Oxford MAT asks: "Which triangle has the biggest area?"
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- čas přidán 6. 05. 2024
- This is a question from the Oxford University Math Admission Test. We would like to find the triangle with the given sides that has the biggest area. Here's the full test: www.maths.ox.ac.uk/system/fil...
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#math #algebra #mathbasics
"which of these choices is the closest to my answer" is goated in exams
Using Heron's formula, I got
area of ans A = ¼√399 ≈ 4.9937
area of ans B = 25/4 × √15 ≈ 24.2061
area of ans C = 25√3 ≈ 43.3013
area of ans D = 75/4 × √7 ≈ 49.6078
area of ans E = 19/4 × √39 ≈ 29.6637
The most area is 10, 10, 14.14 according to Heron’s
You don't need to use the full heron's formula. You can drop the square root and constant terms because order will still be preserved. You then get c^2(400-c^2), and plug in each value of c to find the biggest.
@@nathan87 well actually there is a simpler way to solve it, you can focus on which answer with the angle between two of the lines:10 is closest to 90degree, which my first glance was D and I was right 😊
Awesomeness!!
It’s a non-calculator paper.
Being closer to the maximum doesn't necessarily give you the largest value. If you're pressed for time and can't use a calculator, then 15 is the safest guess, but we don't know for certain that it's correct. (Remove 15 from the list and replace it with 18, and this method will give you the wrong answer).
Thanks. I was trying to find a way to say this.
my dumbass would have used herons formula even if i knew the method 😭
Nothing wrong with that; it's a bit slower, but you only need to look at 8 factors (SP - 10) for all 4 semiperimeters and (SP - X) for the other side
@@dlevi67 yeah but if it were an SAT exam, then i wouldnt have time to do other questions
This is what Excel is for.
@@someonespadreusing Excel in exam . Damn I didn't even knew it is allowed😂
I just plugged Heron's formula into a TI 84, putting s into Y1 in terms of x and A into Y2 in terms of x and Y1. Then I looked at a table, and I graphed it just because I was curious.
My approach was that if you hold the 2 sides constant and equal then the area is 0 when third side lets call x=0 or when the vertex angle is 0, and also when x=20 or when vertex angle is 180, so the maximum area should occur when vertex is 90 degrees and then x can be solved using pythagorean theorem
Thats pretty much what he did lol
I love your videos... I have learnt alot from you... Keep making them❤
Ooh I did this exam last year, was nice to get this one correct!
I’ll attempt to solve it using variational principles, and Heron’s Formula.
suppose the sides of the triangle are a = b = 10, and c is varying which we want to extremise over (0 < c < 20).
the semi-perimeter s = 10 + 0.5c
thus, we want to maximise s(s-10)^2 (s-c) = (10+0.5c) * (0.5c)^2 * (10-0.5c) over c
i.e. c* = argmax c^2 (20+c)(20-c)
this is a negative quartic and hence there exists a unique global maximum over (0,20). In fact, 0 is a doubly repeated root, as is c = 20 (and c = -20). So we can picture the objective function graph in our head (like an M shape), with some T.P maximum over (0,20) (the other two are at 0, and in (-20,0) by the MVT)
This TP indeed occurs at: 2c(20+c)(20-c) + c^2 (20-c) - c^2 (20+c) = 0 (not interested in TP at 0)
==> 800 - 2c^2 + 20c - c^2 - 20c - c^2 = 0
==> 4c^2 = 800 ==> c* = sqrt(200) = 10sqrt(2) which is approximately 14.14, as expected in the right triangle case.
Much longer indeed, but we all love to use Heron’s formula!
No way the "hm ill pick (d) cuz thats the closest to my answer (i got abraham lincoln)" method works!!!
Well, if you don't trust that, you at least know 14.14 is between 10 and 15, so you only need to test c and d
It works due to the law of cosines
All options have sides 10 and 10. Since it's triangles we are talking about, the sides of length 10 have to be adjacent.
One of those sides can function as the base. Then, the other side of length 10 can rotate around either corner of the base. Since the goal is to maximize area and the base is already established as 10, the goal is focused in maximizing height, which is achieved when this other side of length 10 is in 90º degrees with the base.
Then, by Pythagoras the measure of the remaining side is 10*√2, which is ~ 14, and then 15 (option D) is the closest, making it the correct option.
(a previous step would consider checking that the options can actually build a triangle, but since in no option a side is bigger than the sum of the other two, we conclude that all options can be valid)
Did some quick calculus. Triangle with sides (10,10,2x) has area squared (by Heron's) of (10+x)*x*x*(10-x) = 100x^2 - x^4. Differentiating... 200x - 4x^3 has a meaningful zero at sqrt(50); so the biggest area is gonna be 10,10,2*sqrt(50). Triangle d is the closest to that... 2*sqrt(50) is slightly bigger than 14.
It’s a lovely alternative method. Just gotta make sure it is a global maximum at the point where the derivative vanishes, but this is quite obvious anyways
I love counting to three. Ten two three.
And for the answer, even if the other answer were closer to 14.14 or whatever, you can at least eliminate ones that are further away on each side... like you know it won't be 10,10,1 or 10,10,5 or 10,10,19 so you'd only have to do the actual math on 2 of them on either side.
Let x = the side that varies. Draw a perpendicular from that opposite angle to that side x. The resulting triangles are congruent to each other and each have a hypotenuse of 10 and one of its sides as x/2 since the original triangle is an isosceles triangle. So, the height of the original triangle is (100 - (x^2)/4)^0.5. So, A = 0.5*x*(100 - (x^2)/4)^0.5. You can find the area of each of these 4 choices by plugging in the values 1, 5, 10, 15, and 19 to find the answer. You can also take the derivative of the equation for A and solve that for x to get the one that has the maximum value for area and find out that the one with 15 is closer to that. dA/dx = (-x^2)*((100 - (x^2)/4)^(-0.5))/8 + ((100 - (x^2)/4)^0.5)/2 = 0; (x^2)/8 = (100 - (x^2)/4)/2; (x^2)/4 = 100 - (x^2)/4; x^2 = 400 - x^2; 2*x^2 = 400; x^2 = 200; x = +/-10*2^0.5; x = 10*2^0.5; x almost equals 14.14...; The answer with 15 is closest.
If you want to calculate the greatest area without knowing that formula, you can still use Pythagoras, just use a right-angled triangle with a hypotenuse of 10 and a horizontal side of half the variable figure. You can then calculate the height and thus the area. Eg. for an equilateral triangle you get a height of SQRT (10^2 - 5^2) or SQRT 75 and an area of 0.5 * 5 * SQRT 75. Repeat for all triangles.
I thought about that too. If you define x so that 2x is the variable side, the area of the triangle is equal to x.sqrt(10²-x²). From there it's simple.
This is also what I did, I plotted A(x)=√[10²-(x/2)²]⋅(x/2) and quickly saw that x=15 gave the highest area (out of the available options).
This was a pretty cool problem.
I used to suck at these problems back in sixth grade lol. Now I’m pretty fast at these with Heron’s formula in my head, although geometric inequalities also works. Trig also isn’t that bad.
Since they all have two equal legs (10, 10), you can use Pyhagoras: with varying side as base B, height H is √(100 - (½B)²). The surface is now ½•B•H. Since we can rule out a, b and e, we only do this for c and d.
Ex, for (c) this gives √(100-25) • 5 = 25√3. For (d) it's √(100-56.25) • 7.5 = which is clearly close to 7².
Pythagoras only applies if an angle of 90 degrees is involved, which is not the case here. (The only isosceles right triangle is 90-45-45, where the third side would be 10√2.)
Hi Steve, what a wonderful puzzle 🎉. Love it! Please keep it coming!!!
Thank you, I will
It is a totally broken argument. Not the distance from 14*sqrt(2) decides which one has the biggest area. A counterexample:
consider when the third side is 14, it gives area of A1=49.989999.
When the third side is 14.2842 then the area is A2=49.989806.
The A1>A2, but the 14 is farther from 10*sqrt(2) than 14.2842 as:
abs(14-10*sqrt(2))>abs(14.2842-10*sqrt(2)).
The decider is actually the difference between the third side's square and 200. An easier counterexample: 10,10,1 has positive area. 10,10,20 doesn't.
Totally broke might be a bit OTT! 😂 I agree that it is true that he has not definitively proven 15. But this is multiple choice. You use mathematical reasoning to find the correct answer in the shortest time. Personally I think (D) is fairly intuitive given the five options. I just calculated 10² + 10² = 200 and realised 15² was the closest to this. If the other nearby options of 10² or 19² were similar to 200 then more thinking would be required as you say. 😊
I visualized in my head what each looked like, used one of the 10's as an approximate base, and pictured what the altitude would also approximately be. For the first 2, 1 and 5 respectively. ~5 and ~25 for area. 3rd one is equilateral so that's 25√3. 5th is almost colinear, and going to be ~15. 4th one is the closest to a right triangle, and is close to 50. #4 it is. Now to see if I'm right...
I sort of used Heron's Formula. I threw out the 2's in the denominators and the radical sign because I knew that the biggest product without them would still be the biggest product with them.
I went through almost the exact same thought process, Nice
I used Heron's, but didn't go as far as completely calculating the area. Just a quick approximation of the product
Since all triangles are isosceles, you can bisect them into two congruent right triangles. If you let the third side of the original triangle be x, algebra with Pythagoras will quickly give you Area = (¹/₄)x√(400-x²). Dropping the constant coefficients and being a little familiar with maximisation problems you can immediately "see" the expression is maximised when x is equal to the square root term which gives x as 10√2≈14.14. If you don't have that insight, substituting each given option into x doesn't take that long.
Love not using Heron’s formula
The way I did it is that I formed a function for area, given the base size. To get this function, we multiply base (x) by the height of the triangle. Height can be worked out by pythagoras such that sqrt(10^2 - (x/2)^2) is the height. Therefore the area function a(x) = ( x sqrt(100 - (x^2)/4) )/2. I then found the base size in the function that returned the highest value of a(x); this base size will be the x-coordinate of the maximum of a(x). Differentiating a(x) gives us (200 - x^2)/2(sqrt(400 - x^2)). Solving for x when this derivative is equal to 0 gives +- 10sqrt(2) but we eliminate the negative value since in geometry, lengths are only positive. Therefore the base-length with the highest area is 10(sqrt(2)) where I know sqrt 2 is about 1.41 so therefore 10(sqrt(2)) is ~ 14.1, and the closest answer is 15
👍
Thats such a smart way to figure it out lol I was trying weird ways to turn it into a right triangle for each of them and struggling lol 😂❤
After a second look, I realized that this is just a standard high school optimization problem. No trig area function needed.
Start with 10,10,b. A = .5bh. Find h as a function of b (constraint) to get A(b) = .5b*sqrt(100 - b^2/4).
Set A '(b) = 0 to get b = 10sqrt(2) approx 14.1. Done
I find it interesting that you consider calculus and differentiation using the chain rule to be more straightforward than using a bit of trigonometry? I would think that this kind of trig formula would be precalculus? But perhaps that is just in my education?
As a follow up, I think some fairly young children given these options would reason out that 10,10,15 without the need for trig or calculus. The general idea that the area gets bigger but only to a point I think is fairly intuitive even at a young age? I'm fairly sure a lot of my 13 year old students would get the right answer here.
I think it would need a lot more options around the optimal solution to require us to fully solve using trig or calc?
@@72kyle All I said was this is a standard high school calculus optimization problem. This question wasn't intended for 13 yr olds. I don't know what math level of 13 year olds you are teaching, but I guarantee that if you were to give this to a typical Grade 11 class today, 90% of them would not have a clue. Does the U.S. even teach calculus in high school any more?
@@ianfowler9340 I don't teach in the US. I'm in the UK. Students here would learn that area of a triangle ½bh at some point in primary school (so before 10). They would learn trigonometry usually around 14 but the brightest learn it a little earlier. My current year 9 class (13 year olds) learnt some basic trig earlier this academic year. They would learn the ½absinC formula at around 15/16 along with cosine and sine rules. They start learning calculus at 16/17 and would be able to solve the way you did at around 17/18.
I'll try it out with the year 9s tomorrow see how they get on. I think the majority of them would get the question and pick either c or d. Be interesting to see their reasons. The ones who can visualise better will likely get it right. But you never know until you ask eh?
@@72kyle You are lucky as it seems you introduce a lot of your content a year ahead of us. Let me know how it works. Question. Is your group of grade 9s a random sample of the population or do they have to qualify somehow to get into your school?
@@ianfowler9340 Students are set by ability in maths. So this group is the top set of 4 groups. We are a state school in a deprived area so there is a huge range in ability. So in this group you have students who can do trigonometry and then in the bottom sets you'll have a student who can barely add up numbers. Kind of crazy how much they spread out by this point in their education! I agree that the bottom set wouldn't get the question at all! But for those of us watching this channel we were probably the 13 year olds that would have got it! 😊
Well that was much simpler than my basic geometry. Since each triangle is an isosceles triangle I put the variable length on the bottom (that’s the base). I cut base in half and then you have a right triangle with sides of h, base/2, and 10. Solve for h and put that into area calc of 1/2 b h. Set the derivative to 0 to find a maxima and a value for x.
astounding
Another way to solve that is imagine a circle of radius 10. The sides of length 10 are 2 line segments from the center to the edge of the circle. If you have A = (0,0), B = (10,0), and C is some point on the circle, you just need to calculate the point on the circle with the greatest change in y from B.
Since 2 sides are equal and the area of a triangle is base×½height, the largest area is the closest to a right triangle (in this case, that would be half a 10×10 square). The optimal diagonal is then base×sqrt(2). Sqrt(2) is close to 1.414, so the third dide should be the one closest to 14.14.
15 meets this criteria
Area of isosceles triangle would be 0.5(S^2 sinA), where S is the length of one of the equal sides and A is the angle between them
Area = 2 * halfArea = 2 * (0.5 * S sin A/2 * S cos A/2) = S^2 sin A/2 cos A/2 = 0.5 S^2 sin A
Clearly, the isosceles triangle will have maximum area if A==90degrees
therefore length of 3rd side for maximum area: 2 S sin A/2 = 2 S 1/rt(2) = rt(2) S
keeping S=10, we have third side = 14.14
as 15 is the closest, Answer should be (D) 10,10,15
I suspect the intent of the question was to determine whether the questioned knew the alternative forms of the Heron's formula, one of which is
Area = 1/4 Square root ( (4 a^2 b^2) - ( a^2 + b^2 - c^2 )^2 )
Given a and b are both 10, the problem reduces to finding the triangle with the smallest value for the term
( a^2 + b^2 - c^2)^2.
Reducing to
(100 + 100 - c^2)^2
(a) (100 + 100 - 1)^2 = 199^2 sides 10,10,1
(b) (100 + 100 - 25)^2 = 175^2 sides 10,10,5
(c) (100 + 100 - 100)^2 = 100^2 sides 10,10,10
(d) (100 + 100 - 225)^2 = -25^2 sides 10,10,15
(e) (100 + 100 - 361)^2 = -161^2 sides 10,10,19
Making (d) the triangle with the greatest area.
(The reason I suspected an alternative form of the Heron formula was the use of constant sides 10,10 for all the triangles. I suspected a short cut. )
An alternative way of stating this would be, the triangle with the least difference between the sum of the squares of the constant sides, 200, and the square of the varying side, 1,25, 100, 225, and 361 respectively, is the triangle with the greatest area.)
Heron’s formula solution:
Semiperimeter = 10+0.5x
Two sides of 10 -> 0.5x
Third side -> 10-0.5x
This is ((10+0.5x)(0.5x)(0.5x)(10-0.5x))^0.5 or 0.5x*(100-0.25x^2)^0.5
Formula simplifies to 0.25x*(400-x^2)^0.5
There’s probably some integral method, but you get the following:
1 -> sqrt(399) / 4
5 -> 25 sqrt(13) / 4
10 -> 100 sqrt(3) / 4
15 -> 75 sqrt (7) / 4
19 -> 19 sqrt(39) / 4
15 is the answer.
also can use Pythagorean theorem by dropping a perpendicular from the apex of the two identical sides to the odd side. the math to come up with the right answer only requires reasonable approximations.
That was fun.
I imagined a circle of radius 10, into which I imagined triangles with the secant 1, 5, 10, 15 and 19, and saw that the triangle with secant 15 is the closest to a perpendicular triangle, thus that one had to have the greatest area :)
But yes, betting the pythagoras to see it's the one that is closest to 10*sqrt(2) is more exact, as if the numbers given were 13, 14, 15, 16... then my visualization would not have worked due to the angle having low perceptual difference...
5 small diagrams and some approximate pythagoras calcs to get approx heights, and decided on 10x10x15....
1 - (third side)^2/200 is the cosine of the angle where the "10" sides meet
Smallest cosine = Biggest sine = biggest area
1 - 225/200 wins
Consider half the triangle: Base is half of number in third columns say X
Height is Y and hypotenuse is 10.
The locus of the right-angle apex is a circle with diameter 10.
Greatest altitude of triangle is when X=Y hence Maximum area is when X=Y
Thus 2X=10Sqrt(2), closest is 15.
A,B and E are obviously wrong
A 15,10,10 is definitely bigger then a 10,10,10 (just by thinking about what it would look like) so D
Did it in my head. a, b, and e are obviously wrong. A bit difficult to decide the remaining two. d is almost a right triangle with base 10 height 10. c is equilateral with base 10 and height something smaller than 10. They both have the same base, but d is taller. d is biggest.
Just to add a pointless observation, if the triangle has sides a, a, and b, then the area settles down to b * sqrt(a^2 - b^2 / 4) / 2. I was wondering how (10, 10, 13) compared to (10, 10, 15), but 15 has the edge by about .2.
*laughs in Heron’s formula*
Hmm. I used a part of the herons formula. The triangle with the greatest s-c would have the most area. When I checked, the d one had the greatest s-c, and it had the most area.
It’s not the nicest way as s = 10+0.5c is also a function of c, so is different for each option. But yes, it is certainly feasible
@@adw1z how man
@@aneeshbro s changes, so (s-a)(s-b) also change; so it’s not enough to just check s-c, it’s rather lucky that does end up being correct
One thing I somehow do not understand in English maths: How do you differentiate between x as a variable and x as an operator for multiplication? 🤔This is no joke...because in my country we use the • as multiplication operator.
x x x ≠ x^2
x • x = x^2 😉
Me: "I know it's probably d, but how do we be sure?"
Also me: "Oh, wait. 1.732"
10, 10, 15. 10 10 19 would be quite malnourished.
Random buzzwords trying to sound smart
Pythag
10^2 + 10^2 = C^2
200 = C^2
What squared is closest to 200?
15^2 = 225
10^2 = 100
Assume one of the 10s is the base.
Area is base * height. Because second side is 10 max high can at most be ,10
In that case rhe last side would be sqrt(2)*10. 15 is the closest to that so answer needs to be d
My dumbass would have said e. Then i realised the shape of the triangle is flattened compared to d.
One of the restrictions of youtube videos (not just yours) is that posting a hand written photo of a solution to a problem is not allowed. Writing out a solution in text is a real pain. Can anything be done to rectify that?? Just askin'.
🎉🎉🎉 i have solved this question in mind yeaaaaa . 🎉🎉🎉
10*10*10 is my intuitive guess
10,10,19 is sqrt(741/21) TIMES larger than 10,10,1 or about 5.94 times larger.
What about 10-10-321. Don't pay more for long distance.
If you pick a or b they give you a badge and a gun.
*That's easy, just use Heron's formula*
What is smaller
10 x 10 x 1
or
10 x 10 x 19
?
if u assume Triangle 1: a=10, b=10, c=1 ; Triangle 2: a=10, b=10, c=19
The anngle Theta (between a and b) is arccos((a^2+b^2-c^2)/2ab)
Trangle 1: Theta=0.10004 rad(ians) 5.732 deg, Trianle 2: Theta=2.50647 Rad(ians) 143.61 deg
and just calculate ( ab * Sin(Theta) ) / 2
Triangle 1 area = 4.99375 Units
Triangle 2 area = 29.66374 Units
..The answer is Triangle 2 (10,10,19)
..so wich triangle is bigger 1: 10,10,5 or 2: 10,10,10 ... ?
@@Patrik6920 close but you forgot to square the units 🥲
@@pluto9000 True lol ...
10,10,19 is sqrt(741/21) TIMES larger than 10,10,1 or about 5.94 times larger.
An "isosceles" triangle with lengths 10,10,10?
Every equilateral triangle is a special case of an isosceles.
So it is both.
@@BritishBeachcomber yes obviously. It is just peculiar to phrase it as such.
@@richardclegg8027 But beware of the obvious, especially in math. It can blind you to the truth.
Доказательство не строгое..
My approach :
let draw the height of the triangle to the base , since its an isoscele triangle the height will cut the base in half
let name the half base b and the height h so the area of the holl triangle is equal to b*h
and by the pythagrion theorm we have b^2 + h^2 = 10^2 = 100
so let x = b^2 and y = h^2
x + y = 100
wlog assume xi>=yi
i want to proof :
if x1-y1 < = x2 -y2
and xi + yi = 100
then x1y1 >= x2y2
first proof that 50 = xi + yi = 100
2xi >= 100
xi >= 50
now using these
(x1 + y1 = x2 + y2
x1 - y1
sorry the final result should be since 15 is the closet to 2*sqrt(50) then (10,10,15) is the answer
my dumbass split it into 2 triangles
Which is good because sqrts are easier to calculate than sine
why isn't e the largest because it has the longest sides?
It’s like you have to “open those two sides with 10 and results a triangle that’s more flat”
Cos it's base X height and the height is much shorter on the 19 triangle as it's more flat. Draw them and you'll see
oh i see thanks
C is the correct answer. You've over thought the problem
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