Two circles (⊙O and ⊙O', the radius of ⊙O > the radius of ⊙O') intersect at Point A and Point B. Line CD, which is their common tangent line at the bottom, is drawn tangent to ⊙O and ⊙O' at Point C and Point D respectively. Extend CB to intersect AD at E, extend DB to intersect AC at F.\n\nPROVE: DA * DE = DB * DF\n\nDifficult question, isn't it? Try your best.
201089
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
159844
x^2 < 4, swap the < for a = for a sec.  That gives you x^2 = 4, and x = +/- 2.  So on a number line, draw an open circle at -2 and 2 (open circle because the original problem was < and not <=).  Now comes the logical part - if you plug in something to the far left of -2, does it satisfy your original equation?  Nope.  How about far right of 2?  Nope again.  What about between -2 and 2?  Yep.  So fill in or darken the line between the open circles.
582133
AB'C'D + AB'CD = AB'D(C'+C) = AB'D\n\n(AB'+D)(AB'+D+CF')\n= AB' + DAB' + D + AB'CF' + DCF'\n= AB'(1+D+CF')+D(1+CF')\n= AB'+D\n\nI'm not quite sure what you mean by the second question, but the dual of an expression is found using de Morgan's law:\n\nA' + B' = (AB)'\n\nHowever, the expression you're interested in, A(B+C'+1) is trivially equal to A, and so has no interesting dual.
114598
Ok, If A is a point on the circle and C is the middle of the circle then AC is the RADIUS of the circle.  Which means any point on the circle to C WILL ALWAYS be 12.  So if the excess of the circle to B is 8, then Add the RADIUS to the distance to B, and we arrive at the result of 20. Which is AC + (Distance of circle to B)\n\n20
492811
In addition to what “Spoonish1” told you I suggest you to write a program calculating minimal distance between 2 straight lines in 3D. Be this distance =0, then and only then you may state that 2 lines intersect. A classic line equation in 3D looks like:\n(y-y1)/(y2-y1)=(x-x1)/(x2-x1)=(z-z1)/(z2-z1) – just as a matter of fact. Has your desire faded? Or more help?
482553
hi\nthe details which you gave have the answer for this question.\n\nangle E= angle C and angle A= angle B and DA=DB\n\nfrom this given data i will not assume that the triangles are equilateral, I will prove they are.\n\ncalculating the area of the 2 triangles with the formula (1/2)*base*height we can prove these 2 triangles are congruent and AC=BE\n\nsince the angle E=angle C and DA=DB then DB and DA must be the sides opposite these angles E and C respectively. \nhence  the line drawn(at angle of 90 degrees to base DB) from angle E to DB and line drawn(at and angle of 90 degrees to DA) from angle C to DA must be of equal magnitude (otherwise the angles cant be equal and/or sides cant be equal) ,  \n\nhence the length of this line drawn which is called as the height of the triangle is proved to be equal\n\nnow calculating the area with above formula (1/2)*base*height\nheights of 2 triangles are proved to be equal equal and bases are \ngiven to be equal hence areas of these 2 triangles must be equal.\n\nwhen area of 2 trianlges are equal and 2 sets of angles are equal and one side is equal then the 2 triangles must be absolutely congruent. hence AC must be equal to BE
96674
Let the two foci be at (a, b) and (c, d) which we assume are distinct -- if they are identical then you have a circle and a much easier problem. Afterwards we can check the result to see if relaxing that restriction would introduce a division by zero.\n\nLet the sum of the distances from the two foci (in the case of a circle, the diameter) be s.\n\nLet theta be the angle from (a, b) following the usual convention that an angle of zero goes from (a, b) to (a + infinity, b) and theta increases counterclockwise.\n\nThe ray going out from (a, b) in the angle theta direction can be parameterized as the curve\n\n(x(t), y(t)) = (a + t cos theta, b + t sin theta) for t >= 0\n\nWe want to find t such that (x(t), y(t)) satisfies that the sum of its distances from the two foci is s:\n\nThe distance from (a, b) is\n\nsqrt((x(t) - a)^2 + (y(t) - b)^2)\n = sqrt((t cos theta)^2 + (t sin theta)^2)\n = sqrt( t^2 (cos^2 theta + sin^2 theta) )\n = sqrt( t^2 )\n = t\n\nImagine that. :-)  (We don't need to say the absolute value of t because we know t >= 0).\n\nSo we want to find t such that\n\nt + sqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s\n\nsqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s - t\n\n(a + t cos theta - c)^2 + (b + t sin theta - d)^2 = (s - t)^2\n\nt^2 cos^2 theta + 2 (a - c) t cos theta + (a - c)^2 + t^2 sin^2 theta + 2 (b - d) t sin theta + (b - d)^2 = s^2 - 2 s t + t^2\n\nWell look at that, on the left hand side the quadratic term in t is t^2 (cos^2 theta + sin^2 theta) = t^2 whic is also the quadtratic term in t on the right.  So we can drop those and we get\n\nt (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s) = s^2 - (a - c)^2 - (b - d)^2\n\nIf s is at least the distance between the two foci then the right hand side is >= 0 because it is s^2 minus the square of the distance between the two foci.\n\nSo let T = (s^2 - (a - c)^2 - (b - d)^2) / (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s)\n\n(assuming the denonimator is nonzero) and the point you want is\n\n(x(T), y(T)) = (a + T cos theta, b + T sin theta)
255328
define f(x)=d(x,A)/[d(x,A)+d(x,B)].\nThis is a continuous function f:X->[0,1] (use triangle inequality to show x->d(x,A) is continuous for a closed set A). Also f=0 on A and f=1 on B. Let C=f^(-1)[0,1/3) and D=f^(-1)(2/3,1]. Then C and D are open because inverse images of open sets under continuous functions are open.\n\nNote to Adolfo Boy: It is possible for the distance between two disjoint closed sets to be 0. Consider the x-axis and the y=1/x curve. Your argument does work for compact sets, though. The point (a,b) in your argument need not exist if the infimum is not attained.
244909
I know how to do it, but it's going to cost me a lot of work, and sorry I don't have the time!\n\nI can give you some tips to start off though:\n\nDraw an x,y diagram where the center of the circle is the origin. Draw AB as a horizontal baseline which is 21 below the origin.\n\nThen you will have to create functions for the circle and then for the lines that have the tangents with the circle that together form the triangle. Particularly the formulas for the triangle lines will be hard.\n\nYou will have to calculate at which other points than P the triangle lines touch the circle. For that you will need the derivative of the function for the circle.\n\nEnglish is not my native language, so I hoped I used the correct English terminology. Good luck!\n\nP.S.:  aytu_y, what you say about equilateral triangles is correct, but it doesn't work in this case because here point P is not in the middle! (PA = 23 while PB = 27)
351705
Draw the chord. Observe that , unless it goes thru the center of the circle, that it splits the circle into a large part and a smaller part. Now construct a perpendicular bisector on the chord. Note\nthat it's length to the edge of the circle on the larger part is greater than the radius else it would be on the smaller part. \nAlso the distance to the edge in the smaller part is less than\nthe radius. So somewhere on the perpendicular bisector the\ndistance to the edge is equal to the radius. At any\npoint on the bisector draw a pair of lines to the end points of the chord. They make a pair of congruent triangles by S.A.S. therefore the lines are of equal length. At some point the lengths must be equal to the length of the radius. The only place that can be is at the center.
233453
Behind all the nomenclature, this is just a simple minimization problem. You want to find local minima for A(q). Thus, take the derivative of A(q) to look for critical points (places where the graph is locally flat). These will be places where the tangent line has slope 0; so set the derivative equal to 0 and solve for q. You will get two possible choices. Use the second derivative to see what the behavior of the first derivative is at each point (increasing or decreasing). An increasing first derivative around a critical point (where the derivative is zero) implies that the slope locally changed from negative values to positive values. Visualize this to see that it is a minimum point for the graph of A(q). The q that gives you the positive second derivative and has 0 first derivative is thus q of the Wilson lot size formula.\nFor b, simply replace k with k + bq in A(q) and do a similar analysis as above.\n\nPS. The answerer above has forgotten the negative value of q that also satisfies the first equation. Although a negative q does not correspond to the problem, it is unrigorous to say that there is only one answer to the equation. It should at least be explained as being unphysical.
393801
Your first Question:\n\nNow,\n\nImagine a Triangle with vertices ABC. \nAny line from A to BC is AD. D is the point of intersection on BC.\n\nLet E and F be the midpoints of Sides AB anc AC respectively.\n\nJoin E and F to meet AD in G.\n\nNow for the proof:\nTake Triangles AGE and ADB these two are similar.\n\nReason being that EF would be parallel to BC from Mid Point theorem. Therefore, the two triangles will be similar as Angle E and Angle B are equal and Angle A is common. \n\nNow take Triangle ADB, E is the mid point of AB and EG is parallel to BD there for from Converse of Midpoint Theorem G becomes the Midpoint of AD.\n\nHence proved.
509021
a) y= 10^(tan (πθ))\n= e^(ln10*tan(πθ))\n= e^u where u = ln10*tan(πθ\ndy/dθ =dy/du * du/dθ\n= e^u * ln10*πsec²(πθ)\n= ln10*πsec²(πθ)*e^(ln10*tan(πθ))\n= ln10.πsec²(πθ)*10^(tan (πθ))\n\nb) y= √[tln(t^4)]\n= √[4t*ln(t)]\n= 2√[t*ln(t)]\n= 2u^½ where u = t*ln(t)\ndy/dt = dy/du * du/dt\n= u^-½ * (ln(t) + 1)\n= (ln(t) + 1)/√[tln(t)]\n\nc) y = arctan(arcsin (x^½))\n= arctan u where u = arcsin v where v = x^½\ndy/dx = dy /du * du/dv * dv/dx\n= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)\n= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]\n\nd)xe^y = y - 1\ne^y + xe^y * dy/dx = dy/dx\nSo dy/dy - xe^y * dy/dx = e^y\nie dy/dx (1 - xe^y) = e^y\ndy/dx = e^y/ (1 - xe^y)\n= 1/(e^(-y) - x)
461081
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
551560
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
155088
There is an equation for this, but I don't remember what it is.  Guess and check with a calculator is funner anyways...\nI didn't show much work (how much work is there with guess and check), but I think these are correct. (&=infinity)\n\n1. The range of k is [-2,&)\n\n2. When k=3 then the ratio of a:b will be 1:5.  The equation of the parabola will be y=-X^2 + 4X + 5.\n\n3. I wasn't sure exactly what you were asking on Q 3.  If you are looking for where the tangent line of the parabola at x=0 crosses the x-axis, the point will be (-1.25,0).  If you are saying that from (A,0) to (B,0) in Q 2 is the diameter of a circle, and you are looking for where the tangent line of that circle at x=0 intersects the x-axis then the point is (-4.5,0).\n\nI hope that helps.
198876
(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\ngets nasty fast, so remember these rules\ntanX = sinX / cosX\nsin(X+Y) = sinXcosY + cosXsinY\ncos(X+Y) = cosXcosY - sinXsinY\n\n(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\n=[cosAcosBcosC * (tanA + tanB + tanC - tanAtanBtanC)] / [(1 - tabAtabB - tabBtanC - tanCtanA) * cosAcosBcosC]\n=[sinAcosBcosC + sinBcosAcosC + sinCcosAcosB - sinAsinBsinC] / [cosAcosBcosC - sinAsinBcosC - sinBsinCcosA - sinCsinAcosB]\n=[(sinAcosC + cosAsinC)cosB + sinB(cosAcosC - sinAsinC)] / \n[cosC(cosAcosB - sinAsinB) - sinC(sinAcosB + cosAsinB)]\n=[sin(A+C)cosB + sinBcos(A+C)] / [cosCcos(A+B) - sinCsin(A+B)]\n=sin(A+B+C) / cos(A+B+C)\n=tan(A+B+C)