A Very Nice Geometry Problem | You should be able to solve this! 

If a,b,c,d is a fibonacci-type sequence, then ad,2bc,(b^2+c^2) will always be a Pythagorean triple whose incircle radius is ab. In particular, if 'a' is odd and 'a' and 'b' are co-prime, the triple will be fundamental. So, we can make the sequence 5,2,7,9 ----> 45,28,53 as our fundamental triple and 45+28+53= 126. Note that the other possible sequence for a fundamental triple is 1,10,11,21 ---> 21,220,221 has a perimeter of 462. The sequences 2,5,7,12 and 10,1,11,12 produce non-fundamental triples 24,70,74 (perimeter 168) and 120,22,122 (perimeter 264).

7:17 c=53; a m=7; n=2 b=m²-n²=49-4=45 a=2mn=2•7•2=28 😁

73 , 45, and 28 Let vertical line of the triangle = a then from the point of tangency to the top of the triangle is a - 10 and the horizontal = b then from point of tangency to the vertex of the triange is a- 10 then the hypotenuse = (a + b -20) tangent circle theores Hence, the perimeter of the triangle = a + b + a+ b -20 Hence, 2 a + 2b - 20 =126 2a + 2b = 146 a + b =73 Hence, c= 53 (126-73) one get 53, it is basically over given Pythagorean triple 45, 28, and 73 a^2 + b^2 + 2ab = 73^2 (square a + b =73) a^2 + b^2 =53^2 ( Pythag) 53^2 + 2ab = 73^2 2ab = 73^2 - 53^2 2ab =2520 ab = 1260 a = 1260/b Since a + b = 73, then 1260/b + b -73=0 1260 + b^2 - 73b=0 (b-28)(b-45) = 0 b =28 and b=45

Semiperimeter * radius(r) = Area(A) Semiperimeter = 126 /2 = 63 A= 63 * 10= 630 Area of triangle = ½a*b= 630 ===> a*b= 1260 ..... (1) a+b+c=126, a+b-c=20(2r) ===> a+b = 73 ===> b = 73 - a ......(2) c = 126 - 73 = 53 from eq. 1 and 2: a * (73 - a) = 1260 =====> a = (28,45) , b =(45,28)

円の半径などいらない。 3辺とも互いに素な自然数になるなら原始ピタゴラス数を求めるのと同じだ。 ３辺は m^2ーn^2 , 2mn , m^2＋n^2 とおける。( m(>n) , nは自然数 ) 周の長さが126より　(m^2ーn^2) ＋ (2mn) ＋ ( m^2＋n^2)＝126 整理すると m(m＋n)＝63 ｍ,ｎは自然数だから ｍ＝7 , m+n＝9 ∴n=2 よって３辺は　45 , 28 , 53

Hello I am actually a little confused at the 10:43 mark you have 2(a^2-73a+126*10). I think that you forgot to take the 2 out of 126. That would leave you with 2(a^2-73a+63*10) which is equivalent to 2a^2-146a+126*20. I think that that would have made this calculation quicker. I could be wrong.

a+b+c=126, a+b-c=20(2r) ===> a+b = 73 ===> b = 73 - a ===> c= 126 - 73 = 53 b² + a² = c² ===>(73-a)² + a² = 53² ====> (a,b,c) = ( 28,45, 53) or (45, 28, 53)

*Outra maneira:* ab/2=(a+b+c)R/2→ab=1260. Por Pitágoras: a²+b²=c². Por outro lado, (a+b)²=a²+b²+2ab. Como a+b+c=126→a+b=126-c e ab=1260. Logo, (126-c)²=c²+2×1260. Daí, 126²-252c+c²=c²+2520→ 252c=15876-2520=13356 c=13356/252→ *c=53.* Assim, a+b=126-53→a+b=73 e como ab=1260. Facilmente, resolve pela fórmula de equação do segundo grau. Você vai encontrar *a=28 e b=45.*

(10)^2= 100 {18°A+18°B+90°C} =126°ABC {126°ABC/100} =1.26ABC 1^1.2^13 2^13^1 2^1^1 2^1 (ABC ➖ 2ABC+1).

На этот раз все в принципе завязано на Пифагоровой триаде 28-45-53! Легко также вывести и применить формулу 2r = a + b - c. СПАСИБО за красивые задания!🖕

28, 45, 83

❤❤❤amazing math

10+10+x+x+y+y=126 2(10+x+y)=126 10+x+y=63 x+y=53 (10+x)²+(10+y)²=(x+y)² 100+20x+x²+100+20y+y²=x²+2xy+y² 200+20x+20y=2xy 200+20(x+y)=2xy 1260=2xy xy=630 x²+y²+1260=2809 x²+y²=1549 　　　　　　　　 (x-y)²=1549-1260=289 x-y=17 2x=70 x=35 y=18 BC=10+x=10+35=45 AB=10+y=10+18=28 AC=x+y=35+18=53