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11

This is not the case for every triangle, $1^\circ-1^\circ-178^\circ$ triangle, for example, is one of the counterexamples to this claim. However, if all angles are less then $120^\circ$, then the claim is true. To construct such a point; Take any side $[AB]$, find two intersections of perpendicular bisector and circle with radius $\dfrac{|AB|}{2\sqrt 3}$ ...


6

let radius = $r$, then in triangle ACO, using Pythagoras: $$\begin{align} AO^2 &= AC^2+CO^2\\ \\ r^2 &= 12^2+(r-5)^2\\ \\ 144+25-10r &= 0\\ \\ r &= 16.9\\ \end{align}$$ In triangle ACO, $$\begin{align} \sin\theta &= \dfrac{12}{r}\\ \\ \sin\theta &= \dfrac{12}{16.9}\\ \\ \theta &= 45.2^{\circ} (1dp)\\ \end{align}$$


5

Here is an elegant purely geometric solution. The diagram shows the 18-gon that I used to find the solution, but the solution itself does not need the 18-gon. Let $E$ be the image of $B$ under the reflection that maps $A$ to $D$. Then $\angle EAC = \angle EAD + \angle DAC = \angle ADB + \angle DAC = 40^\circ + 20^\circ = 60^\circ$. Also $\angle DEA = ...


4

Consider a pentagon. If you try to add any more edges to the pentagon a triangle will be formed. Thus for graph having a cycle containing 5 vertices ( all vertices that is ) can have at maximum 5 edges without violating the condition. Now consider bi-partite graphs with a total of 5 vertices , say $x$ in one group and $5-x$ in other. Bipartite graphs can't ...


4

Hint: a triangle free graph on $5$ vertices is bipartite unless it contains a $5$ cycle. But if it has a $5$ cycle and is triangle free it is a cycle. Hence you want to find the bipartite graph with five vertices with the most edges.


4

Please see Euclid, Book I, Proposition 6. A very nice web edition of Elements.


4

It's enough to show that $$|BD|^2 + |AC|^2 \stackrel{\color{red}{?}}{=} |CD|^2 + |AB|^2$$ Rearranging we get $$|AC|^2 - |CD|^2 = |AD|^2 = |AB|^2 - |BD|^2$$ which ends the proof. I hope this helps $\ddot\smile$


3

First, see the following image From the figure, $t=\dfrac{x}{2}$ Since, $7<x<7.3$ $\implies 3.5<t<\dfrac{7.3}{2}$ $\implies \dfrac{2}{7.3}<\dfrac{1}{t}<\dfrac{1}{3.5}$ Also, $t\sin\alpha=3.5$ $\implies \sin\alpha=\dfrac{3.5}{t}\in\left(\dfrac{7}{7.3},1\right)$ $\implies \alpha \in (73.52^\circ,90^\circ)$ ...


3

the fact that $DI$ bisects $\angle BDC$ gives ${CD \over DB} = {CI \over BI}.$ same way,${CD \over AD} = {CH \over AH}.$ these two and $AD = BD$ shows ${CI \over BI} = {CH \over AH}.$ now you can conclude that $HI$ is parallel to $AB.$


3

The problem is ill-posed. Notice that $\Delta EBC \sim \Delta AHC$ and $\Delta DCB \sim \Delta AHB$. Thus, $\frac{2}{BC} = \frac{1}{HC}$, and $\frac{3}{BC} = \frac{1}{BH}$. Hence, $HC + BH = \frac{BC}{2}+ \frac{BC}{3} = \frac{5}{6}BC \neq BC.$ (???)


3

All the angles are $120^\circ$ as they add to a full circle. Yes there is such a point as long as the triangle angles are less than $120^\circ$. Imagine having a Y shaped set of sticks at $120^\circ$ angles. If you put one arm through $A$ with the vertex very close to $A$ and a second arm through $B$, the third arm will almost extend $AB$. Put the third ...


3

One neat example is that when considering triangles that lie on a sphere, similar triangles are congruent. That is, it is enough to know the angles of two triangles are the same to determine they are congruent because there is an absolute reference of magnitude, the radius of the sphere you consider.


3

We also know that $a=k(u^2-v^2)$, $b=2kuv$, $c=k(u^2+v^2)$ for some integers $k,u,v$, so $$ 2k^2(u^2-v^2)uv=2009k\cdot(u^2-v^2+2uv+u^2+v^2)$$ i.e. $$ k(u-v)v = 2009.$$ Now match the factors on the left with factors of $2009$.


3

(2) is not unnecessary because if you don't have (2) then you can have that (3) is immpossible (if you have A > $90^\circ$ then a will be largest side) and you have to have (4) because if you don't you can have A = $90^\circ$ witch by (2) is immposible.


2

HINT: $$16\triangle=(b+c+a)(b+c-a)$$ $$\iff16rs=2s(b+c-a)$$ $$8r=b+c-a$$ Using this and $a=2R\sin A$ etc., $$8\cdot4R\prod\sin\dfrac A2=2R\cdot4\cos\dfrac A2\sin\dfrac B2\sin\dfrac C2$$ $$\implies4\sin\dfrac A2=\cos\dfrac A2$$ as $0<B,C<\pi,\sin\dfrac B2\sin\dfrac C2\ne0$


2

$$b^2+c^2-a^2=16\Delta - 2bc \Rightarrow \frac{b^2+c^2-a^2}{2bc}=\frac{8\Delta}{bc}-1 \Rightarrow \cos A=4\sin A-1$$ $$\Rightarrow 2\cos^2\frac{A}{2}-1=8\sin\frac{A}{2}\cos\frac{A}{2}-1 \Rightarrow \tan\frac{A}{2}=\frac{1}{4}$$ Hence, $$\tan A=\frac{2\cdot \frac{1}{4}}{1-\frac{1}{4^2}}=\boxed{\dfrac{8}{15}}$$


2

If the lengths of the sides of the right triangle are $2a, 2b$ and $2c$ (Here $2a$ corresponds to the side in front of area $A$, $2b$ for area $B$ and $2c$ is for remaining side), then the total area of the white space is $\displaystyle\frac{\pi c^2}{2}-2ab$. Now, Area $A$+ Area $\displaystyle B=\frac{\pi a^2}{2}+ \frac{\pi b^2}{2} -$ (Area of white space) ...


2

Basic Answer: You can't. Here's what you can know Given only the length of two sides of a triangle, the length of the third side is not fixed. Let a and b represent the lengths of the two known sides such that $a \geq b$. Let c represent the length of the unknown side, the length of c must fall within $a - b < c < a + b$ Based on the example ...


2

Extend the first line $AP_1$ through the hypotenuse till it reaches a point $D$ such that $\angle{ABD}$ is a right-angle. If you add every odd-numbered segment ($P_0P_1,\; P_2P_3,\;\ldots$) it is the same as length of $AD$. Similarly, if you add every even-numbered segment ($P_1P_2,\; P_3P_4,\;\ldots$) it is the same as length of $BD$. Therefore, ...


2

$$ F(x)=\int\frac{ax}{b}\ dx=? $$ and $$ \int_0^b\frac{ax}{b}\ dx=F(b)-F(0)=? $$


2

Remember: a line that passes through the origin and a given point $(b,a)$ has a slope $\frac{a}{b}$. Hence, the linear equation that has that slope would be $f(x) = \frac{ax}{b}$. Hence, consider the following: $$\int_0^b f(x)dx = \int_0^b \frac{ax}{b}dx= \bigg[\frac{ax^2}{2b}\bigg]_0^b = \frac{ab^2}{2b}=\frac{1}{2}ab$$


2

I've never considered this before and I don't know what such a point is called. But if such a point $O$ exists inside $\triangle ABC$, then $$\angle AOB = \angle BOC = \angle COA = 120^\circ.$$ By the law of cosines, $|Ox|^2 + |Ox||Oy| + |Oy|^2 = |xy|^2$ for every pair $\{x, y\} \subseteq \{A, B, C\}$. (I'm using the bars to mean length.) I'm not sure ...


2

Construct isoceles $\triangle BDF$ with common angle $20^\circ$ and isoceles $\triangle BDG$ with common angle $30^\circ$. Now $\angle BGF=\angle FGD=\angle DGA=60^\circ$, and $\angle GDF=\angle ADG=10^\circ$. So $\triangle GDF\cong \triangle GDA$ and so $\overline{DF}=\overline{DA}$. Since $\triangle ADC\sim\triangle DFB$, we have the desired result.


2

First calculate some angles: $ \angle DAC = \angle DCA = 20^o $ (isosceles triangle theorem https://en.wikipedia.org/wiki/Isosceles_triangle ) $ \angle CAD = 180^o -20^o -20^o = 140^o $ (triangle is $180^o$) $ \angle CAB = 180^o -30^o -20^o = 130^o $ (triangle is $180^o$) $ \angle BAD = \angle BAC- \angle DAC = 130^o-20^o = 110^o $ and the rest is all ...


2

Drop a perpendicular from $D$ to $M$ on $AC$. By congruence of the resultant $\bigtriangleup AMD$ and $\bigtriangleup MDC\,(*)$ we observe that $|AM| = x/2$ and so $|AD| = (x/2)\sec20^{\circ}$. Now we note that $A\hat{D}B = 20\times2 = 40^{\circ}$ since the exterior angle of a triangle equals the sum of $2$ opposite interior angles. So by the angle sum of ...


2

It is not true that this is $5$. Connect one point to three points, and connect the remaining point to the same three points. To proof this optimal you could start distinguishing cases by maximum degree of a vertex. There might be a more elegant argument yet I would not know it momentarily.


2

Hint: compute the area of $PQR$ two ways, one as half a rectangle and one as $\frac 12bh$


2

Here's a hint. The thickness of one layer of the transmitting medium is $3.5$ mm. Call this $t$. The separation of two adjacent reflected rays is the same as half of the path length inside one layer of the material, which is $d = t/\sin \alpha.$ Can you take it from there?


2

Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and: $$e^{\pi i/6}(a+2i) = X $$ so: $$\Im \left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\cdot\left(a+2i\right)\right]=3, $$ or: $$ 1+\frac{\sqrt{3}}{2}a = 3 $$ so $a = \frac{4}{\sqrt{3}}$, and by the Pythagorean theorem: $$ ZY^2 = ...


2

So, we have $$\sin^22A+\sin^22B=\sin^22C$$ $$\iff\sin^22A=\sin^22C-\sin^22B=\sin(2C+2B)\sin(2C-2B)$$ $$2C+2B=2\pi-2A\implies\sin(2C+2B)=-\sin2A$$ $$\implies\sin2A[-\sin(2C+2B)]=-\sin2A\sin(2C-2B)$$ $$\implies\sin2A[\sin(2C+2B)-\sin(2C-2B)]=0$$ $$\implies\sin2A[2\sin2B\cos2C]=0$$ But $\sin2A\sin2B\ne0$ So, we shall get the value of $C$ directly. Then ...



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