Math Olympiad | A Very Nice Geometry Problem | 2 Methods

[marioalb]

Area of right triangle:
A = 2A₁ + A₂ = 2(½bh)+h²
A = b.r + r² = 8.1+1² = 9 cm²
Shaded area :
A = 9 - πr² = 9-π1² = 9-π cm² ( Solved √ )

[ناصريناصر-سب]

The area of triangle ABC is P*r, where P is half the perimeter of the triangle and r is the radius of the circle drawn inside it. Hence, P=(8+8+2)/2=9, so the required area is 9-π

[alexniklas]

r=ab/(a+b+8)=(a+b-8)/2=1;
a+b=2+8=10;
ab=(a+b+8)r=(10+8)×1=18;
S=ab/2-πr^2=18/2-π= 9-π
Thanks sir

[jimlocke]

For completeness, we should compute the value of x to verify that it is a real, positive number. If it weren't, our solution would not be valid. At 7:15, we have found the area of ΔABC to be 9. Its sides are x + 1 and 1 + 8 - x = 9 - x, so [ΔABC] = (1/2)(x + 1)(9 - x) (1/2)(-x² + 8x + 9). However, [ΔABC] = 9, so (1/2)(-x² + 8x + 9) = 9, -x² + 8x + 9 = 18 and x² - 8x + 9 = 0. Solving the quadratic equation, x = 4 - √7 and x = 4 + √7. Both solutions are real and positive, so both are valid. The red area for each solution is the same.

Another solution approach: Let <OCQ = <OCS = Θ, then <SCQ = 2Θ. We get tan(Θ) = 1/(8 - x) and tan(2Θ) = (x + 1)/(9 - x). Using the tangent double angle formula, we can find a third order equation for x. Based on the above, 2 solutions should be valid and 1 solution must be discarded. Let's test the 2 solutions from above. x = 4 - √7 produces Θ = arctan(1/(8 - (4 -√7)) = approx. 8.56° and 2Θ = arctan((4 -√7 + 1)/(9 - (4 -√7)) = approx. 17.11°. x = 4 + √7 produces Θ = arctan(1/(8 - (4 +√7)) = approx. 36.44° and 2Θ = arctan((4 +√7 + 1)/(9 - (4 +√7)) = approx. 72.89°. So, the two solutions are valid to within the precision of my calculations.

[quigonkenny]

Let O be the center of the inscribed circle, M and N the points of tangency between circle O and AB and BC respectively, and P the point of tangency between circle O and CA. Let NC = x.

As AB and BC are tangent to circle O at M and N respectively, ∠BNO = ∠OMB = 90°. As ∠ABC = 90° as well, then ∠NOM must equal 90°, and as adjacent sides ON and OM equal 1, BNOM is a square with side length 1.

As NC and CP are tangents to circle O that intersect at C, CP = NC = x. As BN = 1, BC = x+1. As CA = 8, PA = 8-x.

As PA and AM are tangents to circle O that intersect at A, AM = PA = 8-x. As MB = 1, AB = (8-x)+1 = 9-x.

Triangle ∆ABC:
AB² + BC² = CA²
(9-x)² + (x+1)² = 8²
81 - 18x + x² + x² + 2x + 1 = 64
2x² - 16x + 82 - 64 = 0
2x² - 16x + 18 = 0
x² - 8x + 9 = 0
x = [-(-8)±√((-8)²-4(1)(9))]/2(1)
x = 8/2 ± √(64-36)/2
x = 4 ± √28/2 = 4 ± √7

x = 4 + √7 ≈ 6.65 | x = 4 - √7 ≈ 1.35

The latter fits the diagram better but we'll calculate for both, as both are viable answers.

AB = 9 - x = 9 - (4+√7) = 5 - √7
AB = 9 - x = 9 - (4-√7) = 5 + √7 ✓

BC = x + 1 = (4+√7) + 1 = 5 + √7
BC = x + 1 = (4-√7) + 1 = 5 - √7 ✓

And clearly the answers are symmetrical (as the same equations result if AM = x or if NC = x), so we will calculate assuming the latter result for x (4-√7), as the result will be the same either way.

The area of the shaded region will be equal to the area of triangle ∆ABC minus the area of circle O.

A = bh/2 - πr²
A = (5-√7)(5+√7)/2 - π(1)²
A = (25-7)/2 - π
A = 18/2 - π = 9 - π ≈ 5.86 sq units

[michaeldoerr]

The shaded area is 9-pi units squared. Both methods now seem familiar!!!

[murdock]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = AP + BP = (a - 1) + 1; BC = BQ + CQ = 1 + (b - 1)
8 > a > b; AC = 8 = AS + CS = (a - 1) + (b - 1); sin⁡(ABC) = 1
8 = √(a^2 + b^2) → b - 1 = √(64 - a^2) - 1 → (a - 1) + (b - 1) = 8 →
a + b = 10 = a + √(64 - a^2) → a = 5 ± √7 → a = 5 + √7 → b = 5 - √7 →
shaded area = (1/2)(5 + √7)(5 - √7) - π = 9 - π ≈ 9 - 22/7 = 41/7

[ina-jp]

(1+x)^2+(9-x)^2=64
1+2x+x^2+81-18x+x^2=64
2x^2-16x+18=0
x^2-8x+9=0
x=4+√7
影の領域の面積=(9-4-√7)(1+4+√7)/2-π
=(5-√7)(5+√7)/2-π
=18/2-π
=9-π
=5.858……

[sorourhashemi]

Thanks easy. (X+1)^2+(9-X)^2=64===>x=6.65. Area of triangle= 2.35×7.65/2=8.99~9. 9-π=the area of shaded region

[santiagoarosam]

(a+1)²+(8-a+1)²=8²→ a=4+√7→ AB=5+√7 ; BC=5-√7→ Area sombreada =(AB*BC/2)-π =[(25-7)/2]-π =9-π.
Gracias y un saludo.

[devondevon]

Answer 9 - pi
Let the height = p and the base = h

According to tangent theorem
p - radius + n- radius = hypotneuse
p - 1 + n -1 = 8

p+ n -2 = 8
p+ n =10

let square p+ n
p^2 + n^2 + 2np =100

p^2 + n^2 = 100 - 2np

p^2 + n^2=8^2 = 64 pythagoreasn

Hence,
64 = 100 -2np
36= 2np
np = 18
but the area of a triangle is base * height/2 = np/2 =18

area of the triangle =18

the area of circle is pi r^2 or pi * 1*1 = pi

Hence, the area of the shaded region = 9- pi. Final answer

[AndreasPfizenmaier-yw]

Is it possible to solve by Pythagoras BC^2+AB^2=AC^2 ?

[RealQinnMalloryu]

(1)^2 ➖ (8)^2={1+64}=65 180°ABCP/65=2.50ABCP 2.5^10, 2.5^2^5 1.1^2^1 2^1 (ABCP ➖ 2ABCP+1).