# Tag Info

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For a plane defined by $ax + by + cz = d$ the normal (ie the direction which is perpendicular to the plane) is said to be $(a, b, c)$ (see Wikipedia for details). Note that this is a direction, so we can normalise it $\frac{(1,1,2)}{\sqrt{1 + 1 + 4}} = \frac{(3,3,6)}{\sqrt{9 + 9 + 36}}$, which means these two planes are parallel and we can write the normal ...

4

Huge Hint: If $y=ax+b$ is a linear function, then $y=-\frac{x}{a}+c$ is perpendicular. Where $c$ can be any real number

3

Hint #1: Write $3x + 2y = 7$ in the slope-intercept form $y = mx + b$, where $m$ is the slope of the line. Hint #2: If two lines are perpendicular, then their slopes $m_1$ and $m_2$ satisfy ${m_1}{m_2} = -1$.

3

Let the tangent of circle $O$ at $AB$ be $M$, also let $O$'s tangent at $BD$ be $X$. Let the tangent of circle $O_1$ at $AC$ be $N$, also let $O_1$'s tangent at $DC$ be $Y$. Then obviously $IX=IM$ and $IY=IN$ as triangles $IXB$ and $IMB$ are congruent and $IY$ part similarly. Connect $AI$, since $I$ is the incenter of $ABC$, $AI$ is also an angle ...

3

$$\mathbf Z[\mathrm i]/p\mathbf Z[\mathrm i]\simeq \mathbf Z[x]/(p,x^2+1)\simeq\mathbf Z/p\mathbf Z[x]/(x^2+1) .$$ Thus $p$ is a prime in $\mathbf Z[\mathrm i]$ if and only if $-1$ is not a square modulo $p$. By the 1st supplementary law of quadratic reciprocity, this happens for an odd prime if $p\equiv 3\mod 4$, which is the case when $p$ is not the sum of ...

3

Draw a first right triangle with legs $a$ and $c$, then a second having as a first leg the hypotenuse of the first triangle, and a second leg of length $d$. Finally, draw a third triangle having the same hypotenuse as the second one and a leg equal to $b$. The length of the other leg is your answer.

3

In short, yes, the lines and planes of physical space all appear to have at least some "thickness" to them (we can see them), and they all have finite extent (in our finite universe). The mathematical concept of lines, planes, etc. differ in that they are logical ideals, created to help us reason about geometric relationships. We usually call such concepts ...

3

I take for granted that $x^5-x+1=0$ has Galois group $S_5$ and hence is not solvable by radicals. It can be translated into a geometric problem: Given $A,B$, find all remaining point so that you obtain a figure where triangles $ABC$, $ACD$, $ADE$, $AEF$, $AFG$ are similar $ABCG$ is a parallelogram Note that identifying $\vec{AB}$ with $1\in\Bbb C$ and ...

2

Since $AB || CD$ we have $<BEC =<EBA$ (wherever $E$ is on $CD$). If we also have this common angle equal to $<AEB$ then $\Delta AEB$ is isosceles, with $AE=AB$. therefore: Draw the circle centered at $A$ with radius $AB$ and intersect it with the line $CD$.

2

Given four arbitrary points in the plane, there may or may not be a square such that each of its four (unextended) sides contains one of the four given points. That is why the problem says, "If possible, find a square", rather than simply, "Find a square". How can you know whether the square exists? Perform the construction given as a solution to the ...

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Hint:- Try to solve it from the converse.Imagine a rectangle $ABCD$.Now,draw diagonal $BD$.Now,separate out triangle $ABD$.Taking BD as the diameter draw a circle which will pass through $A$ (why?).Now,if you take any arbitrary point $C$ on arc $BCD$,$\angle BCD=90^0$(why?). Now,you can understand how to relate $90^0$ with the problem.Do the same by ...

2

Parallel planes are level sets of a linear function. In this case, $x+y+2z=c$. The signed distance of $x+y+2z=c$ to the origin is the normalized algebraic value $$\frac{c}{\sqrt{1^2+1^2+2^2}}=\frac{c}{\sqrt{6}}$$ Therefore, the unsigned distance between two planes $x+y+2z=c_1$ and $x+y+2z=c_2$ is $$\frac{|c_1-c_2|}{\sqrt{6}}$$ In your example, $c_1=4$ ...

2

Let $X$ be the intersection of $AM$ and $DH$. The circumcircle $\Gamma_A$ of $B_1 C_1 H$ goes through $A$ since $\widehat{HC_1 A}=\widehat{HB_1 A}$. Its center is the midpoint of $AH$. In a similar way, the circumcircle $\Gamma_{A'}$ of $BCH$ goes through $A'$, i.e. the symmetric of $A$ with respect to $M$. Since $BCB_1 C_1$ is a cyclic quadrilateral, $D$ ...

2

As You have mentioned, by cosine theorem, $AB$ is known. From that, with the given, $\angle A$ can then be found by sine theorem. In $\triangle AOC$, since $AO, OC$ and $\angle A$ are known, apply cosine theorem again to find $AC$. Result then follows.

2

Take the cylinder $x^2+y^2=a^2$ (with $z$ arbitrary), and make a plane passing through the $x$ axis which tilts upward in such a way that the distance from the origin to where the plane cuts the vertical planes $y=\pm a$ is the $b$ of your major axis $2b.$ Now take two spheres of radius $a$ which are to be bounded by the cylinder, one below and one above ...

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This is OEIS A007678, the second sequence when I plugged in your values. It goes $1, 4, 11, 24, 50, 80, 154, 220, 375, 444, 781, 952, 1456, 1696, 2500, 2466, 4029, 4500, 6175, 6820, 9086, 9024, 12926, 13988, 17875, 19180, 24129, 21480, 31900, 33856, 41416, 43792, 52921, 52956, 66675, 69996, 82954, 86800, 102050\dots$ An article (referenced in the OEIS ...

2

We want to prove: If $p$ is not a sum of two squares in $\mathbb Z$, then $p$ is prime in $\mathbb{Z}[i]$. Let's prove the contrapositive: If $p$ is not prime in $\mathbb{Z}[i]$, then $p$ is a sum of two squares in $\mathbb Z$. Indeed, if $p$ is not prime in $\mathbb{Z}[i]$, then we can write $p=\alpha \beta$, with $\alpha, \beta \in ... 1 $$a+b+c = 1000$$ $$a^2 + b^2 + c^2 +2ab +2bc + 2ac = 1000000$$ $$2c^2 +2ab + 2(a+b)c = 1000000$$ $$c^2 + ab + (1000-c)c = 500000$$ $$ab + 1000c = 500000$$ $$c = 500 - \frac{ab}{1000}$$ So $$c < 500$$ is given. Now, can we do better? Well we know from (1) that$ab < ac < c^2 < 250000$so we also know$c>250$, so we're pretty close to the ... 1 Lemma: In triangle$ABC$with$D$on$BC$we have$\frac{BD}{CD}=\frac{AB}{AC}\times\frac{\sin(BAD)}{\sin(CAD)}$Proof: We write Sin Theorem in triangles$ABD$&$ACD$then we have: $$\frac{\sin(BAD)}{BD}=\frac{\sin(ADB)}{AB}\quad \frac{\sin(CAD)}{CD}=\frac{\sin(ADC)}{AC}$$ since $$\sin(ADC)=\sin(\pi-ADC)=\sin(ADB)$$ we have: $$\frac{\sin(BAD)\times ... 1 To repeat the contents of an earlier comment: there is a very tight relation between these points. Indeed, the distance between the orthocenter and the centroid is always twice that between the centroid and the circumcenter. This is not obvious, but neither is the argument terribly difficult. A good reference for it can be found here. 1 For every three points on a line (not necessarily different), does there exist a triangle such that the three points are the orthocenter, circumcenter and centroid ? No, because the three are not independent1 of one another ! You already know that the centroid splits all three medians in a ratio of 1:2, right ? Well, it does the same with the segment ... 1 To my knowledge, the first postulate is that given any two points, there is a line which has them as endpoints; this has little to do with curves, so how could the definition violate that? Actually, a definition in itself cannot violate any statement at all. Even Definition. A pair of two points is called a Jabberwocky if there does not exist a line ... 1 In my solution I will call the tangents of the circles, Q the tangent of the incircle of \Delta ABD to AB and M to BC, N the tangent of incircle of \Delta ADC to BC and S to AC. Fistly notice that AG is the radical axis from the two circles, so it is obvious that AQ^2=AG^2=AS^2, hence AQ=AG=AS. Let now call z=BM=BQ because they ... 1 This can happen if we don't specify whether X is inside or outside interval \overline{BC}. If we take two points X_1,X_2 such that X_2 is between B,C and B is between X_1,C such that X_1B=BC,X_2B=BC/3 then we have X_1C/X_1B=2=X_2C/X_2B. 1 This can be thought of as a reflection problem. That is to say, reflect the point A through the line x=2 (obtaining the point \hat {A}=(0,3)) and then draw the straight line segment connecting \hat {A} to C. That segment crosses the line x=2 at the desired point M. 1 I'll be using the usual standard notation for the elements of \triangle ABC. To solve this problem we need only to find the ratios {QB\over QC},{PA\over PC} and {RA\over RB} in terms of the elements. We simply apply the Angle Bisector Theorem in \triangle ABC to get our first ratio:$${QB\over QC}={c\over b}\tag{i}$$Now just apply simple ... 1 There exist orthonormal basis$(e_1=u,e_2,...,e_n)$and$(f_1=v,...,f_n)$define$A$by$A(e_i)=f_i$. 1 This is actually a well known theorem called the radical axis theorem. Define the radical axis of two circles as the locus of points which has equal power with respect to both circles. ( Show that this locus is a line. To do this show this, take a point$P$which has equal power with respect to both the circles and draw the tangents to both the circles ... 1 You can prove that using the following : Let$O_1$be the circle on which$A,F,D,C$exist. Also, let$O_2$be the circle on which$C,F,B$. Let$X$be the intersection point between$AD$and$CF$. And let$E'(\not =B)$be the intersection point between$O_2$and$BX$. Then, four points$E',A,B,D$are on a circle. This is because we have ... 1 Ignoring the case of parallel (or coincident) lines, suppose$\ell_1$and$\ell_2$meet at the unique point$O$. Any circle about$O$meets the lines at the vertices of a rectangle$\square ABCD$; it also meets the bisectors of the angles formed by those lines at the vertices of a square$\square WXYZ\$ (because the bisectors are necessarily perpendicular). ...

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