@Jack HandMa no you're wrong because there's further steps that MUST be carried out to reach the answer of 1/3 2) BDA is isosceles 3) area of BDA = area of BDC 4) area of BDA = area of the shape on the right aka A'B'CD
This is literally the first step of what you have to do to find the centre of mass of triangle ABC: the areas of the blue triangle on either side of AD must be equal, so the blue triangle is half the area of ABC. Therefore, the answer must be 1/3 (since the white bit left over from the rotated triangle must have the same area as the other white triangle, which is the same as the blue triangle). Still, it's a very nice visual proof of _why_ the areas must be equal (and therefore, why this method actually finds the centre of mass). Both the blue triangle and the white triangle have the same base angles, which makes them isosceles triangles; therefore, the common side and the blue side are equal, and the blue side and the white side are equal, so the white and common sides must be equal. Dropping a perpendicular and using similar triangles shows half the blue area is equal to a quarter of the area of ABC, so the whole blue triangle is half the area of ABC and the centre of mass lies on AD.
I think I did something similar, if a bit simpler. Angle C is congruent with C-prime, because they're the same angle in rotations of the same triangle. Therefore the shaded region is an isosceles triangle, and lines DB and DC are equal. Looking at triangle ADB, we can tell that it is also an isosceles triangle because angle A + angle C + 90 = 180 (a triangle), thus angle A + angle C = 90, and 90 - angle C = angle A. Since C-prime = C, 90 - DBC = angle A, making ADB isosceles, and sides DA and DB equal. So we know DA = DB = DC. Since DA = DC, line DB is the median of triangle ABC, meaning that it divides ABC into two triangles of equal area. That means the shaded portion is half of triangle ABC. Because the two large triangles are rotations of one another, the shaded portion must also be half of triangle (ABC)-prime. The two unshaded halves are equal to one another and to the shaded portion, making the shaded portion 1/3 the area of the entire figure.
This problem was not very hard, but the solution is really beautiful. Im happy that i can solve it myself. And you can solve this easier if you know a theorem that “median of a triangle divides it into two triangles that have a same area”.
I would be more interested in the criteria of when the triangles intersect in that manner (i.e. when point D exists between B and B'. It looks like it might happen when angle ACB does not exceed 60 degrees. For the original problem I just mentally examined the specific case where triangle ABC was isosceles (with two 45-degree angles) where it becomes trivially clear that BB' bisects AC and so cuts the area of ABC in two.
after shwing that BD=DC you could just use the fact that the height of BDC falls at the mid point of BC so the area of this triangle is half of the area of ABC
That sounds like the detail I stumbled across on the path to the answer. Triangle BDC is isosceles, and each symmetric half of it is a right triangle with 1/2 scale similarity to the other right triangles in the diagram. 1/2 scale is 1/4 area, so triangle BDC is 1/2 the area of one large triangle, and 1/3 the area of the pentagon, because 1/(4 - 1)
i don't get why you say the area of BDC is half the area of ABC i agree that the area of BDC *is* half the area of ABC but to arrive at that conclusion from 2 ideas: 1) BD = DC 2) the height of BDC falls at the mid point of BC seems to be a bit of a leap of faith?
@@alwaysdisputin9930 Since △BDC is isosceles (∠DBC = ∠DCB) then altitude DF bisects BC and divides △BDC into two congruent triangles (△DFB and △DFC) △ABC and △DFC are similar since they share a common angle (C) and both have a right angle (B and F respectively) But △DFC has base that is = 1/2 base of △ABC (since DF bisect BC) Thus △DFC has height that is = 1/2 height of △ABC (by properties of similar triangles) → Base of △BCD = BC = Base of △ABC → Height of △BCD = Height of △DFC = 1/2 Height of △ABC → Area(△BCD) = 1/2 * Area(△ABC) The rest follows from this.
Another way of doing it: Flip the second triangle so that B' = C and C' = B still, areas will still be the same, you're left with a rectangle with two diagonals with one resulting triangle removed, each remaining triangle therefore 1/4 of area, 1/4 (shaded) out of 3/4 (whole shape) is 1/3.
You don't even need similar triangles, DB is a median of triangle ABC which divides it into two triangles of equal area, therefore the area of Triangles BDC and BDA will obviously be half the area of Triangles ABC.
Got same result, i figured somehow that AC and BD are diagonals of rectangle what 2 triangles make, so DF = 1/2 of AB, then (AB*BC-1/2*AB*BC/2) / (1/2*AB*BC*2) which gives 1/3
Answer 1/3 , if you let height = 4 inches and base = 3 , then area= 12/2=6 sq inches height of shaded = 2 inches base=3 area= 6/2 =3 as the height of the shaded is 1/2 the height of the triangle Since the area of both triangles without the overlapping = 12 and since the area of the shaded then area of overlapping triangles = 9 Ratio of shaded to to entire triangles = 3/9 = 1/3 or 33.33%
The way the question was worded I assumed the two smaller angles of the triangle could be equal, ie 45°. It is then much easier to visualise the preportions of the triangles that are shaded and the answer becomes obvious.
AC is a diameter of circumscribed circle around ∆ABC, DA, DB, DC - radii R. ∠ACB=α, complementary ∠BAC=90-α, ∠BDC=180-2α, ∠ADB=180-∠BDC=2α. Area of ∆ADB=½*R²*sin(2α) Area of ∆BDC=½R²sin(180-2α)=½*R²*sin(2α) [overlapped region] These areas are equal, so area of ∆BDC (name it x) is exactly one half of area ∆ABC and, of course, one half of area ∆A´B´C´. So area of [DB´A´C] equal to area ∆BDC. So common area is 3*x, and ratio is x/3x=⅓ (one third)
Looks like a long way to solve it, what I did: you can see that the original triangle can be folded horizontally to perfectly overlap the white part on the left, so the height of this triangle is double the height of the blue triangle. Since both triangles have the same base, it means the blue triangle is half the area of the original one. Then the total area is twice the original triangle - the blue triangle. So the ratio is (x/2)/(2x-(x/2))= 1/3.
The way I solved this problem is to reflect the triangle on the right by letting C' stay put and flipping B' on C. This gives us a rectangle with the top edge missing and two diagonals.
Awesome Dear sir honestly I solved this problem without paper, pen and calculator in just a minute That all credit goes to you and your channel because this channel make me so..... Thanks once again
I can't usually do these puzzles I my head, but this one was easy. Add a point D such that abcd is a rectangle. From there it is easy to see that point B' is in the center of the rectangle and h of the small triangle is 1/2 h of the large with the same base. So the area of the small triangle is 1/2 the area of the large. Since the two large triangles are the same size its also 1/2 the area of the other triangle. So you have 3 areas ( 1 shaded and 2 non shaded) each of which is 1/2 the area of the initial large triangle. So it is 1/3 the area of the overall shape. Easy peasy.
a shorter method. the rise of the slope on the left side is the same as the fall of the slope of the first right triangle. Thus ADB is isosceles. Thus The vertical height of D is half the vertical height of A. Thus area of the blue triangle(B) is half the area of ABC. So the total area is 4B-B so the ratio B/3B is 1/3
That was quite easy. The area of a triangle is the base*height/2 . So ABC have the same area with BDC cause AD =DC and they have the same height to B. Usually is hard for me.
@@everydayjokes2321 Haha you're good :b I just remembered it from that mobile game that I played many, many years ago. But it is one of the rather good dad-jokes!
The way the question was worded assumed the angles did not matter, except for it being a right triangle. So I plugged in easy values to work with (30-60-90, which makes the center triangle equilateral) and calculated it out.
Since he didn't give any info about size lengths I just did the problem with an isosceles right triangle with size length of 2 (so AB=BC=2). Therefore, the area of ABC=2 and the area of BDC=1. Total area will be ABC+BB'A'-BDC=2+2-1=3 and then we get BDC/Total area=1/3
Make a square with A, C' B and a new A", realize the shaded area is 1/4 of the square, realize A"DC equals B'DCA', realize the shaded area is therefore 1/3 of the entire area.
Sir your videos are truly awesome as they encase the beauty of mathematics using simple problems, with out of the box, or rather "inside the box" solutions. I am a student of the 11th grade, and I sincerely request you to upload this particular geometry problem in your channel - In triangle ABC, let D be the midpoint of BC. If angle ADB = 45° and angle ACD = 30°, determine angle BAD. This problem appeared in our RMO (Regional Mathematical Olympiad) in 2005, which is a stage in the selection procedure in India for the IMO. I like this problem a lot, as the question seems simple but the solution is really elegant. Hence, mathematics being my favourite subject, it would really mean a lot to me if you upload a video covering this problem. Thanking you, Yours sincerely, Arka Banerjee, From India.
Aha! The two base angles of the blue triangle must be congruent since they are corresponding angles of congruent triangles (similarity being the sufficient condition). It took me a while to spot that. Therefore the corresponding opposite sides of the base angles of the blue triangle must be congruent. It follows that the vertex opposite the base of the blue triangle lies directly above the midpoint of said base. Now consider a similar 1/2-scale version of the original triangle that shares the same vertex on the horizontal axis. Note that this scaled triangle has a right vertex that coincides with the midpoint of the base of the blue triangle, and a final vertex that coincides with the upper vertex of the blue triangle. The scaled triangle is clearly one half the area of the blue triangle, and also one-fourth the area of the original triangle from which it was scaled. Therefore the blue triangle has an area equal to one-half the area of either the original triangle or its rotated and reflected transformation. The entire shape has an area equal to twice the area of the original triangle, minus the area of the doubly-counted blue triangle. Thus the entire shape has an area equal to three times the area of the blue triangle, and the blue triangle has an area equal to one-thirs the entire shape.
With what you are saying, it should make BDC=DB'A'C equal too. So it makes the answer equal to 1/3. Without your comment, I would not have been able to arrive to this answer without watching the answering part of the video.
@@GuillaumePerronNantel I paused the video I gave it a try I solved it Then I watched the solution We met at that point After that my answer was different Thanks for reply
Very easy. The angle at C' is the same as the angle at C so triangle BCD is isosceles. Drop a perpendicular from point D down to point E and triangle DEC is similar to ABC but with half the dimensions so 1/4 the area. Triangle DEB is the mirror image of DEC so the total area of DBC is half that of ABC or 1/3 of the total area.
A good question on similarity of triangles. The problem with these questions is that , most often, there is a particular way of solving these questions. And as always, great presentation skills 👍👍👍❤️❤️❤️
I'm a fan of using co-ordinates for problems like these, so here's my solve: Mark point A as (0, a), B as (0, 0), and C as (c, 0) Therefore, A' is ( sqrt[a^2+c^2], 0 ), since the length from A' to C' is the same as the length from A to C and C'A' is parallel to BC ------------------------------------------------------------------------------------------------------ Because B' has a distance of c from C' and a distance of a from A', we can triangulate B' 's co-ordinates. (X-sqrt[a^2+c^2])^2+Y^2 = a^2 expand: X^2-2Xsqrt[a^2+c^2]+a^2+c^2+y^2=a^2 X^2+Y^2 = c^2, so: 2c^2+a^2-2Xsqrt[a^2+c^2] = a^2 subtract a^2, then add 2Xsqrt[a^2+c^2]: 2c^2 = 2Xsqrt[a^2+c^2] divide by 2sqrt[a^2+c^2]: X = (c^2/sqrt[a^2+c^2]) X^2+y^2 = c^2 subtract X^2: Y^2 = c^2-X^2 expand: Y^2 = c^2 - c^4/(a^2+c^2) share denominators: Y^2 = (a^2 * c^2 + c^4)/(a^2+c^2) - c^4/(a^2+c^2) cancel terms: Y^2 = a^2 * c^2 / (a^2+c^2) take the square root: Y = sqrt[a^2 * c^2 / (a^2+c^2)] B' = ( c^2/sqrt[a^2+c^2], sqrt[a^2 * c^2 / (a^2 + c^2)] ) ------------------------------------------------------------------------------------------------------ After that, to find D, we can just construct two linear formulas. The first to find is BB'. The y-intercept is the origin, so we just need to find the slope. sqrt[a^2 * c^2 / (a^2+c^2)] ------------------------------------------ = m c^2/sqrt[a^2+c^2] multiply by sqrt[a^2+c^2]/sqrt[a^2+c^2]: sqrt[a^2 * c^2 * (a^2+c^2) / (a^2+c^2)] ------------------------------------------------------------ = m c^2 cancel terms in numerator sqrt[a^2 * c^2] / c^2 = m solve square root (ac)/c^2 = m cancel terms a/c = m Therefore, the line aX/c passes through point B'! (I think this is just absolutely crazy, given how simple it is!) Far easier is to find line AC. The slope is -a/c (by rise over run), and the y-intercept is (0, a), therefore line AC has the equation -aX/c + a ----------------------------------------------------------------------------------------------------- Point D is found at the intersection of BB' and AC. aX/c = -aX/c + a multiply both sides by c: aX = -aX + ac divide both sides by a: X = -X + c collect like terms: 2X = c divide by 2: X = c/2 feed X into the equation: Y = a(c/2)/c cancel terms: Y = a/2 Point D is at (c/2, a/2)! ----------------------------------------------------------------------------------------------------- At this point, finding out that the height and width of DCF is half of ABC, I easily figure out that the shaded region is 1/3 of the whole shape.
In my opinion, my solution is not good for proving the result, but as I have to find the ratio of 2 overlapping triangles without knowing any side, I realized I can simplify it with isosceles right triangles, each side with 1cm. So the area of the 1 triangle is 1/2cm2. As the rotated overlapping isosceles right triangle has 45 degrees, it halves the other right triangle in the left corner. So as Presh finds it first, the left side triangle has the same area (in my solution is 1/4-1/4cm2). As the rotated triangle has an area of 1/2cm2 as well, and we found the blue area as 1/4, we can just subtract it, and get 1/4cm again. So we found all three areas are 1/4cm2 so the ratio is 1:3.
Because of the translation and rotation, the overlapping region is an isooscles. Because of symmetry, each half of the isoscles, contains a similar triangle to the original triangle . So area of isoscles is half of the original triangle So 'area of overlapping region' / 'area of total region' = 1/(2+1)=1/3
To proof Triangle ABD equals Triangle BCD, I think the simpler method is to draw a perpandicular line from B to AC. The length (h) of this line would be both the height of triangle ABD and BCD. Since both the bottom length (AD and CD) and the height (h) are the same, their area must be the same.
@3:58 I think you mean "... as long as B' is on the exterior" because I heard you say like "...B" is on the x here". And the subtitle also writes that.
I did not use any algebra or geometry. I drew complementary triangles that extended the bounds of the rotated polygons into rectangles rotated 90 degrees relative to the leftmost corner vertex. I picked the special circumstance where the rotated triangle's corner vertex B prime intersects the hypotenuse of the first triange A B at D. Bisecting each of the triangles in each of the constructed rectangles makes it clear that the intersected area must be 1/4 of the constructed rectangle or 1/3 of the total area of the rotated triangles. I will try to prove the general proof that actually yields the correct answer, since this proof is incomplete.
You can also look at the special case where BA = BC. In this case its obvious that the intersection of the original triangle and the rotated triangle is one third of the total shape.
You can also consider another triangle that is symetrical to ABC by D. It'll share all the same angles and values as A'B'C' but B' will be on C and A' will be opposite to A. You get a rectangle AA'CC'. From then on, problem solved.
BDC is isosceles, descend altitude from D on BC, call that point E. DEC is similar triangle to ABC and EC is 1/2 of BC. So DE also 1/2 of AB, so area of DEC = 1/4 ABC, so area of BDC = 1/2 of ABC. Total area of BADB'A'B = 2 times area of ABC minus area of BDC. ratio of the two areas = 1/2 ABC divided by 2 times ABC minus 1/2 ABC. Simplify => 1/3. No need for trig or whatever. Just reason it through.
Assumed the ratio is the same for all right angled triangles. Then selected an isosceles triangle with sides 1,1, 2^1/2. The result follows 1/4 divided by (1-1/4)=1/3.
I thought I had an easier way However, on rewatching the video, I realised my approach was identical to Nestor's. The only thing is that the discussion of the complementary angles and the equality of AD and DC is not required for the solution; it distracted me.
In the original problem there is NO restriction that B' must be on the exterior of ABC, that visual is just an example. You cannot arbitrarily add that restriction after finding a solution that requires it. I do agree that it would make sense to include that restriction, as the solution for B' inside ABC doesn't seem to be very elegant.
You took the scenic route. Once you determined AD = BD, you know ABD is isosceles. Once you have that, point D is half way between A and B. Area of a triangle takes over. All triangles are bh/2, and the base of the overlapping iso triangle is the same as the base of ABC, and the height is half, the area is half. If it is half of ABC, it is also half of the ABCprime triangle. The non-overlapping areas are each the same as the overlapping area. Therefore, the overlapping area is 1/3 of the total. QED.
It took me 10-20 minutes to analyze the question. I was sure about how to solve it. But, I kept getting 1/4 as the ratio. But later I found out that I was actually overlapping the area of the blue triangle two times. That's how I got 1/3. If you get distracted once in maths, it takes effort to get back to the right path.
Once you get that AD=CD, just imagine rotating the shape so that AC is horizontal. Then ABD and BCD are two triangles with the same height and same base length. So they're equal in area (x). A'B'B has the same area as ABC, and BCD is contained in A'B'B, so B'A'CD also has the same area as ABD. Hence x/(x+x+x)
If it works for any arbitrary right angled triangle, then it works for any nice, convenient right angled triangle, that you can choose to make the problem trivially easy. It's trivially easy.
there is even easier way, because it is not stated what kind of right triangle, you can consider 45, 45, 90, or 30,60,90. and then is more obvious. moreover, it is also obvious that two sub-triangles halve the big one, because they share same perpendicular height(and base is same of half size) - no need to use similar triangles
@@kaleege I took like 20 minutes doing useless stuff with Pythagoras theorem until I realised that the shaded region was an iscoceles triangle and if you divide it by half you'd get 2 similar triangles which are also similar to the big main one and since the ratios are equal I solved it in 5 minutes. I was making it more complicated although it was quite simple. Usually things are not as hard as they seem. We just need to pause a little and understand what we're doing
@@achyuththouta6957 It only works when B' is outside △ABC. There's another case that when ∠C > 60, B' falls inside △ABC, and the shaded region is not even a triangle anymore. but the video didn't discuss that case.
That's assuming the triangles are like the diagram. If the angle at the top of the original right triangle is under 30 degrees they don't form the same intersection triangle and it's instead a square. Complicates things a bit.
As usual, thanks for sharing these problems and thanks for the quote!
5 days ago?
Illuminati!
Presh send him an unlisted link to this vid, so they can comment 5 days ago
@@Eric-dn9bx Some of us just have more foresight. :)
How 5 days Ago 😲😲😲😱😱 ?????
Once you realize that BDC is isosceles, it's over.
My first pass, find a simple case, an isosceles right triangle.
Yup, and I think you can solve it by vector of rotation (trigo?), but I don't have the theorem in head.
There are two isosceles triangles that share a common side. That is what makes it easy, because everything falls into place.
Spoiler alert
@Jack HandMa no you're wrong because there's further steps that MUST be carried out to reach the answer of 1/3
2) BDA is isosceles
3) area of BDA = area of BDC
4) area of BDA = area of the shape on the right aka A'B'CD
This is literally the first step of what you have to do to find the centre of mass of triangle ABC: the areas of the blue triangle on either side of AD must be equal, so the blue triangle is half the area of ABC. Therefore, the answer must be 1/3 (since the white bit left over from the rotated triangle must have the same area as the other white triangle, which is the same as the blue triangle).
Still, it's a very nice visual proof of _why_ the areas must be equal (and therefore, why this method actually finds the centre of mass). Both the blue triangle and the white triangle have the same base angles, which makes them isosceles triangles; therefore, the common side and the blue side are equal, and the blue side and the white side are equal, so the white and common sides must be equal. Dropping a perpendicular and using similar triangles shows half the blue area is equal to a quarter of the area of ABC, so the whole blue triangle is half the area of ABC and the centre of mass lies on AD.
I think I did something similar, if a bit simpler. Angle C is congruent with C-prime, because they're the same angle in rotations of the same triangle. Therefore the shaded region is an isosceles triangle, and lines DB and DC are equal. Looking at triangle ADB, we can tell that it is also an isosceles triangle because angle A + angle C + 90 = 180 (a triangle), thus angle A + angle C = 90, and 90 - angle C = angle A. Since C-prime = C, 90 - DBC = angle A, making ADB isosceles, and sides DA and DB equal. So we know DA = DB = DC. Since DA = DC, line DB is the median of triangle ABC, meaning that it divides ABC into two triangles of equal area. That means the shaded portion is half of triangle ABC. Because the two large triangles are rotations of one another, the shaded portion must also be half of triangle (ABC)-prime. The two unshaded halves are equal to one another and to the shaded portion, making the shaded portion 1/3 the area of the entire figure.
I have also 😅
First problem that I solve in my mind quickly.Really a nice problem.Thanks Presh!
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ....
@@madhukushwaha4578 Thanks!
This problem was not very hard, but the solution is really beautiful. Im happy that i can solve it myself. And you can solve this easier if you know a theorem that “median of a triangle divides it into two triangles that have a same area”.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@@madhukushwaha4578 Ok, thank you.
since the question implies that every right triangle will get you the same answer
you can just start by taking a 90 45 45 triangle to make it simpler
all same
I would be more interested in the criteria of when the triangles intersect in that manner (i.e. when point D exists between B and B'. It looks like it might happen when angle ACB does not exceed 60 degrees.
For the original problem I just mentally examined the specific case where triangle ABC was isosceles (with two 45-degree angles) where it becomes trivially clear that BB' bisects AC and so cuts the area of ABC in two.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...
Right, no restriction on the right triangle so choose 45 degrees, then trivial.
Saw the thumbnail, solved it in 15 sec in my head, then clicked on the video. It might be the easiest problem on your channel ^^
Wow 15 seconds...not gonna lie, that's embarrassingly slow.
after shwing that BD=DC you could just use the fact that the height of BDC falls at the mid point of BC so the area of this triangle is half of the area of ABC
That sounds like the detail I stumbled across on the path to the answer.
Triangle BDC is isosceles, and each symmetric half of it is a right triangle with 1/2 scale similarity to the other right triangles in the diagram.
1/2 scale is 1/4 area, so triangle BDC is 1/2 the area of one large triangle, and 1/3 the area of the pentagon, because 1/(4 - 1)
i don't get why you say the area of BDC is half the area of ABC
i agree that the area of BDC *is* half the area of ABC
but to arrive at that conclusion from 2 ideas:
1) BD = DC
2) the height of BDC falls at the mid point of BC
seems to be a bit of a leap of faith?
@@alwaysdisputin9930
Since △BDC is isosceles (∠DBC = ∠DCB) then altitude DF bisects BC and divides △BDC into two congruent triangles (△DFB and △DFC)
△ABC and △DFC are similar since they share a common angle (C) and both have a right angle (B and F respectively)
But △DFC has base that is = 1/2 base of △ABC (since DF bisect BC)
Thus △DFC has height that is = 1/2 height of △ABC (by properties of similar triangles)
→ Base of △BCD = BC = Base of △ABC
→ Height of △BCD = Height of △DFC = 1/2 Height of △ABC
→ Area(△BCD) = 1/2 * Area(△ABC)
The rest follows from this.
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
@@alwaysdisputin9930 I thought it was a general rule that a perpendicular drawn from an iscoceles triangle divides the base by half
Please upload more often if you can! Cant wait for those videos
Another way of doing it: Flip the second triangle so that B' = C and C' = B still, areas will still be the same, you're left with a rectangle with two diagonals with one resulting triangle removed, each remaining triangle therefore 1/4 of area, 1/4 (shaded) out of 3/4 (whole shape) is 1/3.
thank you for this fun puzzle! I really enjoyed it! :)
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
A beautiful problem!
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun .....
You don't even need similar triangles, DB is a median of triangle ABC which divides it into two triangles of equal area, therefore the area of Triangles BDC and BDA will obviously be half the area of Triangles ABC.
Got same result, i figured somehow that AC and BD are diagonals of rectangle what 2 triangles make, so DF = 1/2 of AB, then (AB*BC-1/2*AB*BC/2) / (1/2*AB*BC*2) which gives 1/3
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Me sees this question
Happy median from the 90 degree vertex noises
Could be generalized tho by similarity
Yay, I solved it myself,
Well I found it quite easier by substituting values
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Same...I just assumed ABC was a 45-45-90 triangle and played around with the angles
Answer 1/3 , if you let height = 4 inches and base = 3 , then area= 12/2=6 sq inches
height of shaded = 2 inches base=3 area= 6/2 =3 as the height of the shaded is 1/2 the height of the triangle
Since the area of both triangles without the overlapping = 12 and since the area of the shaded then area of overlapping triangles = 9
Ratio of shaded to to entire triangles = 3/9 = 1/3 or 33.33%
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
it's a simple question:
the shadowed triangle (isosceles triangle) is 1/2 (in area) of the ABC triangle
so the answer is (1/2) / (2 - (1/2)) = 1/3
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...,
@@madhukushwaha4578 Thanks. I'll visit it later.
Solved it! Very nice.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun .....
The way the question was worded I assumed the two smaller angles of the triangle could be equal, ie 45°. It is then much easier to visualise the preportions of the triangles that are shaded and the answer becomes obvious.
Maind your decisions, aaiam Presh Talwalkar~*goosebumps*
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Finally solved it I was baffled by this problem at first because I was too lazy to get a paper and solve it so I kept trying to do it visually
You can stop at 2:01 and tell answer 1/3. Why? Common area is half of triangle and DB'A'C has same area as BDA.
I mean when CD is half of AC, then BCD is half of area of ABC - common height and half of "a" in equation P=ah/2
There were already 2019 likes. Haven't been able to like the video
Being a 14 year old , I am really happy that I solved it myself..
You have the talent
My boi
Congrats!
You have got mind nothing else
I'm 15, and I solved it too! Highfive XD
Nice job!
Nice, very nice and elegant, but I solved it in my head in about one minute.
Always loving your work 🤗🤗
AC is a diameter of circumscribed circle around ∆ABC,
DA, DB, DC - radii R.
∠ACB=α, complementary ∠BAC=90-α, ∠BDC=180-2α, ∠ADB=180-∠BDC=2α.
Area of ∆ADB=½*R²*sin(2α)
Area of ∆BDC=½R²sin(180-2α)=½*R²*sin(2α) [overlapped region]
These areas are equal, so area of ∆BDC (name it x) is exactly one half of area ∆ABC and, of course, one half of area ∆A´B´C´.
So area of [DB´A´C] equal to area ∆BDC.
So common area is 3*x, and ratio is x/3x=⅓ (one third)
I solved it with trig and much shorter solution. But I am mighty disappointed as Gougo's theorem did not make an appearance again :-(
Disappointed? Really? I was mightily elated!
@@piman9280 Haha its the same feeling, I guess.
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Any math puzzle with triangle in it, yet do not involve Gougo's theorem is not worthy to be called a math puzzle.
@@師太滅絕 Hahah so true , especially for Parsh
Thanks for the video!!
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...
Looks like a long way to solve it, what I did: you can see that the original triangle can be folded horizontally to perfectly overlap the white part on the left, so the height of this triangle is double the height of the blue triangle. Since both triangles have the same base, it means the blue triangle is half the area of the original one. Then the total area is twice the original triangle - the blue triangle. So the ratio is (x/2)/(2x-(x/2))= 1/3.
The way I solved this problem is to reflect the triangle on the right by letting C' stay put and flipping B' on C. This gives us a rectangle with the top edge missing and two diagonals.
i also drew the rectangle yea
This is the simplest and most elegant solution
BD = CD = AD => D is the middle point of AC => (ABD) = (BDC) = (DB’A’C)
Hence,
Ratio = 1/3
Awesome
Dear sir honestly I solved this problem without paper, pen and calculator in just a minute
That all credit goes to you and your channel because this channel make me so.....
Thanks once again
I can't usually do these puzzles I my head, but this one was easy. Add a point D such that abcd is a rectangle. From there it is easy to see that point B' is in the center of the rectangle and h of the small triangle is 1/2 h of the large with the same base. So the area of the small triangle is 1/2 the area of the large. Since the two large triangles are the same size its also 1/2 the area of the other triangle. So you have 3 areas ( 1 shaded and 2 non shaded) each of which is 1/2 the area of the initial large triangle. So it is 1/3 the area of the overall shape.
Easy peasy.
Math in 2020: there is no single number in the question
1/3 in the answer though
a shorter method. the rise of the slope on the left side is the same as the fall of the slope of the first right triangle. Thus ADB is isosceles. Thus The vertical height of D is half the vertical height of A. Thus area of the blue triangle(B) is half the area of ABC. So the total area is 4B-B so the ratio B/3B is 1/3
That was quite easy. The area of a triangle is the base*height/2 . So ABC have the same area with BDC cause AD =DC and they have the same height to B. Usually is hard for me.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ..
Joke of the day:
What do you call an alligator in a vest?
*An in-vest-igator*
Did you really steal that joke from the Bitbook of Tiny Tower? 🤨
I love you
@@Languste nah it's one of my subscriber's jokes, idk where he got it from
@@everydayjokes2321 Haha you're good :b I just remembered it from that mobile game that I played many, many years ago. But it is one of the rather good dad-jokes!
I literally learnt this a decade ago
The way the question was worded assumed the angles did not matter, except for it being a right triangle. So I plugged in easy values to work with (30-60-90, which makes the center triangle equilateral) and calculated it out.
Since he didn't give any info about size lengths I just did the problem with an isosceles right triangle with size length of 2 (so AB=BC=2). Therefore, the area of ABC=2 and the area of BDC=1. Total area will be ABC+BB'A'-BDC=2+2-1=3 and then we get BDC/Total area=1/3
Make a square with A, C' B and a new A", realize the shaded area is 1/4 of the square, realize A"DC equals B'DCA', realize the shaded area is therefore 1/3 of the entire area.
That’s exactly what I did
Sir your videos are truly awesome as they encase the beauty of mathematics using simple problems, with out of the box, or rather "inside the box" solutions. I am a student of the 11th grade, and I sincerely request you to upload this particular geometry problem in your channel - In triangle ABC, let D be the midpoint of BC. If angle ADB = 45° and angle ACD = 30°, determine angle BAD.
This problem appeared in our RMO (Regional Mathematical Olympiad) in 2005, which is a stage in the selection procedure in India for the IMO. I like this problem a lot, as the question seems simple but the solution is really elegant.
Hence, mathematics being my favourite subject, it would really mean a lot to me if you upload a video covering this problem.
Thanking you,
Yours sincerely,
Arka Banerjee,
From India.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@@madhukushwaha4578 Thanks a lot!!
Watch the videos of mind your decisions in hindi
ua-cam.com/video/vcDDc_PjOSU/v-deo.html
Very beautiful problem and solution.... Dear brother
Aha! The two base angles of the blue triangle must be congruent since they are corresponding angles of congruent triangles (similarity being the sufficient condition). It took me a while to spot that. Therefore the corresponding opposite sides of the base angles of the blue triangle must be congruent. It follows that the vertex opposite the base of the blue triangle lies directly above the midpoint of said base. Now consider a similar 1/2-scale version of the original triangle that shares the same vertex on the horizontal axis. Note that this scaled triangle has a right vertex that coincides with the midpoint of the base of the blue triangle, and a final vertex that coincides with the upper vertex of the blue triangle. The scaled triangle is clearly one half the area of the blue triangle, and also one-fourth the area of the original triangle from which it was scaled. Therefore the blue triangle has an area equal to one-half the area of either the original triangle or its rotated and reflected transformation. The entire shape has an area equal to twice the area of the original triangle, minus the area of the doubly-counted blue triangle. Thus the entire shape has an area equal to three times the area of the blue triangle, and the blue triangle has an area equal to one-thirs the entire shape.
Solving these feels so satisfying.
Since you proved that dcb is isosceles
DC =DB
And DB=DA
So BD is a median
Then DBC area =ADB area
And that's lead to the answer
Thanks a lot
With what you are saying, it should make BDC=DB'A'C equal too. So it makes the answer equal to 1/3. Without your comment, I would not have been able to arrive to this answer without watching the answering part of the video.
@@GuillaumePerronNantel
I paused the video
I gave it a try
I solved it
Then I watched the solution
We met at that point
After that my answer was different
Thanks for reply
Very easy. The angle at C' is the same as the angle at C so triangle BCD is isosceles. Drop a perpendicular from point D down to point E and triangle DEC is similar to ABC but with half the dimensions so 1/4 the area. Triangle DEB is the mirror image of DEC so the total area of DBC is half that of ABC or 1/3 of the total area.
A good question on similarity of triangles. The problem with these questions is that , most often, there is a particular way of solving these questions.
And as always, great presentation skills 👍👍👍❤️❤️❤️
Nice video👌👍
Thanks!
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...
One of the very very few problems that i could solve myself from this channel... ahh feels good xD
I'm a fan of using co-ordinates for problems like these, so here's my solve:
Mark point A as (0, a), B as (0, 0), and C as (c, 0)
Therefore, A' is ( sqrt[a^2+c^2], 0 ), since the length from A' to C' is the same as the length from A to C and C'A' is parallel to BC
------------------------------------------------------------------------------------------------------
Because B' has a distance of c from C' and a distance of a from A', we can triangulate B' 's co-ordinates.
(X-sqrt[a^2+c^2])^2+Y^2 = a^2
expand: X^2-2Xsqrt[a^2+c^2]+a^2+c^2+y^2=a^2
X^2+Y^2 = c^2, so: 2c^2+a^2-2Xsqrt[a^2+c^2] = a^2
subtract a^2, then add 2Xsqrt[a^2+c^2]:
2c^2 = 2Xsqrt[a^2+c^2]
divide by 2sqrt[a^2+c^2]: X = (c^2/sqrt[a^2+c^2])
X^2+y^2 = c^2
subtract X^2: Y^2 = c^2-X^2
expand: Y^2 = c^2 - c^4/(a^2+c^2)
share denominators:
Y^2 = (a^2 * c^2 + c^4)/(a^2+c^2) - c^4/(a^2+c^2)
cancel terms: Y^2 = a^2 * c^2 / (a^2+c^2)
take the square root: Y = sqrt[a^2 * c^2 / (a^2+c^2)]
B' = ( c^2/sqrt[a^2+c^2], sqrt[a^2 * c^2 / (a^2 + c^2)] )
------------------------------------------------------------------------------------------------------
After that, to find D, we can just construct two linear formulas. The first to find is BB'. The y-intercept is the origin, so we just need to find the slope.
sqrt[a^2 * c^2 / (a^2+c^2)]
------------------------------------------ = m
c^2/sqrt[a^2+c^2]
multiply by sqrt[a^2+c^2]/sqrt[a^2+c^2]:
sqrt[a^2 * c^2 * (a^2+c^2) / (a^2+c^2)]
------------------------------------------------------------ = m
c^2
cancel terms in numerator
sqrt[a^2 * c^2] / c^2 = m
solve square root
(ac)/c^2 = m
cancel terms
a/c = m
Therefore, the line aX/c passes through point B'! (I think this is just absolutely crazy, given how simple it is!)
Far easier is to find line AC. The slope is -a/c (by rise over run), and the y-intercept is (0, a), therefore line AC has the equation
-aX/c + a
-----------------------------------------------------------------------------------------------------
Point D is found at the intersection of BB' and AC.
aX/c = -aX/c + a
multiply both sides by c: aX = -aX + ac
divide both sides by a: X = -X + c
collect like terms: 2X = c
divide by 2: X = c/2
feed X into the equation: Y = a(c/2)/c
cancel terms: Y = a/2
Point D is at (c/2, a/2)!
-----------------------------------------------------------------------------------------------------
At this point, finding out that the height and width of DCF is half of ABC, I easily figure out that the shaded region is 1/3 of the whole shape.
In my opinion, my solution is not good for proving the result, but as I have to find the ratio of 2 overlapping triangles without knowing any side, I realized I can simplify it with isosceles right triangles,
each side with 1cm. So the area of the 1 triangle is 1/2cm2. As the rotated overlapping isosceles right triangle has 45 degrees, it halves the other right triangle in the left corner. So as Presh finds it first, the left side triangle has the same area (in my solution is 1/4-1/4cm2). As the rotated triangle has an area of 1/2cm2 as well, and we found the blue area as 1/4, we can just subtract it, and get 1/4cm again.
So we found all three areas are 1/4cm2 so the ratio is 1:3.
Because of the translation and rotation, the overlapping region is an isooscles. Because of symmetry, each half of the isoscles, contains a similar triangle to the original triangle .
So area of isoscles is half of the original triangle
So 'area of overlapping region'
/ 'area of total region'
= 1/(2+1)=1/3
Amazing problem !! Thank you !!
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
To proof Triangle ABD equals Triangle BCD, I think the simpler method is to draw a perpandicular line from B to AC. The length (h) of this line would be both the height of triangle ABD and BCD. Since both the bottom length (AD and CD) and the height (h) are the same, their area must be the same.
Complex bashed (or you could also coordinate bash but using complex is easier)
@3:58 I think you mean "... as long as B' is on the exterior" because I heard you say like "...B" is on the x here". And the subtitle also writes that.
I heard "X Ear" twice. :D Took a moment to figure out it was "exterior".
Thanks presh, after many days got a beautiful geometry prob....solved it...ans is 1/3.....
This is realll amazing! Elegant!
I did not use any algebra or geometry. I drew complementary triangles that extended the bounds of the rotated polygons into rectangles rotated 90 degrees relative to the leftmost corner vertex. I picked the special circumstance where the rotated triangle's corner vertex B prime intersects the hypotenuse of the first triange A B at D.
Bisecting each of the triangles in each of the constructed rectangles makes it clear that the intersected area must be 1/4 of the constructed rectangle or 1/3 of the total area of the rotated triangles. I will try to prove the general proof that actually yields the correct answer, since this proof is incomplete.
At 0:34, what does "exier" mean? This is a term I'm not familiar with. Maybe I am misunderstanding the narrator. Thank you.
"exterior"
@@marktilford5825 Thanks!
You can also look at the special case where BA = BC. In this case its obvious that the intersection of the original triangle and the rotated triangle is one third of the total shape.
Scaling on x or y, makes the rule still apply. There might be some general restriction that makes this to be true.
Love to have more challenging. This is cool but quite simple question
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Double thank you!
You can also consider another triangle that is symetrical to ABC by D. It'll share all the same angles and values as A'B'C' but B' will be on C and A' will be opposite to A. You get a rectangle AA'CC'. From then on, problem solved.
BDC is isosceles, descend altitude from D on BC, call that point E. DEC is similar triangle to ABC and EC is 1/2 of BC. So DE also 1/2 of AB, so area of DEC = 1/4 ABC, so area of BDC = 1/2 of ABC. Total area of BADB'A'B = 2 times area of ABC minus area of BDC. ratio of the two areas = 1/2 ABC divided by 2 times ABC minus 1/2 ABC. Simplify => 1/3. No need for trig or whatever. Just reason it through.
Assumed the ratio is the same for all right angled triangles. Then selected an isosceles triangle with sides 1,1, 2^1/2. The result follows 1/4 divided by (1-1/4)=1/3.
Yayy! I solved it.... Well it was not too tough for me thou....
BTW, your channel is the best math channel on UA-cam
I had to run the video at half speed and rewind a few times so I could follow along. Safe to say I've been out of school for a while... :)
I thought I had an easier way
However, on rewatching the video, I realised my approach was identical to Nestor's. The only thing is that the discussion of the complementary angles and the equality of AD and DC is not required for the solution; it distracted me.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ....
Calculated the areas using tan(ACB). Work far enough and the tan cancels out and you end up with 1/3
*TIP:* ΔBCD being half of ΔABC is an incredibly powerful hint! ⭐
In the original problem there is NO restriction that B' must be on the exterior of ABC, that visual is just an example. You cannot arbitrarily add that restriction after finding a solution that requires it. I do agree that it would make sense to include that restriction, as the solution for B' inside ABC doesn't seem to be very elegant.
A pretty easy question this time but interesting... Thank you 😊
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...
@@madhukushwaha4578 thank you for your help 😊
You took the scenic route. Once you determined AD = BD, you know ABD is isosceles. Once you have that, point D is half way between A and B. Area of a triangle takes over. All triangles are bh/2, and the base of the overlapping iso triangle is the same as the base of ABC, and the height is half, the area is half. If it is half of ABC, it is also half of the ABCprime triangle. The non-overlapping areas are each the same as the overlapping area. Therefore, the overlapping area is 1/3 of the total. QED.
It took me 10-20 minutes to analyze the question. I was sure about how to solve it. But, I kept getting 1/4 as the ratio.
But later I found out that I was actually overlapping the area of the blue triangle two times.
That's how I got 1/3. If you get distracted once in maths, it takes effort to get back to the right path.
Wow ! A problem with two right triangles...
and no need for the PythaGougu theorem ?
Absolutely fabulous ! I'm flabbergasted ! 🤩
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Once you get that AD=CD, just imagine rotating the shape so that AC is horizontal. Then ABD and BCD are two triangles with the same height and same base length. So they're equal in area (x). A'B'B has the same area as ABC, and BCD is contained in A'B'B, so B'A'CD also has the same area as ABD. Hence x/(x+x+x)
hello! I like your decisions so much! please tell me what programs do you use? if not secret!
When I saw 1/2 I was gonna go into a rant about how it's 1/3
Very good video
I’he solved it ! Thanks for sharing this problem!
Beautiful
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If it works for any arbitrary right angled triangle, then it works for any nice, convenient right angled triangle, that you can choose to make the problem trivially easy. It's trivially easy.
This one was a mind-bender.
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What video editing software do you use? It looks really good.
there is even easier way, because it is not stated what kind of right triangle, you can consider 45, 45, 90, or 30,60,90. and then is more obvious. moreover, it is also obvious that two sub-triangles halve the big one, because they share same perpendicular height(and base is same of half size) - no need to use similar triangles
When you realized that ΔBCD is half of ΔABC, the answer is immediately obvious.
I used a trick: i used a 90 45 45 triangle and then it was simple
those are tricks we use in JEE exams lol
I was waiting to see how you solve it when B' is inside the triangle. and then that's the end of the video... I was like: What? That was it?
Will this ever happen ?
@@amitavaparui3280 When The angle C is larger than 60 degrees.
@@kaleege glad to know I wasn't the only one to think of that.
@@kaleege I took like 20 minutes doing useless stuff with Pythagoras theorem until I realised that the shaded region was an iscoceles triangle and if you divide it by half you'd get 2 similar triangles which are also similar to the big main one and since the ratios are equal I solved it in 5 minutes. I was making it more complicated although it was quite simple. Usually things are not as hard as they seem. We just need to pause a little and understand what we're doing
@@achyuththouta6957 It only works when B' is outside △ABC. There's another case that when ∠C > 60, B' falls inside △ABC, and the shaded region is not even a triangle anymore. but the video didn't discuss that case.
Truly remarkable one 😍🥰
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That's assuming the triangles are like the diagram. If the angle at the top of the original right triangle is under 30 degrees they don't form the same intersection triangle and it's instead a square. Complicates things a bit.
would prefer if little square thingy was there to indicate right triangle
"I think that's quite beautiful"
Similarity concepts & concepts regarding right triangles always make up brilliant & incredible things, aren't they?
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Analysing question how beautiful is more special than just solving it by learning formula thats why i love presh channel
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html