c-a}&=\frac{x+y}{2x}+\frac{y+z}{2y}+\frac{z+x}{2z}\\ &=\frac{3}{2}+\frac{1}{2}\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right) \end{align} Now use AM-GM inequality.


3

Hint: Divide the triangle into 4 triangles.


3

When something violates an estabilished theorem, the reason is always that it doesn't respect its hypothesis. So, what does Pythagoras' Theorem say? From Wikipedia In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the ...


3

All answers are ok, but on a sphere of radius $\frac{2}{\pi}$, you can have an equilateral triangle of sides $1$ with 3 right angles. But the Pythagorean theorem only applies in euclidean geometry, not in spherical geometry. So you may violate one or another hypothesis (this answer is just to illustrate that "a right angle" is not the only hypothesis of ...


3

Hint: let $x,y$ be the area of $\triangle ADO, \triangle AEO$ respectively. Then: $O$ is the midpoint of $BE$ (why).So: $x+3 = y$.And $\dfrac{x}{y+7} = \dfrac{3}{7}$. Can you continue? Once you solve this system of equations , you know what $x,y$ are and the area of the quadrilateral is $x+y$.


3

The angle aren't multiples of 30 degree. It can be any value you like. If $D$ is moved upwards then $A$ and $C$ contract inwards (and move upwards as well) and vice versa. Some examples:


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If you prefer a purely trigonometric approach: Let the equilateral triangle be ABC with centre O and side $2$ units. The vertex of the pyramid is V. Let N be the foot of the perpendicular from A to line VB, so that $\angle ANC=70$ as given and the triangle ACN is isosceles. Then $$AN=\frac{1}{\sin 35}\Rightarrow\angle ABN=\arcsin\left(\frac{1}{2\sin ...


2

If the vertex of the pyramid is at the center of a unit sphere, then the sides of the pyramid would intersect the sphere in an equilateral triangle with angles of $\newcommand{\degree}{{\unicode{x00B0}}}70\degree$. The Spherical Law of Cosines says $$ \cos(A)=-\cos(B)\cos(C)+\sin(B)\sin(C)\cos(\alpha)\tag{1} $$ Plugging in $A=B=C=70\degree$, we get the ...


2

The absolute size of the pyramid is irrelevant. Therefore we may assume that the vertices of the base triangle are $${\bf a}_0=(1,0,0),\quad{\bf a}_1=\left(-{1\over2},{\sqrt{3}\over2}\right),\quad {\bf a}_2=\left(-{1\over2},-{\sqrt{3}\over2}\right)\ ,$$ and the top vertex is at ${\bf s}=(0,0,h)$ with an unknown $h>0$. The outward normal of the triangle ...


2

Call the angles of the three outer triangles $a, b, g$; $c, d , h$; and $e, f, i$, respectively. Then $$a + b + c + d + e + f = (a + b + g) + (c + d + h) + (e + f + i) - (g + h + i) = 3 \times 180^{\circ} - 180^{\circ} = 360^{\circ},$$ where we've used the fact that $g, h, i$ are also the angles of the inner triangle.


2

You already noted $A = \frac{ab}{2} $ and $a^2+b^2 = c^2$. Now note $ab = 2A$ and $$(a+b)^2 = a^2+2ab+b^2 = c^2+4A$$ $$\Rightarrow a+b = \sqrt{c^2+4A}$$ Can you go from here?


2

Hint: find a suitable rotation around $C$.


2

Hint: If $M_C$ is the midpoint of $AB$, then $A M_CS$ and $M_CBS$ have equal bases and share the same height. Can you proceed for the first part? For the second part, a similar approach, except consider the triangles $CM_CB$ and $CM_CA$. Are these areas equal? Express each of the areas in terms of $X, Y$ and $Z$. Can you finish up for the second part?


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I will follow Lucian's hint and show that $O$ is the orthocentre of triangle $CPQ$. It is enough to show that $PO$ is perpendicular to $CB$ and that $QO$ is perpendicular to $CA$. We will prove the first of these statements, and the other will follow by symmetry. Since $OB = OA$, the arcs $OA$ and $OB$ are opposite. Hence by the inscribed angle theorem, ...


2

I understand that a point $P$ is chosen at random inside a triangle $ABC$ according to a uniform probability distribution, and you want the expected value of the sum of the distances from $P$ to the sides of the triangle. The distance from $P = (x,y)$ to one of the sides is a linear function $ax + by + c$ of the coordinates $x, y$. Thus the sum of the ...


2

I thought the universal family is the space of triangles before equivalence. No, the universal family is supposed to be a family (over the moduli space) which contains every isomorphism (i.e. similarity) class exactly once. More precisely, if $M$ is the moduli space and $U$ is the universal family, there is a map $$ \pi: U \rightarrow M$$ such that ...


2

You have $AC=CD=a$, because $CO\perp AM$. Also $\angle HCD=90-\beta$ $\frac{CH}{CM}=\cos(90-\beta+\beta-\alpha)=\cos (90-\alpha)=\sin\alpha \quad\quad(1)$. Also $\frac{CH}{a}=\sin\beta\quad\quad (2)$ From $(1)$ and $(2)\Rightarrow CM=\frac{CH}{\sin\alpha}=\frac{a\sin\beta}{\sin\alpha}=\frac{a^2}{b}$, because from the $\sin$-theorem we have ...


2

Your error comes from the fact that the area of the triangle is $$|\triangle ABC| = rs = \frac{1}{2}bh,$$ so there is a missing factor of 2. We know $|\triangle ABC| = 27 \sqrt{15}$, $r = \sqrt{15}$, $s = 27$, $b = 24$, so $h = \frac{9}{4}\sqrt{15}$. Consequently, the ratio of similitude of the larger to the smaller triangle is $$\frac{h}{h-r} = ...


2

You may write $$ x=x-y+y $$ giving, by the triangle inequality, $$ |x|\leq |x-y|+|y| $$ or $$ |x|-|y|\leq|x-y| $$ then do the same starting this time with $$y=y-x+x.$$


1

This is one approach. I don't pretend it's the nicest. Let $u = \tan\alpha$. Then $$ \tan2\alpha = \frac{2u}{1-u^2} $$ $$ \tan3\alpha = \frac{u(3-u^2)}{1-3u^2} $$ Since $\tan3\alpha = 2\tan2\alpha$, we have $$ \frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2} $$ $$ (1-u^2)(3-u^2) = 4(1-3u^2) $$ $$ 3-4u^2+u^4 = 4-12u^2 $$ $$ 1-u^4 = 8u^2 $$ Now let $y = BE$. ...


1

(1) The brown dotted line is the extension of $BE$. (2) The green dotted line ($AC’$) is the angle bisector of $∠BAE$ such that $\beta_1 = \beta_2 = \beta$. Another obvious fact is $\beta ‘ = 2\beta$. (3) The blue dotted line is the perpendicular bisector of $AB$ cutting $AB$ and $AC$ at $P$ and $Q$ respectively. By intercept theorem, $AQ = QC$. By ...


1

This can be solved using the Sine rule. The area of the triangle can be expressed as, $$Area=\frac12 ab\sin C$$ where $a,b$ are the lengths of the sides opposite vertices $A,B$ respectively. Here, you have, $$Area=8.4$$ $$a=5.2$$ $$C=29°$$ Using these, you can find $b$, which is the length of the side $AC$.


1

The questioner is right to suspect that it must be quite easy: Length $OA$ = length $OB$. Triangle $AOB$ is isosceles. Now use the angle you have. So it is "hence"


1

You don't need to use the answer to part C. The vector from $A$ to $B$ is $$\mathbf{u}=\langle6-2,4-4,2-6\rangle=\langle4,0,-4\rangle$$ and the vector from $A$ to $O$ is $$\mathbf{v}=\langle-2,-4,-6\rangle$$ The angle between these is $$\theta=\arccos \left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)=\arccos ...


1

You can always parameterize a triangle with $u,v$, where $u,v>0$ and $u+v<1$. This always leads to an integral $$\int_0^1 \int_0^u\ldots dv\,du$$ Convex interpolation inside the triangle with vertices A,B,C is then $$T=Au+Bv+C(1-u-v)$$ Imagine that "u" is the variable that tells you how much you move from C to A, and "v" the same from C to B. ...


1

Yes, it can be done using two double integrals. Here you are dividing the triangle into two parts by the vertical line $x=2$. The equation of the line joining $(1,2)$ and $(2,1)$ is $x+y=3$, and that of the line joining $(1,2)$ to $(3,3)$ is $x-2y+3=0$. You can set $x=1$ to $x=2$, and then you will have the range of $y$ as $y=3-x$ to $y=\frac{x+3}{2}$. ...


1

It is straightforward. $\angle A+\angle B+\angle C=\pi$ But given that $CM=AM=BM$, hence creating two isocele triangles $AMC$ and $BMC$ you have $\angle A+\angle B=\angle C$ Hence the same as in your book $2\angle C=\pi$...



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