A Very Nice Geometry Problem | Math Olympiad | 2 Different Methods
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- Опубліковано 7 лип 2024
- A Very Nice Geometry Problem | Math Olympiad | 2 Different Methods
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Let BC=a ; AB=b
a^2+b^2=64 (1)
AC=AD+CD
8=b-1+a-1
So a+b=10 (2)
b=10-a
(1) a^2+(10-a)^2=64
a^2+100-20a+a^2=64
2a^2-20a+36=0
a^2-10a+18=0
a=5-√7
b=10-5+√7=5+√7
Shaded srea=1/2(5+√7)(5-√7)-(π)(1^2)=9-π.❤
Dear, thank you.for letting know the values of a and b sides of triangle
Dear,
circle + blank area = triangle area(9) =
Pi + (58+ square root of 2 - 4Pi)/8 =9
TRUE Pi can be derived from the above equation
x+y=8...(1+x)^2+(1+y)^2=64...x^2+2x+1+81+x^2-18x=64...x^2-8x+9=0...x=4-√7...y=4+√7...Atr=(5-√7)(5+√7)/2=9....Ared ..
Because r = 1, 2S = a + b + c (of triangle)
Square this gives (c=8):
a^2 + b^2 + 8^2 + 16(a + b) + 2ab = 4S^2
First two terms give 8^2.
a + b = 2S - 8
Then you have
S(S - 9) = 0, S = 9,
etc.
Forgot
2ab = 4S
Dear,
circle + blank area = triangle area(9) =
Pi + (58+ square root of 2 - 4Pi)/8 =9
TRUE Pi can be derived from the above equation
Love 🎉❤❤❤
AC=x+y=8, semiperimeter s of ABC=x+y+1, triangle area=s*radius=9. Red area=9-π
BC=X=1+a, AB=Y=1+(8-a) =9-a -->
(1+a) ^2 + (9-a) ^2 = 8^2 = 64 -->
1+2a+a^2 + 81-18a+a^2 = 64 -->
2a^2 - 16a + 18 = 0 --> a^2 - 8a + 9 = 0 -->
a = [8 +/- sqrt (8^2 - 4*9)]/2 = [8+/- sqrt (28)]/2 = [4+/- sqrt (7)] = 4-sqrt(7) since a is the smallest tangent. The other one, b = 8-a = 4+sqrt (7) --> X = 1+4-sqrt (7) = 5-sqrt (7), Y = 1+4+sqrt (7) = 5+sqrt (7) --> [ABC] = X*Y/2 = (5+sqrt (7)) * (5-sqrt (7)) /2 = (25-7)/2 = 18/2 = 9 sq. units
This video is an absolute game-changer.
S=9-π≈5,86
Math Booster solves the problem without computing the lengths of sides AB and BC! In method #1, the two sides have lengths (x + 1) and (7 - x), the hypotenuse has length 8 and the Pythagorean theorem can be applied to produce an equation that is solved for x (x = 4 + √7). In method #2, prossvay8744 did compute the lengths a and b as (5 - √7) and (5 + √7). The radical √7 goes away when the product is taken to compute the triangle's area, leaving an integer for the area.
I understood BOTH methods and both methods relied on a DIFFERENT circle method. And much better than the comments.
_|AP| = |AS| = x |BP| = |BQ| = y_
_x + y = 8_ (side _AC)_
⇒ _(x + y)² = 8²_
⇒ _x² + y² = 64 - 2xy_
_(x + 1)² + (y + 1)² = 8²_ (Pythagoras)
⇒ _x² + y² + 2(x + y) + 2 = 64_
⇒ _64 - 2xy + 2(8) + 2 = 64_
⇒ _xy = 9_
Area of *_ΔABC_*_ = (½)(x + 1)(y + 1) = (½)(xy + x + y + 1) = (½)(9 + 8 + 1) = _*_9_*
Answer 9 - pi
Let the longest base of the triangle = n
and the shortest base = p
then n^2 + p^2 = 8^2 = 64 Equation 1
Construct a square from the radius of the circle by drawing a vertical and horizontal line from the circle's center to
each base of the triangle. Hence, the length of the square = 1. Hence, the distance above the square on the longest base =n-1
and the distance from the square on the shortest base = p-1
Hence, the length of the hypotenuse = (n-1) + (p-1) (circle theorem)
Hence, n-1 + p-1 = 8
n+ p -2 =8
n+ p = 8+2
n+ p =10 Equation 2
(n+p)^2 = 10^2 square on both sides
n^2 + p^2 + 2np = 100 Equation 3
64 + 2np =100 substitute the value of Equation 1 into Equation 2
2np = 100 -64
2np = 36
np =18
np/2 = 9, so the area of the triangle = 9
since the radius of the circle =1 , then its area = 1^2 pi = 1 pi = pi
Hence, the area of the shaded region = 9 - pi Answer
Though Math Booster didn't ask to find the sides of the rectangle, I did so to verify if the answer was correct
n =7.64575 and p=2.35425
and 7.64574 + 2.35425 = 10
and 7.64574 * 2.35425 = 18
and 7.64574^ 2 + 2.35425^2 = 64
I always try to work it out first before i look st your solution. The original diagram did not have 90 degrees. 😢
Don't try from thumbnail, click on the video and then pause the video. You will always get all the informations there.
9 - pi
How did you know in advance that x/2 will cancel with -x/2?
Any number - the same number = 0
@@Grizzly01-vr4pn I mean, why did you calculate the area of the 3 triangles as a function of x? Did you know in advance that x/2 and -x/2 will cancel from the calculation? Thanks
(1)^2A/O/Coso° =1A/O/Coso°(8)^2^ A/O/Tano° =64A/O/Tano° {1A/O/Coso°+64A/O/Tano°=65A/O/Coso°Tano° 180°/65A/O/Coso°Tano° =2.50A/O/Coso°Tano° 2^1.2^25 2^1.2^2^5^5 1^1.1^21^1 2^1 (A/O/Coso°Tano° ➖ 2A/O/Coso°Tano°+1)
Do you have to say "By Pythagoras Theorem" in every episode? Isn't it obvious that everybody already knows that?