Quick and cute summer Sangaku #7

Geometry Level 2

The diagram shows a red circle with radius 1 and a purple circle with radius 4 tangent to each other and to the same black line. We inscribe a light green circle tangent to the black line and to the two circles. We use the three tangency points to draw a triangle. At last, we inscribe a black triangle inside the bigger triangle. The black triangle has the smallest perimeter possible.

  • Question : The ratio between the blue area and the black area can be expressed as a b \boxed{\frac{a}{b}} where a \boxed{a} and b \boxed{b} and coprime positive integers
  • Calculate a + b \boxed{a+b}


The answer is 65.

Valentin Duringer by Valentin Duringer , 30, Belgium

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Valentin Duringer
Jul 14, 2020
  • First we need to get the radius of the green circle. We can use Descartes' Circle Theorem or the Pythagorean theorem : x \boxed{x} is the radius of the green circle:
  • α 2 = 4 x \boxed{α^{2}=4x} and β 2 = 16 x \boxed{β^{2}=16x}
  • Also α + β = 4 \boxed{α+β=4} < = > <=> 2 x + 4 x = 4 \boxed{2\sqrt{x}+4\sqrt{x}=4} < = > <=> x = 4 9 \boxed{x=\frac{4}{9}}
  • Now we can use coordinate geometry to calculate the coordinates of the 3 intersection points.
  • M ( x M ; y M ) \boxed{M(x_M;y_M)} is the intersection point between the green circle and the red circle
  • N ( x N ; y N ) \boxed{N(x_N;y_N)} is the intersection point between the green circle and the purple circle
  • O ( x O ; y O ) \boxed{O(x_O;y_O)} is the intersection point between the green circle and the black line
  • x 2 + ( y − 1 ) 2 − 1 = 0 \boxed{x^{2}+(y-1)^{2}-1=0} is the equation of the red circle
  • ( x − 4 ) 2 + ( y − 4 ) 2 − 16 = 0 \boxed{(x-4)^{2}+(y-4)^{2}-16=0} is the equation of the purple circle
  • ( x − 4 3 ) 2 + ( y − 4 9 ) 2 − 16 81 = 0 \boxed{(x-\frac{4}{3})^{2}+(y-\frac{4}{9})^{2}-\frac{16}{81}=0} is the equation of the green circle
  • y = 0 \boxed{y=0} is the equation of the black line
  • To determine the coordinates of M ( x M ; y N ) \boxed{M(x_M;y_N)} we need to equalize the equation of the red circle and the equation of the green circle : x M 2 + ( y M − 1 ) 2 − 1 = ( x M − 4 3 ) 2 + ( y M − 4 9 ) 2 − 16 81 \boxed{x_M^{2}+(y_M-1)^{2}-1=(x_M-\frac{4}{3})^{2}+(y_M-\frac{4}{9})^{2}-\frac{16}{81}} < = > <=> y M = 12 x M − 8 5 \boxed{y_M=\frac{12x_M-8}{5}}
  • Now we take one of the equations of the circles and substitute y M \boxed{y_M} with 12 x M − 8 5 \boxed{\frac{12x_M-8}{5}} and find x M \boxed{x_M}
  • x M 2 + ( 12 x M − 8 5 − 1 ) 2 − 1 = 0 \boxed{x_M^{2}+(\frac{12x_M-8}{5}-1)^{2}-1=0} < = > <=> x M = 12 13 \boxed{x_M=\frac{12}{13}}
  • To get y M \boxed{y_M} we use the equation y M = 12 x M − 8 5 \boxed{y_M=\frac{12x_M-8}{5}} < = > <=> y M = 8 13 \boxed{y_M=\frac{8}{13}}
  • We use a similar method to get the coordinates of N ( x N ; y N ) \boxed{N(x_N;y_N)} and O ( x O ; y O ) \boxed{O(x_O;y_O)} and we get :
  • M ( 12 13 ; 8 13 ) \boxed{M(\frac{12}{13};\frac{8}{13})}
  • N ( 8 5 ; 4 5 ) \boxed{N(\frac{8}{5};\frac{4}{5})}
  • O ( 4 3 ; 0 ) \boxed{O(\frac{4}{3};0)}
  • Now we can use the distance formula to get N M \boxed{NM} , N O \boxed{NO} and O M \boxed{OM} :
  • N M = 4 130 65 \boxed{NM=\frac{4\sqrt{130}}{65}}
  • N O = 4 10 15 \boxed{NO=\frac{4\sqrt{10}}{15}}
  • O M = 8 13 39 \boxed{OM=\frac{8\sqrt{13}}{39}}
  • By Heron's formula the area of M N O \boxed{MNO} is 16 65 \boxed{\frac{16}{65}}
  • Now we find the area of the black triangle. Since it must have the smallest perimeter possible, this triangle must be the orthic triangle. We then need to find the intersection points between the sides of the triangle and their relative altitude. We shall determine the equation of each side of the triangle M N O \boxed{MNO} . We use this general formula to obtain the equation of a line using two points A \boxed{A} and B \boxed{B} : y = y A + y B − y A x B − x A . ( x − x B ) \boxed{y=y_A+\frac{y_B-y_A}{x_B-x_A}.(x-x_B)}
  • The line of side MN is y = 3 x 11 + 4 11 \boxed{y=\frac{3x}{11}+\frac{4}{11}}
  • The line of side OM is y = − 3 x 2 + 2 \boxed{y=\frac{-3x}{2}+2}
  • The line of side ON is y = 3 x − 4 \boxed{y=3x-4}
  • Since an altitude is perpendicular to the side and the triangle, the slope of its line is inverse et opposite to the side's slope. Moreover, if the altitude passes trough the opposite vertex we can obtain the y intercept of the altitude's line.
  • The altitude of side MN has equation y = − 11 x 3 + 44 9 \boxed{y=\frac{-11x}{3}+\frac{44}{9}}
  • The altitude of side OM has equation y = 2 x 3 − 4 15 \boxed{y=\frac{2x}{3}-\frac{4}{15}}
  • The altitude of side ON has equation y = − 1 x 3 + 12 13 \boxed{y=\frac{-1x}{3}+\frac{12}{13}}
  • Now we can equalize the equation of the altitudes with their relative sides to obtain the coordinates of the intersection points, we get :
  • F ( 96 65 ; 28 65 ) \boxed{F(\frac{96}{65};\frac{28}{65})}
  • G ( 224 195 ; 44 65 ) \boxed{G(\frac{224}{195};\frac{44}{65})}
  • H ( 68 65 ; 28 65 ) \boxed{H(\frac{68}{65};\frac{28}{65})}
  • By Heron's formula the area of triangle F G H \boxed{FGH} is 224 4225 \boxed{\frac{224}{4225}}
  • Finally : 16 65 − 224 4225 224 4225 = 51 14 \boxed{\frac{\frac{16}{65}-\frac{224}{4225}}{\frac{224}{4225}}=\frac{51}{14}}
  • 51 + 14 = 65 \boxed{51+14=65}
0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...