Sweden Math Olympiad | A Very Nice Geometry Problem
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- เผยแพร่เมื่อ 10 ก.ย. 2024
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By liberal application of the "2 tangents from a point" equality, and "the radius of a circle which intersects the circle at point X is perpendicular to the tangent line through point X", we can get the following:
1) Assign the length _a_ to the 2 tangents from point A. Assign the length _b_ to the 2 tangents from point B.
2) #1 results in assigning the lengths (2√5 - _a_ ) to the 2 tangents from point D, and (3√5 - _b_ ) to the 2 tangents from point C.
3) Construct the other 3 radii, each intersecting at the other 3 tangent lines of the circumscribed quadrilateral.
4) Note that each of these radii forms the altitude(s) of 4 pairs of identical right triangles, the corresponding bases of which are the lengths set out in #1 and #2.
5) Note that the sum of the areas of these 4 pairs of triangles constitutes the area of the quadrilateral, and that if we find that area ([ABCD]) and then subtract the area of the circle (4π), we will have the answer to the challenge.
6) For example, if, in the original drawing of this question, we designate the intersection point of the radius as point E, then, based on the assignments from #1 and #2, the area of △DEO would be given by:
2 • (2√5 - _a_ )/2, or just
(2√5 - _a_ )
Further, we recognize that there is an identical triangle formed by the radius that intersects segment AD of the quadrilateral. Therefore, the area of, let's call it the lower-left quadrant, is twice the area just calculated, or:
2 • (2√5 - _a_ )
7) Now let's extend that pattern to the other 3 quadrants, adding as we go, and we have (proceeding clockwise):
2 • (2√5 - _a_ ) + 2 • _a_ + 2 • _b_ + 2 • (3√5 - _b_ ), or
4√5 - 2 • _a_ + 2 • _a_ + 2 • _b_ + 6√5 - 2 • _b_ , or
10√5
8) Now, back to the assertion in #5, which is that the answer to the problem is given as the area of the quadrilateral minus the area of the circle, or
[ABCD] - 4π
We just calculated [ABCD] to be 10√5, so the answer is... drum roll...
10√5 - 4π
Now to watch the video to see if that's correct.
Cheers!
2a+2b+2(3√5-b)+2(2√5-a)= Área del trapezoide =10√5 → Área sombreada =10√5-4π ud².
Gracias y saludos.
Area of the quadrilateral = s*r where s is the semi parameter of the quadrilateral and r is the radius of the inscribed circle. Hence, the area of the shared area = area of the quadrilateral = (2|/5 + 3|/5)*2 = 10|/5 subtracted by the area of the circle = 4pi = 10|/5 - 4pi. Knock out!
ABCD → AB = AP + BP = y + x; BC = BQ + CQ = x + (3√5 - x); CD = CS + DS = (3√5 - x) + (2√5 - y)
AD = AT + DT = y + (2√5 - y); PO = TO = QO = SO = r = 2 →
yr + xr + (3√5 - x)r + (2√5 - y)r = 10√5 → shaded area = 10√5 - 4π = 2(5√5 - 2π)
Aq=(somma di4 triangoli)=2√5*2/2+3√5*2/2+AB*2/2+DC*2/2=5√5+(AB+DC)=5√5+5√5=10√5...Ared=10√5-4π
Yay, we think Alike : )
I was wondering if there was an entire other way to work it with a formula.
The answer is 2[5sqrt(5)-2pi]. I think that a there is a another property regarding right angles: no matter how different the sides are, if right angles can be formed according the circle's radius, the quadrilateral is divided into 4 triangles that always sum up to the two opposite sides. I could be wrong.
Usando o teorema de Pitot:
diz que um quadrilátero convexo ABCD é circunscritível se, e somente se, a soma dos lados opostos é a mesma.
A questão sai mais rápida.
Smart 👌
(5)2=25 (5)^3=125 {25+125}= {150 ABCD➖ 360STO°}= 210ABCDSTO/2= 100.5 ABCDSTO 10^10.5 2^52^5.5 1^1^2^1.5^1 2.1 (ABCDSTO ➖ 2ABCDSTO+1).
@@RealQinnMalloryu4 Not even Froyd can explain this madness! 🤣😂🤣😂
Тhe drawing with real dimensions does not correspond to what is shown in the conditions !