Math probelm?

Question

Theres a tryangle one sie id 17in other 18in and the 3 one 10in. theres a circle in it as big as it can be whats its area

Answer 1

This sounds like a first question on a Putnam examination. The center of the circle in question lies on the intersection of the angle bisectors of the triangle. To convince yourself of this, consider that for any acute angle, a circle with center along its angle bisector can be made to be tangent simultaneously to both rays of the angle. Just adjust the radius the circle. By a transitive argument, where two angle bisectors of the triangle intersect, a circle centered at that point can be made to be tangent to all three sides of the triangle simultaneously.

I'll lay the longest side along the X axis, with a point at (0,0). Then the other point is (18,0), and the remaing point (x,y) I'll put in the first quadrant [x>0, y>0] so that it satisfies being distance 17 from (0,0) and distance 10 from (18,0). In algebraic form:

dist[ (x,y), (18,0) ] = 10
(x-18)^2 + y^2 = 100

and

dist[ (0,0), (x,y) ] = 17
x^2 + y^2 = 17^2

to solve simultanouesly, consider (-1) multiplied by the first equation plus the second equation to eliminate the y^2 term:

x^2 - (18-x)^2 - x^2 = 17^2 - 10^2 = 289 - 100 = 189

expanding and simplifying

x^2 - ( x^2 - 2*18*x + 18^2 ) = 189
x^2 - x^2 + 2*18*x - 18^2 = 189
2*18*x = 189 + 18^2 = 189 + 324 = 513 = 9 * 57
x = (9 * 57) / (2 * 18) = 57/4 = 3*19/4 = 14 + 1/4

now a point y along the vertical line x=57/4 satisfies being distance 17 from (0,0), so

dist[ (0,0), (57/4, y) ] = 17
57^2/4^2 + y^2 = 17^2
y^2 = 17*17 - (3*19)^2/4^2 = 17*17 - 3*3*19*19/4*4
y^2 = 289 - (203 + 1/16) = 85 + 15/16 = 1375/16 = 25*55/16
y = 5/4 * sqrt(55) =~ 9.270248

as a check, is the point (57/4, 5*sqrt(55)/4) of distance 10 from (0,18)?

(57/4 - 18)^2 + 25*55/16
= 57^2 / 4*2 - 2* 57/4 * 18 + 18^2 + 25*55/16
= 203 + 1/16 - 513 + 324 + 85 + 15/16 = 100 = 10^2, Yes.

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Now we need to bisect the angle { (57/4, 5*sqrt(55)/4), (0,0), (18,0) }. Since (57/4, 5*sqrt(55)/4) is distance 17 from (0,0), consider the line segment from (17,0) to (57/4, 5*sqrt(55)/4). Using vector notation, this is described as (17,0) + k[ 57/4 - 17, 5*sqrt(55)/4 - 0 ]. for k=0, we have (17,0); and for k= 1, we have (57/4, 5*sqrt(55)/4). the midpoint is where k= 1/2, so the midpoint from these two points of distance 17 from (0,0) is

(17,0) + 1/2[ 57/4 - 17, 5*sqrt(55)/4 ]
= ( 17+ 57/8 - 17/2, 5*sqrt(55)/8 ) = (17/2 + 57/8, 5*sqrt(55)/8)
= ( (17*4+57)/8, 5*sqrt(55)/8 ) = ( 125/8, 5*sqrt(55)/8 )

Thus the circle in question is along the line that contains both (0,0) and (125/8, 5*sqrt(55)/8), which can be expressed as

(0,0) + k( (125/8, 5*sqrt(55)/8)
= k(125/8, 5*sqrt(55)/8) = k/8( 125, 5*sqrt(55))

since k is a variable of proportionality, exchanging k with 8*j does not lose generality, so

= j( 125, 5*sqrt(55) ).

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now we need to bisect the angle { (57/4, 5*sqrt(55)/4), (18,0), (0,0) }. Since (57/4, 5*sqrt(55)/4) is distance 10 from (18,0), consider the line segment from (8,0) to (57/4, 5*sqrt(55)/4). Using vector notation, this is described as (8,0) + k[ 57/4 - 8, 5*sqrt(55)/4 - 0 ]. for k=0, we have (8,0); and for k=1, we have (57/4, 5*sqrt(55)/4). The midpoint is where k= 1/2, so the midpoint of these two points of distance 10 from (18,0) is

(8,0) + 1/2[ 57/4 - 8, 5*sqrt(55)/4 ] = ( 8 + 57/8 - 4, 5*sqrt(55)/8 )
= ( (4*8 + 57)/8, 5*sqrt(55)/8 ) = (89/8, 5*sqrt(55)/8)

Thus the circle in question is also along the line that contains both (18,0) and (89/8, 5*sqrt(55)/8), which can be expressed as

(18,0) + k(89/8 - 18, 5*sqrt(55)/8 )
= ( 18 + 89*k/8 - 18*k, 5*k*sqrt(55)/8 )

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To intersect these two lines, we have two linear equations in two unknowns. The X values satisfy

125*j = 18 + (89/8 - 18)*k

and the y values satisfy

5*j*sqrt(55) = 5*k*sqrt(55)/8
j = k/8

back substitute j=k/8 into the first equation

125*k/8 = 18 + (89/8 - 18)*k
k*( 125/8 + 18 - 89/8 ) = 18
k*( 36/8 + 18 ) = 18
k*( 9/2 + 36/2 ) = 18
k*( 45/2 )= 18
k = 36/45

This, the intersection of the two angle bisectors is the center of the circle that can be made to touch all three sides of the triangle simultaneuosly. By the second equation it is

(18,0) + (36/45)*( 89/8 - 18, 5*sqrt(55)/8 )
= ( 18 + (36/45)*(89/8) - 36*18/45, 5*36*sqrt(55)/(8*45) )
= ( 18 + (12*3*89)/(9*5*8) - (9*4)(9*2)/(9*5), 5*36*sqrt(55)/(8*9*5) )
= ( 18 + (89)/(5*2) - (9*4*2)/(5), 5*36*sqrt(55)/(8*9*5) )
= ( 18 + 89/10 - 144/10, 4*sqrt(55)/8 )
= ( 18 - 55/10, sqrt(55)/2 )
= ( 12 + 1/2, sqrt(55)/2 )

This is the center of the required circle. Its radius is equidistant and tangent to each of the sides of the triangle. Since we set the triangle so that one of the sides lies along the x axis, and the y value gives the vertical distance to the x axis, the radius of the largest circle is simply this y value, sqrt(55)/2. The area of the circle is pi * r^2 = pi * (sqrt(55)/2)^2 = 55*pi/4 = (55/4)*pi. Since the unit is inches, the area is more properly stated as (55*pi)/4 square inches.

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To verify that the same circle touches another side of the triangle, consider the line from (0,0) to (57/4, 5*sqrt(55)/4). In normalized form it can be expressed as

(0,0) + k/17( 57/4, 5*sqrt(55)/4 )

and a unit perpendicular vector is (1/17)*( -5*sqrt(55)/4, 57/4 ). The point from the center of the circle in question, of direction of that perpendicular and distance of the radius should lie on the line (satisfy the equation of the line).

(25/2, sqrt(55)/2) + (sqrt(55)/2)*(1/17)*( -5*sqrt(55)/4, 57/4 )
= (25/2 + (sqrt(55)/34)*( -5*sqrt(55)/4 ), sqrt(55)/2 + (sqrt(55)/34)*(57/4) )
= (25/2 + 55*(-5)/(34*4), sqrt(55)*( 1/2 + 57/(4*34) ) )
= ( (25*2*34 - 5*55)/(4*34), sqrt(55)*( (2*34 + 57 )/(4*34) ) )
= (1/(4*34))*( 25*2*34 - 5*55, sqrt(55)*( 2*34 + 57 ) )
= (1/(4*34))*( 1700 - 275, sqrt(55)*125 )
= (1/(4*34))*( 1425, sqrt(55)*125 )
= (25/(4*34))*( 57, 5*sqrt(55) )
= (25/34)*( 57/4, 5*sqrt(55)/4 )

it should be clear that the line is satisfied with k = 25/2.

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Having found the area of the largest circle that the triangle contains, I ask the following followup question: what is the area of the smallest circle that contains the triangle?

Answer 2

You want the area of the circle inscribed in the triangle. If p is the semiperimeter of the triangle and a,b, c are their sides, then the circle's radius is r = sqrt(p(p-a)(p-b)(p-c))/p. So, p =(17 + 18 + 3)/2 = 19
and r = sqrt(19*2*1*16)/19. And it's area is A = pi*r^2

Answer 3

78.5 inches