Can you solve these geometry problems?
ฝัง
- เผยแพร่เมื่อ 2 มิ.ย. 2024
- Can you solve these interesting questions?
0:00 problems
1:39 solution 1
6:00 solution 2
Problem 1 different solution by @AndyMath
• Tricky Geometry Challenge
Problem 2
/ 1730904461449785733
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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).
Commented 2 days ago? How
@@TheHYPFNHis videos are initially private for his patreons, only made public days later.
@@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.
I used the centre and got the same answer. Could go SLIGHTLY left/right of the centre and x won't change significantly.
Not only significantly, it shouldn't change anyway.😅
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
Thank you for these wonderful teasers, you make TH-cam a pleasure to peruse.
This was informative.
Great Work
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.
Why is the diameter 2rk
@@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.
@@NateHays alright thanks
Very nice questions! Learned two things I never knew
That was a good one. I learned stuff from the 2nd one for sure, so this was a 'good day' :)
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
Very nice!
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.
Also works when the chosen point is at the intersection of the diameter and the circumference, Easy to figure in one’s head.
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.
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.
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.
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)
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.
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!
Gotta agree, both problems were really fun to solve 😊
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.
yes it is from RMO
Have u guys given RMO 2023
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.
nice work
Saw the same problem in VDP being solved twice in a day. Cheers
(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°. ∎
1:23
It is not on the center, but you CAN put it on the center, so the value is equal to the radius
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.
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)
At 7:16 you have a periferiangle of 30 degree to the cord X, and you know the radius is 8, then the cord X = 2*radius*sin(30 = 2 *8*sin(30) = 8 unit.
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.
all respect from Morocco genious Mathematician guy
Can someone explain 7:28?? Could not follow that part. How do we conclude arc represents 60°??
Though how arrogant of the geometer to assume that such grand ideas await them at such convenience.
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
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.
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
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.
Just curious, at the conclusion of #2 you said, "and that's the answer!"
I just wanted to confirm, that that that [the key] wasn't hanging out...
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.
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
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.
I love how enthusiatic he gets as we get closer to it lmao
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!
For the first problem, is it always numerically 2 times the diameter? Or was that just a coincidence?
Q1 : did I miss the part where you showed the midpoint of the long edge was on the semicircle?
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
Triangle construction by draw line ...one of h= 8 root2...similarly 8-4root2...h=...then 120 degree....cosine formula
Tank you
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
(It may seem kind of complex from the lengthiness of my reply but it's actually very straightforward, I just don't know how best to convey it)
@@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 solving problem 1 , we could use the solution of problem 2.
I think that's easier to solve.
very interesting... :D
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 first prblm, how are we assuming that doing the rectangle in half will make it a square?
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.
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?
0:26
For the first problem (after about 3:50), I would drop a perpendicular from the point on the left, note the side ratios and solve from there.
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.
@@shaventalz3092 Sorry, mental typo.
Love from India 🇮🇳
1. problem Area(blue rectangle)=20
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*
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
How do you know that splitting the rectangle in half, will create 2 squares on problem 1?
Yeah, that's what I'm wondering, too
Yeh, me too. And why must the adjacent corners lie on the semicircle?
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.
@@neuralwarp Everything that lies on the semi-circle was given in the question
@@shashankmaurya1263 But why will adjacent sides be equal? It wasn't mentioned in the question and never got explained in the solution
we can solve the problem 2 by proving cyclic quadrilateral. Then we have two angles which are equal and finally, we have the answer
I don’t get why the midpoint of the rectangle must be a point of the circumference
It is a condition given by question
Somehow I also missed that. But indeed it is there
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.
Where are u from
@3:40 why do they meet at a 45 degree angle?
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
3:35 please highlight it. I had to replay it multiple times so I could understand what the heck you were talking about
Hello, how can I give you a math problem?
Problem #2…
16cos(60degrees)
8; (for problem 2); x/sin60 = 8/sin60; x = 8
I did not understand the move at 7:20 - 7:35, could someone explain?
Definition of Inscribed Angle - an inscribed angle in a circle corresponds to half of the respective arc
Presh, first exercise please change task. Find area of each square. 🐶
Referring to an arclength as 60 degrees is nonsensical. Just use radians
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...
🙄🙄🙄🙄
I am from india love you
Edit :: just because of one word missing no one is able to understand what i was telling 😅😅😅😅
You are not India .
You are Indian 😂
Hello india I am love
@@hermit25 you don't exist in India bro
@@hermit25 And I am You
You are the whole country?
Is this cheating: 99+(9-9)!
99+ (9/9)
I love pizza , 🍕
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
For the first problem i saw an easier solution by "andy math" ... .. also sharing link..
th-cam.com/video/VMiS7G8xqQs/w-d-xo.htmlfeature=shared