Spain Math Olympiad | A Very Nice Geometry Problem
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- Опубліковано 14 чер 2024
- Spain Math Olympiad | A Very Nice Geometry Problem
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From Figure mentioned by Mathbooster
length of OE=2*OC Cos70°=10Cos70°=OP(as angle OEP=angle OPE=70°).
Area of shaded region=Area of (∆BOD-∆EOP)=1/2*5*5*Sin40°-1/2*10Cos70°*10Cos70*Sin40°=25/2 sin40°(1-4Cos^2 70°);
•Let us simplify trigonometric expression sin40°(1-4Cos^2 70°)=
2Sin20°Cos20°(1-4sin^2 20°) and now put 1=cos^2 20°+Sin^2 20°
Following expression becomes
2Sin20°Cos20°(Cos^2 20°-3Sin^2 20°)=
2Sin20°(Cos^3 20°-3Cos20° Sin^2 20°)=
[ Cos3§=Real parts of(Cos§+i Sin§)^3=Cos^3 §-3Cos§Sin^2 §(From De -Moivre theorem).]
2Sin20°(Cos60°)=Sin20°.
Area =25/2 Sin20°
Remark :
This expression could also be simplified in following way :
2Sin20°Cos20°(1-4sin^2 20)
=2Sin20°Cos20(1-4(1-cos^2 20°)
=2Sin20°Cos20(4Cos^2 20°-3)
=2Sin20(4Cos^3 20°-3Cos20°)[from well known formula of Cos 3§=4Cos^3 §-3Cos§)]
=2Sin20°(Cos60°)
=Sin20°.
One thing that might be useful for math Loving people
That Cos3§=Cos^3 §-3Cos§ Sin^2 §...
& Sin3§=. -Sin^3 §+3Cos^2 § Sin§.
As ∠COE = ∠OEC = 30°+40° = 70°, ∆ECO is an isosceles triangle and EC = OC = 5. ∠ECO = 180°-(70°+70°) = 40°. As OB = OD = 5 and ∠DOB = 40°, ∆DOE is congruent with ∆ECO and ∠OBD = ∠BDO = 70°.
OE² = EC² + OC² - 2(EC)(OC)cos(40°)
OE² = 5² + 5² - 2(5)(5)cos(40°)
OE² ≈ 50-50(0.766) ≈ 50 - 38.302
OE ≈ √11.698 ≈ 3.420
BD = OE ≈ 3.420 as ∆DOB and ∆ECO are congruent.
Let P be the intersection point of EC and OD. As ∠POE = 40° and ∠OEP = 70°, ∠EPO = 180°-(40°+70°) = 70°. ∆POE is therefore isosceles and similar to ∆ECO and ∆DOB. OP = OE ≈ 3.420.
The area of shaded area PEBD is equal to the area of ∆DOB minus the area of ∆POE.
[PEBD] = OB(OD)sin(40°)/2 - OE(OP)sin(40°)/2
[PEBD] ≈ (sin(40°)/2)(5²-3.420²)
[PEBD] ≈ (0.643/2)(25-11.698)
[PEBD] ≈ (0.321)(13.302) ≈ 4.275 sq units
Solution in exam answer format:
1. Let P be the intersecting point of lines CE and OP.
2. Angle OCE = 180 - 30 - 40 - 70 = 40 (angle sum of triangle)
3. Hence triangle OCE is isosceles (base angles = 70). Hence OC = EC = radius = 5.
4. Triangles OCE and OBD are congruent. (SAS: OC = EC = OD = OB = radius, inclusive angles = 40)
5. OE = (2)(5)sin 20 (Use perpendicular line form C to OE as angle and line bisector for isosceles triangle OCE.)
6. OP = OE = (2)(5)sin20 (Triangle OEP is isosceles as base angles = 70)
7. Area of triangle COP = (1/2)(OC)(OP)sinOCP = (1/2)(5)(2)(5)sin20sin30 = (25/2)sin20
8. Equal area of congruent triangles OCE and OBD - area of triangle OPE = equal area => area of triangle COP = area of DBEP.
Hence area of DBEP = (25/2)sin20.
Shaded area is the difference of areas of two Isosceles triangles
Area of ODB is
1/2*5*5*sin(40)
Area of OEF is
1/2*|OE|*|OF|*sin(40)
OCE is isosceles so let draw the height CG
From sine in right triangle CGO we have that
sin(20) = |OG|/5
sin(20) = |OE|/10
|OE| = 10sin(20)
Angle OFE = 180 - 40-70 = 70
so triangle OFE is isosceles
Triangle OFE is isosceles and angle OEF = angle OFE so
|OF| = |OE|
1/2*5*5*sin(40) - 1/2*100*sin(20)^2*sin(40)
1/2*25*sin(40)*(1 - 4sin(20)^2)
I do not know why but something went wrong in my solution
But If we rewrite my solution above we will get
25/2*(sin(40)-4*sin(40)*sin(20)^2)
25/2*(sin(40)+2(-2sin(40)*sin(20))sin(20))
25/2*(sin(40) + 2(cos(60) - cos(20))sin(20))
25/2*(sin(40) + 2*1/2*sin(20) - 2*cos(20)sin(20))
25/2*(sin(40) + sin(20) - sin(40))
25/2*sin(20)
I expected other solution but it seems to be correct
^OEP =70°
^OBD =70°
PE //BD.
∆OEP simular ∆OBD
OE = BD
At 3:45, Math Booster has found that ΔCOE and ΔOBD are congruent. Drop a perpendicular from C to OB and label the intersection F.
Nessa questão, deveria dar o valor aproximado do sin20°, fazia mais sentido. Uma vez que não conseguimos expôs o sin20° em termos de radicais.
This is something that I never thought of. That the are of a trapezoid can be computed by computing the area of the triangle that shares a vertical angle. If that is the case, what other similar problems are there may I ask?
Área de um trapézio? Você falou? Você deve olhar a definição do que é um trapézio. Pois, o que foi calculado foi a área de um quadrilátero que é equivalente em termos numéricos a área do triângulo.
I agree. I mis-spoke. I probably should have just said rectangle. Tbh, the quadrilateral looked like a trapezoid. No offense. And I agree with what was calculated. May I ask what part of my comment struck you as obtuse?@@imetroangola4943
I could have just said, quadrilateral. Based on the thumbnail, I thought it looked like a trapezoid. And agree with what was the lesson.
Great.
Rabbit out of the hat 😊