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71

Your question has nothing to do with right triangles. If $c=a+b$ in any triangle, then the vertex $C$ must lie on the segment $AB$, and this is called a "degenerate" triangle.


64

(Addendum: As noted as well by others, this should highlight that your question has little to do with right triangles, as this is in fact a property of triangles in general)


32

From the pythagorean theorem, $a^2+b^2=c^2$. Suppose $a+b=c$, then $$ a^2+b^2=(c)^2\iff a^2+b^2=(a+b)^2 $$ Can you take it from here?


25

Substituting $c = a + b$ into $a^2+b^2=c^2$ gives us $a^2+b^2=(a+b)^2$. Multiplying that out gives us $a^2+b^2=a^2+2ab+b^2$ which means $2ab=0$ So to satisfy both $c = a + b$ and $a^2+b^2=c^2$ either $a=0$ or $b=0$. The question then becomes a bit more philosophical. We have a result that satisfies the Pythagorean equation but can we really consider it a ...


9

There is another way to formally answer this question, namely using the cosine law. For any triangle, it is true that $a^2+b^2-2ab\cos\gamma = c^2$ where $\gamma$ is the angle between the $a$ and $b$ sides. On the other hand, if $c=a+b$ then $c^2=(a+b)^2=a^2+2ab+b^2$, thus $$a^2+2ab+b^2 = a^2+b^2-2ab\cos\gamma,$$ which is possible only if $\cos\gamma=-1$ ...


8

The general rule that relates the sides of a triangle is called the law of cosines. It regards the following situation, where you know an angle of the triangle and want to relate the side lengths: It is stated as follows: $$C^2=A^2+B^2-2AB\cos(\alpha).$$ Note that this has a very similar form to the Pythagorean theorem - just with an extra adjustment for ...


5

Sorry for the short answer. Cosine Rule, whose special case is the Pythagoras theorem. $$ c^2= a^2 + b^2 - 2 a b \cos C$$ when $C = 90^0. $


5

Like Alex M. said, this doesn't have much to do with right triangles, but it's worth knowing the more general fact, and there's an easy visual! Imagine you have two sides of a triangle, like an adjustable protractor. As you increase the angle between them, the other side is getting longer. But it won't equal their sum until you have your protractor fully ...


5

The triangle with angles $\frac \pi \beta$, $\frac \pi \beta$ and $\frac \pi \gamma$ is an isosceles triangle. Tracing the height to the uneven side, two triangles with angles $\frac \pi 2$, $\frac \pi \beta$ and $\frac \pi {2 \gamma}$ are obtained. And the assertion follows.


4

HINT: Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ and $\sin(A+B)=\sin(\pi-C)=?$ and Sine Law


4

By dimensional analysis, we know the result has dimension $\text{length}^{-1}$. This immediately rules out choice $B$ and $C$. If this is an exam, I will probably just write down the result as $A$, move on and revisit the problem when I have time. When one need to really solve this, one can use choice $A$ as an ansatz. One should express everything in terms ...


4

No. If we are talking about a Euclidean plane, anyway. Without even mentioning triangle inequality, it's a straight-forward proof right from the axioms of the Euclidean plane that $a+b=c$ would imply that $C$ lies on $AB$.


4

The first thing is to show that A, M N, C are con-cyclic. Added:- If $\angle B = \theta$, then $\angle BAM = 90^0 - \theta$ and therefore $\angle MAN = \theta$ because of complementary angles. Note that $\angle BCN = \theta$ also because of alternate angles. “$\angle MAN = BCN$” implies $MACN$ is a cyclic quadrilateral [converse of angles in the same ...


3

A simple formula for the area $A$ of a triangle given the lengths of two of its sides $a$ and $b$ and the angle 'between' them $C$ is $$ A = \frac{1}{2} ab \sin C$$ In your situation, you'll have $a=b$ and you know $A$ and $C$.


3

Flip the whole image through the symmetry median of the $20-80-80$ triangle. The outer part forms a large equilateral triangle. Construct a smaller equilateral triangle outside the $80-80$ edge. Find the reflective and rotational symmetry, and check that $\alpha$ is half of $20^\circ$. Edit: changed diagram to name points


3

we have $$h_a=\frac{2A}{a}$$ etc and we get $$\frac{1}{2A}(a\cos(\alpha)+b\cos(\beta)+c\cos(\beta))=\frac{2R}{abc}\left(\frac{a(b^2+c^2-a^2)}{2bc}+\frac{b(a^2+c^2-b^2)}{2ac}+\frac{c(a^2+b^2-c^2)}{2ab}\right)=$$ $$\frac{R\cdot16(a+b-c)(a+b+c)(a+c-b)(b+c-a)}{16abc}=\frac{16R}{(abc)^2}\cdot A^2=\frac{16R}{(abc)^2}\cdot \frac{(abc)^2}{16\cdot R^2}=\frac{1}{R}$$ ...


3

It's the subgroup $\langle xyx,y \rangle$ of $\langle x,y \mid x^2,y^\beta,(xy)^{2\gamma} \rangle$. The index of this subgroup $2$, so checking that the subgroup has the presentation $\langle z,y \mid z^\beta,y^\beta,(xy)^\gamma \rangle$ (with $z=xyx$) is routine.


3

By the law of cosines, $$a^2=b^2+c^2-2bc\cos A=\left(\frac{(m+1)a}{2}\right)^2+c^2-2\cdot\frac{(m+1)a}{2}\cdot c\cdot \frac 12\sqrt{\frac{(m-1)(m+3)}{m}}$$ from which we have $$\frac{(m-1)(m+3)}{4}a^2=-c^2+\frac{(m+1)ac}{2}\sqrt{\frac{(m-1)(m+3)}{m}}$$ Now dividing the both sides by $a^2$ and letting $x=c/a$ give ...


2

Let $\;O\;$ be the center of the circle with the red line and $\;M\;$ the one of the upper circle, with $\;A\;$ the point where these two lines intersect. If the radius of the circles is $\;R\;$, then in the straight triangle $\;AOM\;$ we have $$MA^2+AO^2=OM^2\implies MA+R^2=(2R)^2\implies MA=\sqrt3\,R$$ so $$\frac{MA}{2R}=\frac{\sqrt3}2$$


2

Suppose the circles all have radius $1$ (and hence diameter $2$). Draw a line segment from the top of the blue line to the midpoint of the red line. You have a right triangle (along with the red and blue lines) with an angle of $60^\circ$; thus the blue line has length $\tan(60^\circ)=\sqrt3$.


2

First of all, your transformation does not correspond to any particularly well behaved geometric transform, as it is not invariant to rotation. Component-wise product is not a vector operation. Consider a "product" of a line segment with a point. The point here represents a stretch in both directions by some factor. So edge × point=edge. So consider edge × ...


2

$$\begin{align} |\triangle ABC| &\;=\; |\triangle BDC| + |\triangle ADC| \\[6pt] \implies \qquad \frac{1}{2}\,a b &\;=\; \frac{1}{2}\,a d \sin 30^\circ \;+\; \frac{1}{2}\,b d \sin 60^\circ \\[6pt] \implies \qquad \frac{1}{2}\,a b &\;=\; \frac{1}{2}\,a d\cdot\frac{1}{2} \;+\; \frac{1}{2}\,b d\cdot\frac{\sqrt{3}}{2} \\[8pt] \implies \qquad 2 a b ...


2

Let the equal sides be of length $a$ and the third side be of length $c$. Let the angles be $A,B,C$, where $C$ is the angle opposite side $c$. $$\text{Area}=\frac12a^2\sin C$$ Using this, find $a$. Now, using the cosine rule, $$\cos C = \frac{a^2+a^2-c^2}{2a^2}$$ $$\cos C = 1 - \frac{c^2}{2a^2}$$ Using this, and the value of $a$, find $c$.


2

Hint: A triangle is the convex envelope of its vertices $A,B,C$, i.e. the set of linear combinations $$ \alpha A+\beta B+\gamma C $$ with $\alpha,\beta,\gamma\in[0,1]$ and $\alpha+\beta+\gamma=1$. The boundary is given by the set of the previous linear combinations in which $\alpha,\beta$ or $\gamma$ equals $0$. If we have two triangles, our locus is given ...


2

We have $\frac{MB}{MC}=2=\frac{BA}{CK}$ (because $K$ is the centre of the circle $ABC$).Also $\angle MBA=\angle MCK$ because triangle $KBC$ is isosceles.


2

It is easily seen that the red and the blue $\beta$s in the following figure are equal. On the other hand, since $AM$ is the height of the parallelogram $P:=ABCD$ with respect to the base $BC$, and $AN$ is the height of $P$ with respect to the base $CD$, or $BA$, we have $$|BC|\cdot|AM|={\rm area}(P)=|BA|\cdot|AN|\ ,$$ and therefore ...


1

One of the principles that this question illustrates is that in Euclidean geometry any path from point A to point B is not shorter than a straight line from A to B, i.e. a straight line segment A->B is the shortest path from A to B. Any deviation made through a point C that is not on the segment A->B makes a path longer and the only way that (A->C)+(C->B) ...


1

True. Let $H -$ orthocenter and $O -$ circumcenter, $R -$ radius, $F -$ midpoint $HO$. Then nine-point circle has a center point $F$ and a radius $\frac 12 R$


1

For each of the vertices $V_i$, find the angle $OV_i$ makes with the $x$-axis. Compare them and you should be able to find the vertices that make the largest and the smallest angles respectively. From the vertex that makes the largest angle, counting anti-clockwise the edges of the triangle until the vertex that makes the smallest angle, and these edge(s) ...


1

Since you've included the “cross-product" tag, here's a way to do this with cross products: Sort the vertices of the triangle in order of increasing angle. Slope will do since all of the points are in the first quadrant. Call these points, in order, $P_1$, $P_2$ and $P_3$. Side $P_1P_3$ is the only one visible if it is closer to the observer than $P_2$, ...



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