A Very Nice Geometry Problem | Math Olympiad | 2 Different Methods

A Very Nice Geometry Problem | Math Olympiad | 2 Different Methods
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[@guyhoghton399]

_|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_*

[@giuseppemalaguti435]

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 ..

[@plamenpenchev262]

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.

[@plamenpenchev262]

Forgot 2ab = 4S

[@sarvajagannadhareddy1238]

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

[@prossvay8744]

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.❤

[@sarvajagannadhareddy1238]

Dear, thank you.for letting know the values of a and b sides of triangle

[@sarvajagannadhareddy1238]

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

[@jimlocke9320]

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.

[@nunoalexandre6408]

Love 🎉❤❤❤

[@Antony_V]

AC=x+y=8, semiperimeter s of ABC=x+y+1, triangle area=s*radius=9. Red area=9-π

[@juanalfaro7522]

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

[@AmirgabYT2185]

S=9-π≈5,86

[@michaeldoerr5810]

I understood BOTH methods and both methods relied on a DIFFERENT circle method. And much better than the comments.

[@devondevon4366]

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

[@Buy_YT_Views.293]

This video is an absolute game-changer.

[@devondevon4366]

9 - pi

[@harrymatabal8448]

I always try to work it out first before i look st your solution. The original diagram did not have 90 degrees. 😢

[@MathBooster]

Don't try from thumbnail, click on the video and then pause the video. You will always get all the informations there.

[@tomripleyro]

How did you know in advance that x/2 will cancel with -x/2?

[@Grizzly01-vr4pn]

Any number - the same number = 0

[@tomripleyro]

@@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

[@brettgbarnes]

@@tomripleyro I would say it's a known technique he has used many times.

[@RealQinnMalloryu4]

(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)

[@danmimis4576]

Do you have to say "By Pythagoras Theorem" in every episode? Isn't it obvious that everybody already knows that?