Q156 | Math Olympiad | Geometry | Constructing Squares | Isosceles Right Triangles

Q156 | Math Olympiad | Geometry | Constructing Squares | Isosceles Right Triangles | Polygon Area

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Күн бұрын

Пікірлер: 18

@Ikkarson 5 ай бұрын

For a complete proof, you still need to show that the smaller square always fits inside the larger one, which you do by computing its diagonal and show that it is smaller than the half-side of the larger square.

@Patrik6920 2 ай бұрын

..actually u dont have to... as the squeare can be in any angle, saying a square with side 21 are inside a square with side 32 is enugh , or a square of side 32 can fit any square with the diagonal 32/√2 in any angle [aprox. 22.627 units]

@Ikkarson 2 ай бұрын

@@Patrik6920 in the general case, you still need that bit of proof, since exhibiting one example is not enough. And in this specific case, well, you just proved it anyway 😉

@Darisiabgal7573 3 ай бұрын

32 = r * SQRT(2) thus r = 16 * SQRT(2) the area of the triangle is 16*32/2 is 256. Since the triangle is anchored to the center of the Square we would like to know how much space to rotate it . The chord of 90° is SQRT(2) on the unit circle the bisector and halfchords are r SQRT(2)/2. Thus thus is bisector SQRT(2)/2 * 16 *SQRT(2) = 16. While we are here the area between a chord and the origin is bisector * halfchord = 256. That’s rather obvious but it proves our math so far is sound. If we inscribed the square into a circle the the radius is 21/SQRT(2) or 10.5 * SQRT(2) So this is smaller than 16. This means we can rotate the triangle so that its sides are orthogonal to the sides of the square, done. The square is 10.5 x 10.5 = 21^2/4 The shaded area is 256 - 21^2/4 = 21x21/4 = 441/4 = 110.25 256-110.25 = 145.75 This is the complete answer.

@hamshlam 5 ай бұрын

Good solution but it isnt provable because the right angle of the triangle isnt labeled as the middle of the square

@Patrik6920 2 ай бұрын

bugged me too, but we can atleast prove the area outside the smaller square are 583 and on avrage one triangle is 583/4, from the given information as u say it isent provable any of the areas outside the triangle is 583/4 we can also prove each of the triangles(larger) is (((32√2)/2)²)/2 =256 wich is equal to 32²/4

@Grizzly01-vr4pn Ай бұрын

It is stated to be the case in the voiceover, but yes, always annoying that it's not explicitly written or indicated as such on screen. Seems to be a particular problem with many of these maths channels.

@ironhawk929 5 ай бұрын

My method is to rotate the triangle around the center until the right-angled sides are parallel to the square, and then the area is equal to the triangle minus the small square.

@KipIngram 4 ай бұрын

Notice that the angle the triangle is rotated to is NOT SPECIFIED in this problem. And yet we are asked for a NUMBER. There is only one way for this to be possible - the shaded area must be INDEPENDENT of that angle. If complete the triangle into a new square with side length 32, it has an area of 32^2. We subtract the inner square area of 21^2. We are interested in a quarter of that area, so the shaded area is (32^2 - 21^2)/4 = 583/4. This will be the result regardless of the orientation of the triangle, provided that the 21-square doesn't extend beyond the 32-square side.

@markp7262 4 ай бұрын

I extended the legs of the isosceles triangle to the edges of the square and then went off the rails. I proved that the four resulting quadrilaterals were congruent based on perpendicular/parallel line - angle theorems. This also means that the area of the quadrilateral is always 1/4 of the area of the square. So I rotated the triangle until the legs passed through the corners of the square. This turned the shaded area into a trapezoid with bases of 21 and 32. I calculated the height of the trapezoid as 5.5, and used the area formula ((b1+b2)*h/2). I sometimes have trouble seeing the easy solution. 😁😁😁

@reachthestars98 4 ай бұрын

lol

@iceberg988 5 ай бұрын

Very beautiful solution👍

@quigonkenny 4 ай бұрын

It's actually not necessary to add the entire square construction to solve the problem. Simply drawing perpendiculars from the center of the square to the right and bottom midpoints will allow you to show via congruent triangles that the overlap area will always be 1/4 the area of the square. As long as the distance from the top triangle vertex to the midpoint of the triangle base is greater than the distance from the square center to its corners. In this case that's no problem, as 16 > 21/√2 (≈14.8). As for determining the area of the triangle, for an isosceles right triangle, that's simply the base squared and divided by 4. That can be proven by either constructing a full square as in the video (where the base length would become the square side length in A = s² ==> s²/4) or by mirroring the triangle about the base, where it becomes the diagonal of a square, whereupon you halve the area formula with respect to the diagonal (A = d²/2 ==> d²/4).