For problem 2, @franolich3 sent me an interactive Desmos illustration so you can try it for yourself. Do check it out! Here is the link: www.desmos.com/calculator/3veoetqzgm
For the second problem. From the statement of the problem, it seemed obvious that (as Professor Talwalkar said at the end of the video) that the length would be an invariant, for any point on the diameter. So take the point being at the center of the circle, and it is obviously an equilateral triangle, with the answer being equal to the length of the radius of the circle.
It's definitely a good way to "find" the answer or to check. (I did say "not the center" to avoid this easy trick ;) I also caution that some problems are just wrongly stated--I get emails about this every week. So if we can prove it is invariant, then we can use this method. But we can't use this idea and assume it's invariant...just a small since the fun is the journey to get the answer, and not necessarily the answer itself! (I'm also not a professor--too many politics in academia, and your career is in the hands of people who may think MindYourDecisions isn't even a math channel).
@@MindYourDecisions If you wanted to take away the easy trick, one way would be to state the problem in a way which makes it appear that x may be variable. “Determine x as a function of the distance of the intersection point from the center of the semicircle” or similar. The trick is only relevant if the problem indicates x is constant. Which, while not explicitly stated, is certainly implied. And I think “Professor Talwalkar” was meant as a well-deserved honorary title for all the educating you do here, not that he actually thought that was your profession… 😉
I came to the same conclusion by the way the problem was set up. No matter where the point was picked, the answer would be the same. I then forgot the part about not picking the midpoint of diameter and reasoned it would be true from the center and result in an 8-8-8 triangle. Therefore the chord would be 8 from any two points at 60⁰. While you can't use that method if the random point is picked for you, between the nature of this channel and the way the problem was set up, a certain level of logic can be inferred and included as additional information to the plain statement of the problem. True, even then it's not foolproof, but the better one knows the opponent, the better shot one has solving their riddles.
Problem 2 - the problem suggests x is always the same. So I put the point all the way onto the circle. This makes it clear that the chord is just one side of an inscribed regular convex hexagon.
Yea. They should have said that the point is, idk, half the radius away from the center, so you'd have to get that the position doesn't matter and prove it
For 1 you can do a couple simplifying transforms: First, pull out a scaling factor so we can work with easier numbers: I set the rectangle as 1k x 2k, and the diameter as 2rk so: A = 2k^2 2rk = 10 => k = 5/r, then A = 2(25/r^2) = 50/r^2 (1) I can then solve in the scaled system for r using a 1 x 2 rectangle and scale it back to A later. Second, pick a coordinate system that yields simple points: I shifted the figure so the highest point moves to the origin and then rotated so that we have a circle that passes through the three points {(1,0), (2,0), (0,1)}. Let (x, y) be the center then we get three equations: r^2 = (1-x)^2 + (0-y)^2 = 1 - 2x + x^2 + y^2 (2) r^2 = (2-x)^2 + (0-y)^2 = 4 - 4x + x^2 + y^2 (3) r^2 = (0-x)^2 + (1-y)^2 = 1 - 2y + x^2 + y^2 (4) Solving this easy system (the squared terms fall out), we find: (2) - (3): 0 = 1 - 2x - 4 + 4x = 2x - 3 => x = 3/2 (2) - (4): 0 = 1 - 2x - 1 + 2y = 2y - 2x => y = x = 3/2 Plugging x, y into (2): r^2 = 1 - 2(3/2) + (3/2)^2 + (3/2)^2 = 5/2 Plugging r^2 back in to our scaling equation (1), A = 50/(5/2) = 20.
@@FatFatty k is the scaling factor for all givens. The circle has some radius, call it R. Then I can scale the circle's size by setting R = rk and then divide by k to scale it up/down. I do the same scaling on the rectangle, dividing the sides by k, so the geometry stays the same. Once I'm done calculating in the scaled figures, I undo the scaling to get back to the actual size where 10 is the only thing specified. So eqn. (1) is a preview of how I will un-scale later. This is a common technique in physics where there are all kinds of annoying constants to carry around like the speed of light in meters per second, the number of seconds in a year, the charge on an electron, the mass of an electron in grams, etc... All of these constants can be made to equal 1 if you choose the right scaling factors. After all, a second is a purely arbitrary unit of time for example.
60deg semi circle problem. As you get to chose the point, the unknown must remain constant for any position of the apex along the diameter. The instruction "except the centre" was simply to prevent examination of the trivial case. Seeing no reason why a relationship should skip only one point along a line length, I took it as a clue, and decided 8cm in my head from that example.
Problem 1 : Diagonal(say c) of the Rectangle is the longest side (whose opposite angle is 135 which is formed between sides say a and b) of the triangle which is circumscribed and Circumradius is 5. Area of Δ is (1/2).a.b.Sin135 = abc/(4.5) So c = 10/√2 If sides are rectangle are x and 2x then diagonal is √5x = 10/√2 and Area is 2x² = 2(10/√10)² = 20
(1): Consider a circle K centered at the origin, passing through the points (±1,±3) and (±3,±1). This circle has radius √(3²+1²) = √10. Then K and, say, the rectangle ABCD, with A = (-1,-1), B = (-1,3), C = (-3,3), D = (-3,-1), will meet exactly as the rectangle and circle given in the problem. Now ABCD has area 2·4 = 8. But since the circle in the problem has radius 5, we need to scale things up by a factor of 5/√10. Thus the rectangle in the problem has area 8(5/√10)² = 20. ∎ (2): Let U:= the upper half circle, AB:= the diameter spanning U, O:= (A+B)/2 the center, and V ∈ AB\O. We want points P( = P(V)) and Q( = Q(V)) ∈ U to be such that ∠BVP = ∠QVA = 60°. Meta solution: the problem tacitly states that x(V) ( = |PQ|), as a function of V, is just some constant c. It's also clear that x(V) is continuous on AB, so that x(O) = c. Letting V ∈ {A, O, B} immediately yields x = 8. ∎ Non-meta solution: ignoring the giant hint that x(V) is constant, we'll proceed by "angle-chasing" to show that ∠POQ = 60° from which x = 8 follows immediately. Let S the reflection of P across AB and X the intersection of PV and QO. Note that Q, V, S are collinear. Let α = ∠VQO. Then, since QOS is isosceles, ∠OSV = α. So by symmetry across AB, we have ∠VPO = α. Also, ∠ OXP = ∠QXV = sup(∠XVQ+∠VQX) = 120°-α. Therefore ∠POQ = sup(∠OXP+∠XPO) = 60°. ∎
The angle-chasing part of Problem 2 is a special case of RMO-2023 Problem 2. (a). Let be a semicircle with AB as the bounding diameter and let CD be a variable chord of the semicircle of constant length such that CD lie in the interior of the arc AB. Let E be a point on the diameter AB such that CE and DE are equally inclined to the line AB. Prove that the measure of CED is a constant.
For the 1st problem, you can also solve it by analyzing the inscribed angle formed by the leftmost square's top side and one of the rightmost square's diagonals. Using the rectangle's diagonal also helps.
For Problem 1, I used coordinate geometry. Making the smaller side of the rectangle "x", I made the lowest point where the rectangle intersects the semicircle to be (0,0), and the others (0,x) and (x,2x). I called the origin of the radius (a,b). Knowing that the distances between (a,b) and (0,0), (0,x), and (x,2x) are all 5, you can set up equations and solve for a and b in terms of x. (They are 1.5x and 0.5x.) Then, since (a-0)^2 + (b-0)^2 = 25, plug in x instead and use your solution of x to determine the area of the rectangle.
I applied the concept from the second problem (that the chord is the same length no matter where the point is on the diameter) to solve the first. Move the point to the middle. Draw the diagonal that is also the chord of the circle. We have a right triangle with two sides = 5 (the radius). Since we have a 45-45-90 right triangle the radius (the length of the chord) must be 5sqrt(2). Thus in the rectangle where we have one side x and the other side 2x, the diagonal of the rectangle (the chord of the circle) is 5sqrt(2). This means that x^2 + (2x)^2 = (5sqrt(2))^2 => 5x^2 = 50 => x^2 = 10 => x=sqrt(10). The two sides are sqrt(10) and 2sqrt(10). Thus the area is sqrt(10)*2sqrt(10) = 20.
Problem 1 is super easy to solve with the Intersecting Chords Theorem. You could have used either of the 2 chords drawn to solve it. r = radius of circle = 5 Left Chord Equation: (x)(x) = (r + 3x)(r - 3x) Top Chord Equation: (x√2)(x√2) = (r + 2x√2)(r - 2x√2)
it exists a fairly easy way to solve problem 1: we can consider the rectangle as two squares: let's say that square 1 is the square inside the semi circle and square 2 the other square. Say that length of square is x, therefore, the diagonal measures xsqrt(2). Consider vertical diagonal of square 1 and horizontal diagonal of square 2 (demonstration is trivial). It is easy to construct the inner square of the semi circle of xsqrt(2) sidelength. Now, notice that horizontal diagonal of square 2 is a circle's cord. Therefore, it is split in the middle by circle's radius (r) that is perpendicular to it and thus parallel to vertical diameter of square 1. Now we can construct a rectangle triangle of r as hypothenus and xsqrt(2) and xsqrt(2)/2 as opposit sides. according to pythagorian theorem x² = 2/5r² and 2x² =4/5r² (sanity check: r = 5 ==> 2x² = 20)
There is a much easier way to solve Problem 1. Consider the chord connecting two vertices of the rectangle (the diagonal line of the rectangle), and the triangle formed by this chord and two radius. Prove this triangle is a right angle equilateral triangle (hint: it is 135 degree at the other crossing point from these two vertices) From here the problem can be done easily without even using algebra.
The solution to the second one in this video is unnecessarily complicated, the easier solution according to me would be that if the radius is 8, then the distance between the point of the triangle touching the semicircle on the top and the semicircle's bottom centre would also be 8. Therefore joining those points give us a right angle triangle. By using basic trigonometry, using Sin60, we get 16/root3, again using Sin60, we get X times root3 upon 16 equal to root3/2, which on solving gives x=8 👍 By this method, I solved this problem mentally within 5 seconds.
i have re-created the solution for puzzle#1, just rotate the rectangle so much you can tell the coordinates and then calculate 3 equations with 3 unknown numbers. finally, the hardest part was the rotation for the graphic display: 10 print "mind you decisions-2 nice geometry problems":@zoom%=@zoom%*1.4:nu=59 15 a11=1:a12=1:a13=11:a21=1:a22=-1:a23=0:gosub 50:print xl,yl 20 dim x(4),y(4):la=1:x1=0:y1=la:x2=la:y2=la:x3=2*la:y3=0 25 x(0)=0:y(0)=0:x(1)=2*la:y(1)=0:x(2)=x(1):y(2)=la:x(3)=la:y(3)=la:x(4)=0:y(4)=la 26 xc=x(0):yc=y(0):goto 100 30 a11=2*(xu2-xu1):a12=2*(yu2-yu1):a13=xu2^2+yu2^2-xu1^2-yu1^2 40 a21=2*(xu3-xu2):a22=2*(yu3-yu2):a23=xu3^2+yu3^2-xu2^2-yu2^2 50 ngl1=a12*a21:ngl2=a22*a11 60 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 70 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 80 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 90 xl=zx/ngl:yl=zy/ngl:xm=xl:ym=yl:return:rem print "x=";xl;"y=";yl 100 xu1=x1:yu1=y1:xu2=x2:yu2=y2:xu3=x3:yu3=y3:gosub 30:r=sqr((xu1-xm)^2+(yu1-ym)^2):print r: 115 rr=sqr((x3-xm)^2+(y3-ym)^2):fe=(1-r/rr)*100:print"der fehler=";fe;"%":rem st0str$err 120 dim xu(4),yu(4):for a=0 to 4:xu(a)=x(a):yu(a)=y(a):next a 130 sw=.1:w=sw:goto 190 140 for au=0 to 4:xvx=(xu(au)-xc)*cos(rad(w)):xvy=(yu(au)-yc)*sin(rad(w)):xv=xvx-xvy+xc 160 yvx=(xu(au)-xc)*sin(rad(w)):yvy=(yu(au)-yc)*cos(rad(w)):yv=yvx+yvy+yc 170 x(au)=xv:y(au)=yv:next au:xu1=x(4):yu1=y(4):xu2=x(3):yu2=y(3):xu3=x(1):yu3=y(1) 180 gosub 30:dg=ym:return 190 gosub 140 200 dg1=dg:w1=w:w=w+sw:if w>90 then stop 210 w2=w:gosub 140:if dg1*dg>0 then 200 220 w=(w1+w2)/2:gosub 140:if dg1*dg>0 then w1=w else w2=w 230 if abs(dg)>1E-10 then 220 240 print w,"%",xm,"%",ym:xmin=xm-r:ymin=0:mass=1E3/2/r:goto 260 250 xbu=(x-xmin)*mass:ybu=(y-ymin)*mass:return 260 x=x(0):y=y(0) 280 gosub 250:xba=xbu:yba=ybu:for a=1 to 5:ia=a:if ia=5 then ia=0 290 x=x(ia):y=y(ia):gosub 250:xbn=xbu:ybn=ybu:goto 310 300 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 310 gosub 300:next a:x=xm+r:y=ym:gosub 250:xba=xbu:yba=ybu:gcol 9 320 for a=1 to nu:wa=a/nu*180:x=r*cos(rad(wa)):x=xm+x:y=r*sin(rad(wa)):gosub 250:xbn=xbu:ybn=ybu 330 gosub 300:next a mind your decisions-2 nice geometry problems 5.5 5.5 1.58113883 der fehler=0% 45% 0.707106781 % 7.35601263E-11 > run in bbc basic sdl and hit ctrl tab to copy from the results window.
For problem 1, since the center of the large circle lies on the smaller circle containing the four corners of the rectangle with center the intersection of the diagonals, we can write 2R^2=AB^2 where AB is the length of the diagonal of the rectangle. Also, since the sides of the rectangle are x and 2x, we have 5x^2=AB^2. Equating the two values, solving for x, and multiplying by 2 (to get a formula for the area) we get that A=2x^2=4/5 R^2 and since R=5, A=20.
puzzle number one was posted earlier by "andy math". however concernung puzzle 2 it's remarkable that the result won't depend on what has been chosen for "l3" in line 20: 10 print "mind your decisions-2 nice geometry puzzles":nu=59 20 dim x(2),y(2):r=8:l3=r*.57:sw=.1:w1=60:w2=60:l2=sw:goto 40 30 dl2=2*l2*(r-l3)*cos(rad(w2)):dg=((r-l3)^2+l2^2-dl2)/r^2:dg=dg-1:return 40 gosub 30 50 dg1=dg:l21=l2:l2=l2+sw:if l2>10*r then stop 60 l22=l2:gosub 30:if dg1*dg>0 then 50 70 l2=(l22+l21)/2:gosub 30:if dg1*dg>0 then l21=l2 else l22=l2 80 if abs(dg)>1E-10 then 70 90 l1=sw:goto 120 100 dl=2*l1*(r-l3)*cos(rad(180-w1)):dg=(l1^2+(r-l3)^2-dl)/r^2:dg=dg-1 110 return 120 gosub 100 130 dg1=dg:l11=l1:l1=l1+sw:if l1>10*r then stop 140 l12=l1:gosub 100:if dg1*dg>0 then 130 150 l1=(l11+l12)/2:gosub 100:if dg1*dg>0 then l11=l1 else l12=l1 160 if abs(dg)>1E-10 then 150 170 print l1,l2 180 dlx=2*l1*l2*cos(rad(180-w1-w2)):lx=sqr(l1^2+l2^2-dlx):print "x=";lx 190 x(0)=l3:y(0)=0:x(1)=l2*cos(rad(w2)):x(1)=x(1)+l3:y(1)=l2*sin(rad(w2)):x(2)=l1*cos(rad(w1)) 200 x(2)=l3-x(2):y(2)=l1*sin(rad(w1)):mass=1E3/2/r:goto 220 210 xbu=x*mass:ybu=y*mass:return 220 x=x(0):y=(0):gosub 210:xba=xbu:yba=ybu:for a=1 to 3:ia=a:if ia=3 then ia=0 230 x=x(ia):y=y(ia):gosub 210:xbn=xbu:ybn=ybu:goto 250 240 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 250 gosub 240:next a:x=2*r:y=0:gosub 210:xba=xbu:yba=ybu:gcol 9:for a=1 to nu 260 wa=a/nu*180:x=r*cos(rad(wa)):x=x+r:y=r*sin(rad(wa)):gosub 210:xbn=xbu:ybn=ybu 270 gosub 240:next a mind your decisions-2 nice geometry puzzles 5.704607739.14460773 x=8 > run in bbc basic sdl and hit ctrl tab to copy from the results window
problem 1 has a simple solution though it can be problematic to think of it. The line from the bottom corner of the rectangle to the point where the side touches the circle is parallel to the median of the top triangle which also passes through the center of the circle because it is a segment divider. Now the first line can be extended to touch the circle on the other side if a full circle was drawn. A similar line can be drawn on the right side such that it is parallel to the left line and it is trivial that the three lines are parallel (two diagonals of the squares meeting on the circle are perpendicular by Pythagoras theorem). We get an inscribed rectangle and its diagonal passes through the center. It is a rectangle because, for the second segment, we can use the reverse reasoning to show that it is perpendicular to the left line. The diagonal is sqrt(10)x of the starting side of the rectangle. so x is sqrt(10). Hence the area is 20. The solution doesn`t require any paper and just 10-20 seconds of thinking.
We can solve problem 2 by using a locus method, i.e. an idea of a moving rod and continuity. Rotate a chord AC (congruent to the radius) about the centre O, At any point (during this rotation), if you join A to a point B on the diameter so that the segment AB is inclined at 60 degrees to the diameter. Note that ACBO is cyclic (by noting angle ABO =60 and angle ACO = 60). This shows we have angle ABC = 60 degrees and thus we have the configuration given in problem 2. As you rotate the chord through the circle, you hit every point on the diameter (you need an application of intermediate value theorem and continuity for the construction here). Continuity can be proven by analyzing the steps in a coordinate construction. The technique I have described compared to Presh's technique is very similar to the following: If f:[0,1] -> [0,1] is given by f(x)=x^2. And the question was to show f(x) = 1/4 has a solution, we can either construct it explicitly as x=1/2 OR we can say f is continuous with f(0)=0 and f(1)=1, so somewhere in between every value from 0 to 1 must be hit. So in particular 1/4 must have been hit. The second method is quite powerful since you just have to check it at convenient points but you need to show continuity.
If they wanted to make the second one seem more difficult, they could have added information, such as the position of the vertex. Knowing that you don't need that information makes the problem seem easier due to the concept of moving the vertex to either the center or exterior of the circle snd solving it there.
yes, as people have pointed out, the second problem is trivial. move the point on the diameter to be either the center of the circle or on one of the endpoints of the diameter and the answer pops out!
At 2:20 of the first problem don't you need to show that when dividing in to 2 squares that the midpoint of the top part of the rectangles is in fact also a point on the circle?
With the invariance found in problem 2, solving problem 1 is much easier. Shift the bottom left corner to the edge of the circle and find that the diagonal of the rectangle is 10root2, etc
I solved both these problems using simpler methods (which admittedly played a bit fast-and-loose and depended on the drawings not being misleading). For Problem #1, I more or less guessed that a line drawn down from the topmost point of the rectangle to be perpendicular with the diameter of the circle would intersect the center of the circle, and would also bisect the lower-right side of the rectangle. This also suggested that the lower-left point of the rectangle is halfway through a radius of the circle. These two statements (if they are indeed true) would give us a 45-45-90 triangle, with legs length 2.5 (1/4 of the diameter) and hypotenuse 2.5*sqrt(2). The rectangle then has area 2.5*sqrt(2) times 5*sqrt(2), or 20. If I were trying to prove this rigorously, I'd actually validate my assumptions, but they seemed good enough for playing along with a RU-vid video. For Problem #2, I noted that you said the point could be freely slid along the radius of the circle without changing the length x. Thus, although you said the point wasn't located at the circle's center, I knew it could be moved arbitrarily close without altering the length of x. So I simply moved the point to the center, noted that I now had an equilateral triangle, and so x had to equal the radius, 8. I should mention, based solely on the video's thumbnail I thought there might be a constraint that x be one leg of a right triangle, but I also worked out (using the-picture-looks-right logic similar to with Problem #1) that if it were a right triangle, we could draw a similar triangle and prove that x is the same as the radius. Which gave me more confidence that there wasn't weird behavior where x was the radius only when the point was the center (or along the circumference) of the circle.
I came up with an interesting way of thinking about the second one. After constructing the second chord, notice that the two triangles are similar. So, the inscribed angle adjacent to x would be equal to 120. We know that the interior angles in a regular hexagon are equal to 120, so x is the side length of the inscribed hexagon. Now you can say x is equal to 8 because the side lengths of a regular hexagon are equal to its radius.
1) First, Call the rectangle ABCD, S.T AB = 2x and AD = x. Then, call M the midpoint of AB. Now construct the radii OA and OC, as well as the segments AM and MC. Observe that BMC is an equilateral right triangle; therefore, the angle BMC = 45° and AMC = 135°, so by the inscribed angle theorem, the larger angle AOC = 270° and the smaller angle AOC = 90°. Notice that AC is the hypotenuse of both ABC and AOC; therefore, AC = root(5)*x = 5*root(2). Therefore, x = root(10) and area = 20 2) First, call the triangle ABC, S.T A is the point within the circle. Construct the radii OB and OC. Notice that the angle BAC = 60°. Now observe that BAC = BOC = 60° because they observe the same circumference arc. Therefore, BOC is an equilateral triangle because, since it's isosceles, its height is also the angle bisector, so tracing the height OM creates two right triangles where MOB = MOC = 30° and MBO = MCO = 60°. Therefore, BC = 8 Edited for clarity
Problem 2, agree that the point on the diameter may be at any point. Please expand the reason that one of the angles does not become undefined (Zeno's Paradox?) when the point is on the circumference.
I haven't seen the answer to the second one yet, but the hypotenus of the right triangle formed by the line segments is right on top of the point in the center. If you construct a line from there to the center point, it will have a lengrh equal to the radius, 8. Then since each triangles are 30-60-90 triangles, x is equal to y√3 and so is the 8 because theu are botht the longer legs. This means they are the same
For Problem 1, there is a much simpler and more straightforward solution. Draw the vertical diagonal of the lower square and the horizontal diagonal of the upper square. Then draw a line straight down from the rightmost vertex of the original rectangle, perpendicular to the diameter. By simple angle chasing we can see that the three drawn segments and the diameter meet each other at right angles; and since their lengths are the same, they hence form a square inscribed within the semicircle. From here we can do some simple calculations to get area=20
@@brycenyair6481 you haven`t shown that the lines meet at 90, look at my solution above though. I think it proves the solution and the proof is straightforward. My solution is similar but some things are adjusted a bit to show why statements are true.
For problem 1, let the smaller side of the rectangle be s and the radius of the circle be r. The diagonal of the rectangle subtends an angle of (π/2+π/4) on the upper portion of the circle and so it subtends an angle of π/2 at its centre, forming a right triangle with sides r, r and s√5. So, r²+r² = 5s² and the area of the rectangle is 2s² = 4r²/5. In this problem, r=5; so 2s² = 20.
But without going through the steps to prove where that rightmost point is, you don't _know_ that a perpendicular dropped from there will intersect the center of the circle.
Problem 2: If one set the point (that's on the diameter,) to the fare left, then you get a 90, 60,30 triangle with hypotenuse 16, and one get x = 8 on one of the catheter
Can you please elaborate your approach here. CoZ it really seems like that triangle will be 30,60,90° ∆ , although I failed to prove it. So I need your help buddy :) *Please*
x = 8 because we can say that the height of the semi ciccle is 8 then it becoms a right angled triangle then , by using Trigonometry we find the hypotenous which is 16/√3 then we can easily find X which is 8.
Problem 1 - here's the most simple solution i.e. We can just project the diagonal of the square created by halving the rectangle to the other side of the semicircle, and we can clearly observe that a square has been created with one side along the diameter with the center of semicircle as the center of that side , then just use pythagorus theorem as (√2x)²+(x/√2)²=5² We get x² as 10 . Where x is taken as the smaller side of rectangle
Problem 1: I am wondering about the distance of the point of the rectangle on the diameter of the circle from the center of the circle and the two outer angle at this point of the rectangle with the diameter. Your sketch makes it look like they were both 45°, but I think that's not correct, isn't it? Any solutions to this? Thanks!
Second problem is easier, there are other ways to get the answer. But it would be interesting to know if any engineering application has been made of this particular invariant.
For the second problem, we can tell it is an invariant problem from the description, but are told not to pick the easy center point. So pick the other extreme. Choose a point on the circle itself.
I see below that others asked my question and that this is given as a condition of the problem. Now I am trying to determine if this condition were not provided would it be possible to construct a rectangle that satisfies all the other conditions of the problem but no that one
It is given in the question, though you don't actually need it. I missed it too, and managed to find a solution, though granted it was a little trickier.
As geometry doesn't come easy to me and that co-ordinate geometry makes solving far easier with defined steps (though it includes more steps) , I solved for Problem 2 with co-ordinate geometry ---> *Found the equations of the lines, the circle, the co-ordinates of the pts, and used distance formula to find the length of the chord* . I wish I could have that kind of eyes to see the connections with simple geometry, like you found there to be congruent traingles and deduced an equilateral triangle etc 😢 Btw do you have a discord server where I can upload my solution? Thx.
Why is it that people are going so crazy about chord problems So let’s pretend we don’t know anything but 360° system Problem #1 If we see that we have a square half outside we can duplicate a mirror image on the other side. This indicates the top blue superposed squares are on an axis of symmetry. This means we can inscribe a rectangle over the origin that is three sides long by one side, it’s diagonal is also a diameter. Thus 10 = SQRT(3^2 + 1) s = 10 = SQRT(10) s s = 10/SQRT(10) = SQRT(10) the box is thus 2 * SQRT(10)^2 = 20 My answer is better😎 Problem #2 Again a chord problem but this one is a floating problem. So to understand this problem it helps to know this. If you have a chord that you start at 1,0 on the unit circle, then position of the chord is 2 * half chord times bisector. So the inside 60° and the diagonal we can float from O to -r. So let’s put it at -r. Fair enough. It’s not at the center🤣. The inside line is now x. X is the chord of 60°, that’s fine but we should notice that a chord is always an isosceles triangle with respect to the radii that attach the end to the center, and an isosceles triangle of 60° for any angle is an equilateral. So one answer is 1. Next let’s the intercept of one end be 1,0. So 1 = h * SQRT(3)/2 h = 2/SQRT(3) and thus diagonal intercept is -1/SQRT(3) So the other line (left and up at 60°) travels out at -SQRT(3)/2 / 1/2= -SQRT(3) It’s y intecept is 0 + 1/SQRT(3) - SQRT(3) = -1 So the equation for the line is -SQRT(3)x -1 = -(x SQRT(3) + 1) = y and the equation for circle in this case is x^2 + y^2 = 1 Thus x^2 + (-(SQRT(3)x+1))^2 = 1 x^2 + 3x^2 + 2SQRT(3)x + 1 = 1 4x^2 + SQRT(3)x = 0 4x + 2SQRT(3) = 0 4x = -2SQRT(3) x = -SQRT(3)/2 Thus y = -SQRT(3/4)*-SQRT(3) - 1 = SQRT(9/4) -1 = 3/2 - 2/2= 1/2 Thus x = SQRT (SQRT(3/4)^2 + (1-1/2)^2) = SQRT(3/4 + 1/4) = 1. And of course the deception is at the center the chord is the chord of a hexagon of sides 1. The answer is a simple rescaling to 8 Can we do the same thing with a 90° angle. If we have it pointing strait up the chord of 90° at center is SQRT(2) if we move it to the left side then it’s x is the chord of the side which is 90° with respect to the center.
Because adjacent sides will be equal and since it was split using a parallel line,the resulting figure would be a rhombus.But since one angle is 90° it must be a square.
Gee... my solution to no 1, although correct was pretty agricultural.... I modelled three of the points of the rectangle as a triangle on the circle circumference, then assumed a length, found the circumcentre and the radius that fit that side length, then scaled it up!
El primer jugador puede poner la x en 9 posiciones, el segundo en 8, la siguiente jugada del primer jugador habrá 7 posiciones... O sea 9 factorial xd Además son menos porque el tablero no debe estar completo para terminar la partida. Sí dividimos entre 4 para rotaciones y entre 2 para reflexiones el número es muchísimo más bajo En resumen Alva Mayo no sabe matemáticas
Given the stated information on the second problem, you should have STARTED with fact that the length of x was invariant! If it wasn't then the problem would have been UNSOLVABLE! Since the point could be ANYWHERE on the diameter, just place it at the center and the solution is automatic. Stop being needlessly complex when it is NOT required.
This Video on Second problem: Create a triangle that does NOT use the center point of the circle. Also This Video on Second Problem: Now redraw the triangle using the center point of the circle... 🙄🙄🙄🙄
You took the simplest problem that could be solved by using the sin formula twice and turned it into an ancient Greek tragedy!!! Who are you, my man, Euripides? For the love of God....stop! I teach these to eighth graders
At about 7½ minutes into the video, your presentation undergoes a partial eclipse for 15 seconds when trailer ads for other videos are shown. This is very annoying and disturbing. Can you do something about it?
For the first problem i saw an easier solution by "andy math" ... .. also sharing link.. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-VMiS7G8xqQs.htmlfeature=shared
