Math Olympiad | A Very Nice Geometry Problem
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- Опубліковано 22 чер 2024
- Math Olympiad | A Very Nice Geometry Problem
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The 1st solution:
A = 6 * 8 = 48
Now, the area will be maximum if we have 4 equal rectangels.
48: 4 = 12
The 2nd solution:
8:2 = 4
6:2 = 3
A = 4*3 = 12
The 3rd solution:
Pythagorean triangle.
6, 8, 10
The max area will be if we have two equal area
So
three sides divided by 2
3, 4, 5
Area = 3*4 = 12
There can be an another approach,
AB=6, BC=8, So, AC=10
ABC~ADE, AD=6x, DE=8x (x is +ve)
So, AE=10x
ABC~EFC, EF=6y, FC=8y (y is +ve)
So, EC=10y
Therefore, 10(x+y)=10 or, x+y=1
So, the max value of xy be ¼.
DEFB= 6x.8y= 48xy
max DEFB= 48×¼= 12
Sir, please give me a feedback, if you like it...❤
Great solution using calculus 😊 However, as this fuction is a quadratic, no differentiation is needed. Just rewrite the fuction in vertex form. Using the canonical way of choosing coordinates (x horizontal, y vertical, (0,0) at B) we get
A(x) = -0.75x² + 6x = -0.75(x² -8x + 16 - 16) = -0.75( (x - 4)² - 16) = -0.75(x - 4)² + 12.
So the vertex is (4, 12) and therefore the maximum area is 12 which corresponds to a side length x of 4.
An excellent solution using calculus! An approach for those who have not studied calculus. We find early on that ΔADE and ΔEFC are similar, with height equal to 3/4 of the base. A good educated guess might be that the base and height of the rectangle are half the base and height of the triangle, so b = 4, h = 3 and area = bh = 12. A quick check finds that the similar triangles have the same base and height as each other, thus are similar (but are also congruent), so this case is valid. Now let's increase the base of the rectangle by a small amount, let's say 0.001. CF has decreased by 0.001, so ΔEFC's height must decrease by (3/4)(0.001) to maintain the same ratio for similarity, so FC = 2.99925 and the area of the rectangle is 11.99999925. Now let's decrease the base of the rectangle by 0.001, leaving 3.999. The base of ΔEFC is 4.001, so its height is 3.00075 and the rectangle area is again 11.99999925. So, either adding to the base or trimming it by this small amount reduces the area. We have strong evidence that our educated guess is the correct answer. We can try even smaller increments if we like.
Is this saying that the max rectangle dimensions for any right angled triangle will be in the same ratio as the base and height of that triangle?
4 *3 = 12
A different approach
Let the length of the rectangle = x, then AD = (6/8 )x since all the triangles are similar.
Hence, x + (6/8)y = 8 (since 8 is the longest leg of the triangle)
Let the width of the rectangle = y, then FC = (8/6) y
Hence, y + (8/6)x = 6 (since 6 is the shortest leg)
x + (6/8)y = 8
y + (8/6)x = 6
-------------------------
(14/8)x + (14/6)y = 14 Add both equations
7/4 x + 7/3y = 14
21x + 28 y = 168
3x + 4y = 24
This is a Linear Diophantine equation
The rectangle will have the Maximum area when the length is close to the width. I.e., why a square has a maximum area because the difference between its length and width =0
Using Congruences to solve the linear diophantine equation.
3x + 4y = 24 mod 3
0 + y = 0' mod 3
y =0 mod 3
y = 0 + 3k
y =3k
3x + 4(3k) = 24 substitute y=3k into the original equation (3x + 4y =24)
3x + 12 k = 24
3x = 24-12k
3x = 3 (8- 4k)
x = 8 -4k
Start with 0,
when k =0 x=8 and y =0
when k =1 x =4 and y= 3, only this one can work, which has two positive values
when k =2 x=0 and y =6
when k=3 x= -4 and y=9
x and y cannot be zero or negative
So, the answer is when k =1 as y=3k = y=3, and x = 8 -4k = 8-4(1) = 8-4 =4
so the Answer when x =4 , and y =3 , the area of rectangle is at the its maximum
4 * 3 = 12 Answer
Let DB = y and BF = x. Area of rectangle BFED is xy. As the rate of change of the position of E is -6y for every 8x, the slope is -6/8 = -3/4. The y-intercept is at y= 6, so the equation for CA is:
y = 6-3x/4
Substituting into the equation for the area gives us:
Aᵣ = xy = x(6-3x/4)
Aᵣ = 6x - 3x²/4
As the area is clearly at 0 when x = 0 and when x = 8, and positive between, so we can take the derivative of the equation and where it is zero within that range will be the maximum:
Aᵣ = 6x - 3x²/4
Aᵣ' = 6 - 6x/4 = 6 - 3x/2
6 - 3x/2 = 0
3x/2 = 6
x = 6(2/3) = 4
y = 6 - 3x/4 = 6 - 3(4)/4
y = 6 - 3 = 3
Rectangle BFED:
Aᵣ = xy = 4(3) = 12 sq units
It's a quadratic formula, which is symmetric, so the maximum is halfway: 3*4 = 12
The formula for A is formally nonsymmetric (i.e., the axis of reflexion is not x=0) so how do you find the axis of symmetry? You could figure A = 0 = x(6 - 3x/4) @ x=0 or 8. Or, just notice that at x=0 and and x=8 the rectangle has a size of 0. So max(A) is when x=4.
we can prove this just by proving that AE=EC and we get 6-b =b and 8-a =a
This is another example easier than it looks and that was facillitated by HL similarity and simple Calculus. And explained MUCH BETTER than the comments. I shall practice that in order to feel better abt geometry and Calculus!!!
If AD = x, DE = BF = y, then BD = 6- x, FC = 8- y, ...=> x / y = (6-x)/ (8-y), ...=> y = 4x/3, S = y (6-x), ...=> S(x) = 4x(6-x)/3, ...=> Smax = S(3) = ...= 12. Thanks for the beautiful assignment!
We can solve by using AM-GM inequality.
I expect most people will look at this, figure the equation of the line AC, and find the maximum of xy.
Here, you do effectively that, except you swap the normal axes. A bit confusing; was that the intention?
So it's 1/2 the ABC area. Another way to phrase the problem: "Prove that the rectangle's area is Max when DF || AC or, even funnier: when BE = AC/2
dA/dx is 0 even for minimum area
In this problem, x or y can tend to 0 so minimum area is not possible. That's why dy/dx will be 0 for maximum area.
d2A/dx2 is negative for x = 3, so it is maximum.
12 ( 4*3)
φ = 30°; ∆ ABC → sin(ABC) = sin(3φ) = 1; AB = AD + BD = (6 - x) + x; BC = BF + CF = y + (8 - y)
AC = AE + CE → sin(FED) = 1 → DE = BF = y → BD = EF = x; FCE = DEA = δ →
tan(δ) = (6 - x)/y = x/(8 - y) → y = (4/3)(6 - x) → xy = g(x) = 8x - 4x^2/3 →
dg(x)/dx = 8 - 8x/3 = 0 → x = 3 → y = 4 → xy = 12; tan(δ) = 3/4
6/2*8/2 -- 12
🎉
4 *3 = 12
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