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Since $\triangle CBA = \triangle CDB + \triangle CAD$, we have $$\frac{1}{2} a \cdot CD \cdot \sin 45 + \frac{1}{2} b \cdot CD \cdot \sin 45 = \frac{1}{2} ab$$ This gives $$CD=\frac{\sqrt{2}ab}{a+b}$$

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Let's call the center of one fixed circle $O_1$ and the other $O_2$. Say the radius of the variable circle is $r$. Because the variable circle maintains tangency with the fixed circles we have the following equations: $$\overline{CO_1}=a+r$$ $$\overline{CO_2}=b-r$$ And adding the two we have:$$\overline{CO_1}+\overline{CO_2}=a+b$$ Hence we conclude that ...

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Using the Evan Chen's mixtilinear incircle article in here, the results become trivial. I will change some notations. In $\triangle ABC$, let the incircle hit $BC$ at $D$ and the $A$-mixtilinear incircle hit the circumcircle of $\triangle ABC$ at $E$. Prove that $\angle DEB = \angle ABC$. Since (9) holds, we have $\angle DTM_A = \angle AFB=180-\angle B - ... 0 Let$v(x,y)=e^{-(x^4+y^2)}.$Then$P_t$is the curve defined by the equation $$x^4+y^2=-\ln t$$ The eccentricity is determined by the intersections of that curve with the coordinate axes. It intersects the$X$-axis at$|x|=\sqrt[4]{(-\ln t)}$and the$Y$-axis at$|y|=\sqrt{(-\ln t)}.$The eccentricity is thus$(-\ln t)^{1/4}$which becomes arbitrarily ... 0 Given$u^TBv = u^TCv\forall u,v\in U\subset\mathbb{R}^3$Consider$A=(B-C)$, let$v\in U$, then $$\|Av\|_2^2 = (Av,Av)=v^tA^TAv=0$$ we got that$\|Av\|_2=0,\forall v\in U$, therefore$A=0$and hence$B=C$. This could only be true for an open subset (of$\mathbb{R}^3$in this case), because otherwise$U$could be a discrete set, e.g. subset of the ... 0 Distance from$(x,y)$to$(4,4)$is$|\sqrt{(x-4)^2+(y-4)^2}| = 1$Distance from$(x,y)$to$(4,1)$is$|\sqrt{(x-4)^2+(y-1)^2}| = |\sqrt{10}|$Square both sides of both these equations and subtract one from the other to get$(y-1)^2-(y-4)^2=10-1=9$. So you get, $$y^2+1-2y-(y^2+16-8y)=9\Longrightarrow 6y-15=9\Longrightarrow y=4$$ If you put$y=4$an any ... 1 The trick is the following. Let$ON \cap AB = P'$. It suffices to show that$P'$lie on the circumcircle of$BOM$. After that, we will have$P'=P$, and then we will have$P$lie on$ON$. Let$\angle BAM = a$. We begin angle chasing. We have $$\angle BP'O + \angle BMO = (a+ \angle ANO) + (\angle BMA - \angle OMA) = a+90+(180- \angle B - a) -(90 - \angle ... 0 HINTS: Use standard polar coordinates ellipse equation:$$ p/r = 1 - \epsilon \cdot \cos \theta $$using scale factor \frac12 conveniently and then convert it to Cartesian:$$ p/2 = \sqrt{x^2+y^2} - \epsilon \cdot x $$Next, shift the origin to center (0,0), Done. 1 Here's part of a solution based on a comment from @Blue. Here's the picture showing a few additional points and lines. Angle equivalence As per the suggestion from @Blue, we first reflect the point F about the perpendicular bisector of \overline{GH}. Then we draw a segment \overline{F'D}. Since \overline{F'G} = \overline{FH}, the law of sines ... 1 You already know that smaller base is \frac{a}{3}, since this is a third of the larger base's length. Now set up an equation using the area of a trapezoid. \frac{\frac{4a}{3}a\sqrt{6}}{2}=a^2\sqrt{3} Note that the a^2 will cancel out. \frac{2}{3}\sqrt{6}=\sqrt{3} This is a false statement, so such a trapezoid does not exist. 0 Hint: A=\frac{a+b}2\cdot h=2bh. 3 No. For example a straight line (a Bezier curve) might be conformally mapped to a circle (a non-Bezier curve). However, Bezier curves are invariant under affine linear transformations. 0 Let b_a and b_b be the respective bases of the heights h_a and h_b The area of the triangle is given by the formula$$A=\frac{1}{2}*(base)*(height)$$Therefore,$$A=\frac{1}{2}b_ah_a=\frac{1}{2}b_bh_b\frac{b_a}{b_b}=\frac{h_b}{h_a}\frac{b_a}{b_b}=\frac{9}{8}\text{as$h_a=8m$and$h_b=9m$}P=30=b_c+b_a+b_bb_c=13m$$Heron's formula ... 2 You know it's a circle with center O, so you just need a point to find the radius. You can take a tangent the the first circle: y=1 and a perpendicular tangent to the other circle: x=\sqrt{7} , the intersection (1,\sqrt{7}) is on the circle: the radius is$$\sqrt{1^2+\sqrt{7}^2}=\sqrt{8}=2\sqrt{2}$$2 Let AE and BF be perpendicular to AB, with E and F lying on CD. Let N be the reflection of P, across the midpoint of EF. It follows that CN=CE+EN=EP+PF=NF+FD=ND, so N is the midpoint of CD. It follows that ABFE is a rectangle, hence it is similar to BAEF. Applying power of point M to both circles, it can be seen that ... -1 There are infinite solutions. Please see the GeoGebra.org figure that I constructed to demonstrate my answer here. (the following text is repeated there, so no need to read it twice) Here are the steps I took to construct this figure, under the assumption that the given figure in the original post is not to scale: I drew arbitrary points A1 and B1, ... 0 try this approach to solve the problem. Let me know if you were successful. 2 Notations: Write a:=GH, b:=HF, c:=FG, and s:=\frac{a+b+c}{2}. Let \Omega and \omega be the circumcircle and the incircle of FGH. The circle internally tangent to FG, FH, and \Omega is denoted by \Gamma, centered at X. Denote by \omega_a the excircle opposite to F of FGH, which touches GH at T. Extend FT to meet ... 0 Truth of matter is that we're not actually told what to do, nor is there a method to be sure of what to do -- at all. I'm sure that even the publisher would state that "doing this and that" will hopefully give us the correct solution to whatever p^2 q^2 is. A.k.a, telepathy dependent questions. The equation determines the foci i presume - maybe even the ... 3 Euclid certainly did not use real numbers. He showed that the diagonal and side of a square have no "common measure", i.e. there is no segment that can be laid end-to-end an integer number of times to get the length of the side and also to get the length of the diagonal, and he knew how to define what it means to say the ratio of lengths of segment A to ... 11 I'll be repeating some stuff that has already been said, but I have my own spin on it. Can we deduce, from some axiomatic geometries, an algebraic structure? Yes. As you can find in Hartshorne's book, or in Hilbert's Foundations, the idea is an "algebra of segments" for any ordered Desarguesian plane in which you construct an Archimedian, ordered ... 20 Hilbert's Foundations of Geometry did more or less precisely what you are asking for. Starting from an extension of Euclid's Axioms, Hilbert proves that any model of the axioms is isomorphic to \mathbb{R}^2 with the usual definition of line. Later, Tarski gave a first-order axiomatization of plane geometry. Because of built in restrictions in the ... 11 This question is a little broad, and so this certainly doesn't answer all of it, but hopefully you find it interesting - an example of a situation where a geometric construction is equivalent to an algebraic property. Given any division ring R (that is, a set with two binary operations, \oplus and \otimes, such that \oplus and \otimes are ... 2 The pictures I made with Cinderella Geometry show that the problem is stated correctly. The answer is that indeed x=y. (Of course, a proof is needed.) A different triangle, angles again are equal. 0 In vector terms consider three vectors 0,a,b. Let c=a-b. The median-vectors are \{m_1,m_2,m_3\}=\{b-a/2,a-b/2,(a+b)/2\}. We have$$\sum_{i=1}^{i=3}\|m_i\|^2==\frac {3}{2}(\|a\|^2+\|b|^2-a\cdot b)==\frac {3}{4}(2\|a\|^2+2\|b\|^2)-2 a\cdot b)==\frac {3}{4}(\|a\|^2+\|b\|^2+\|a\|^2+\|b\|^2-2a\cdot b)==\frac ... 0 Let us say that the square and the circle are centered around the origin in a coordinate plane. It then follows that since the side length of the square is$5$, the radius of the circle is$\frac{5}{2}$. We can say the circle's equation is: $$x^2 + y^2 = \left(\frac{5}{2}\right)^2$$ $$x^2 + y^2 = \frac{9}{4}$$ We cannot express the square in terms of an ... -2 If you want a circle of center$O(x_O, y_O)$and radius$r$, then you need to remove all points$M(x,y)$such that :$$(x-x_O)^2 + (y-y_O)^2 > r^2$$ 1 Short answer: The question boils down to whether$k$is a zero divisor, modulo$m$. This happens precisely when$k$and$m$are not coprime, so that$\{m/k\}$is indeed connected (it takes$m$full steps of size$k$to get back to where we started) whenever$m$and$k$are coprime. But if you're not sure about some of this, keep reading. Let's look at star ... 0 See emilios figure so i continue from there sorry dunno how to add pictures.$|DM|+|MC|=|DC|$.(1).(triangle law of vectors. And$|AM|+|MB|=|AB|$..(2) but$|AB|=|DC|$. .. opposite sides of parallelogram are equal thus$|DM|+|MC|=|AM|+|MB|$.. from 1,2 but$|AM|=|MC|$M is the midppint thus$|DM|=|MB|$which implies M is also the midpoint of diagonal BD. 2 It is impossible to make five 2D shapes (on a flat plane) which all touch each other. This is a consequence of the four color map theorem. 5 This is NOT an answer but is an as-accurate-as-possible re-sketch of the original figure after guessing. Please let me know if there is any mis-interpretation. [Note: The previous diagrams have been incorrectly drawn and were therefore deleted. The one below is the most updated version. Sorry for giving some misleading info.] This time Geogebra shows ... 2 Hint (from the figure) $$DM=MB \qquad and \qquad CM=AM$$ 0 You can find the slope of line$ON$from point (c) and the slope of line$PQ$: since those two lines are perpendicular, the product of their slopes must be$-1$. From that you get$p^2q^2=1$. 1 Suppose your 3D points and the projections are given in homogeneous coordinates$X_1, X_2, X_3,..., X_N \in \mathbb{R}^4$and$x_1, x_2, x_3,..., x_N \in \mathbb{R}^3$. Then they have the simple relationship $$x_i=PX_i$$ with the camera matrix $$P= \begin{pmatrix} p_{11} & p_{12} & p_{13} & p_{14}\\ p_{21} & p_{22} & p_{23} ... 1 Use coordinates. Let the y axis be the perpendicular from O to l, so that O has a positive y. Let the x axis be l. Let P(a,0), O(0,h), Q(0,t), and the radius of the circle be r. The length of the tangent from P to the circle is \sqrt{PO^2-r^2}=\sqrt{a^2+h^2-r^2}. The length of PQ is \sqrt{a^2+t^2}. For the two values to be ... 3 I was staring at this \space\downarrow for nearly a week, and couldn't come up with anything because I was searching for similar triangles. Then I drank a few cups of coffee and remembered something: 0 If you take the sum of i, f, g, it will be normal to the plane you want by symmetry. So the normal is parallel to (2, 2 \sqrt{3}/3, \sqrt{2/3}). Then your equation for the plane is$$2x + 2\sqrt{3}y/3 + \sqrt{2}z/\sqrt{3} - 2= 0,$$where we used point-slope form with the point as (1,0,0). I'm not sure what your answer key says, but this equation ... 0 The cross-product (B-M)X(O-M) will produce a vector perpendicular to MB and MO. Adjust its length to make \frac{|PM|}{|MO|} =\tan(\alpha) . 2 The sequence 1,6,20,48,90,132,132 can be generated as \displaystyle {5+i \choose i-1}\dfrac{8-i}{7} with i running from 1 to 7, though it would be slightly more conventional to write \displaystyle {n+k \choose k}\dfrac{n-k+1}{n+1} with n=6 and k running from 0 to 6. The right hand part of the first diagram is known as Catalan's ... 1 Imagine connecting your "inlets" together to the left of your city and the "outlets" at the bottom, like this (shown for 4 of each instead of 7 of each): * | +-->: | ::. +-->:::: | :::::. <- dotted area corresponds to your drawing +-->::::::: | ::::::::. +-->:::::::::: v v v v | | | | +--+--+--+--* Then each route ... 1 I don't think this is exactly what you are asking for, but an isometric embedding of a flat unit torus has the property that if you pick a direction at random and start walking, with probability 1 you will never intersect your own path. Any line with an irrational slope will extend forever in length without meeting itself. 9 There's not enough information in that picture to find any horizontal distance. If you have a situation that matches the information there, you can stretch or shrink it horizontally by any (nonzero) factor you'd like, and it would still match the information. 3 One possibility is a flat 3-torus. This is the shape you get by taking a cube (or more generally, a parallelopiped) and declaring that the opposite faces of the cube are actually equal to each other. This means that when you walk far enough in one direction, you "wrap around" back to the other side. In particular, this means you can keep walking ... 0 The center of the circle is (3, b). 3 came from mid point between A and B, and it's also the radius of the circle. And$$b^2=r^2-1^2=3^2-1=8$$So P=(0,2\sqrt2). 1 All that calculation was really unnecessary. Here's a fast way: The area of the quadrilateral PQRS is equal to the sum of the areas of the triangles \triangle RQS and \triangle PQS. And since the area of the triangle \triangle PQS is fixed, we need only to maximise the area of the \triangle RQS Additionally since the base QS of the \triangle RQS ... 2 This answer uses complex numbers. We want to find a,b,\theta\in\mathbb R where 0\lt\theta\lt \pi such that$$(16+9i)-(a+bi)=\sqrt 2(8+2i)(\cos\theta+i\sin\theta)(16+11i)-(a+bi)=\sqrt 2(9+3i)(\cos\theta+i\sin\theta)$$Then, solving$$16-a=\sqrt 2(8\cos\theta-2\sin\theta)9-b=\sqrt 2(8\sin\theta+2\cos\theta)16-a=\sqrt ... 1 We have $$\angle FI_1E = \pi - \angle I_1EF - \angle EFI_1 = \pi - \frac{\angle FEA}2 - \frac{\angle AEF}2 = \pi - \frac{\pi - \angle FAE}{2} = \frac{\pi + \angle BAC}{2}$$ and $$\angle EDF = \pi - \angle FDB - \angle CDE = \pi - \frac{\pi - \angle DBF}{2} - \frac{\pi - \angle ECD}{2} = \frac{\angle CBA}{2} + \frac{\angle ACB} 2 = \frac{\pi - \angle ... 1 Your transformation contains translation (2 parameter), rotation (1 parameter) and stretching, which I hope means scaling (1 parameter). This in general is no linear but an affine transform, except for the case that the origin gets mapped to the origin, which I doubt here. The transform would be like this, using homogeneous coordinates:$$ T = \left( ... 0 This is Desargues' theorem: Corresponding sides of the two triangles are parallel, hence intersect in three points of$\ell_\infty$, the line at infinity. It follows that the three lines connecting two corresponding vertices intersect in a point$P$. The extra assumptions made here guarantee that$P\notin\ell_\infty$. 1 We know that$\triangle{OFA}$and$\triangle{FXA}$are similar, so we have $$AO:AF=AF:AX,$$ i.e.$$AO\times AX=AF^2$$ Suppose that$(1)\$ is true. Then, we have $$AI_1=AF$$ This is impossible.

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