Russian Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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- Опубліковано 6 кві 2024
- Russian Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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I appreciate your methods : I probably wouldn't have found either. Still, I think you could also say:
sin(t) = (b/2)/5 = b/10, (t = theta), Eq. 1
sin(2t) = b/6 Eq.2
[sin(2t)]/sin(t) = [2sin(t)cos(t)]/sin(t) = 2cos(t) =[ b/6]/[b/10] = 5/3, so cos(t) = 5/6,
then a = 5(5/6) = 25/6, and
(b/2)^2 = 5^2 - (25/6)^2 =(900 - 625)/36, = 275/36 , so b/2 = [5{sqrt(11)}]/6,
Area = ab = (25/6)[5{sqrt(11)}/3] = [125 sqrt(11)]/18
Even if in my opinion the trigonometric solution (BC = 6*sin 2theta, BC/2 = 5*sin theta) is better, here it is a non trig. solution:
If from point F we trace the perpendicular FG to line DE we get two congruent right triangles: DFC and DFG for ASA criterion. Then :
FC = FG = X
DG = DC
If we draw a circle with center in F and radius = FC, this circle will be tangent to line DE in point G and for the two tangent theorem GE = BE = Y
Now DG + GE = DE
DG + Y = 6 => DG = DC = 6 - Y
So we are able to write two pythagorean identities:
On DFC:
5² = X² + (6 - Y)² => X² = 12Y - Y² - 11
On DEH (EH being the perpendicular to line DC)
6² = (2X)² + (6 - Y - Y)² => 4X² +4Y² - 24Y = 0
substituting the first equation in the second
48Y - 4Y² - 44 + 4Y² - 24Y = 0
Y = 11/6
DC = 6 - 11/6 = 25/6
X = (5/6)√ 11
BC = 2*5/6√ 11 = 5/3√ 11
Area rectangle = 25/6 * (5/3)√ 11 = (125/18)√ 11
Try simpler: sin∆=x/5 and sin(2∆)=2x/6=x/3
sin(2∆)=2sin∆cos∆
Therefore: 2(x/5)cos∆=x/3
x and x cancel out, so cos∆=5/6 and ∆=33.9°
Knowing that you may find directly 2x and a.
I think that the problem will be solved much faster if you drag the parallel of BC from E. Then using the same equations you' ll come up to ab=125(sqrt(11))/18!
Если продлить EF до пересечения с продолжением DC (пусть это точка P) то треугольник EDP равнобедренный - в нем DF биссектриса и медиана одновременно (очевидно, что EF = FP). То есть DF перпендикулярно EF, треугольники DEF и DFC подобны, откуда сразу находится DC, а затем FC, что полностью решает задачу. Что-то из этого есть в ролике, но там проще.
Square....right angel triangle form by drawing lines...there after triplets...then sin area formula
3rd method
FC = 5Sin@ (@=theta)
therefore
BC = 10sin@
by the way
BC also = 6sin2@
so equation 10sin@ = 6sin2@
and sin2@ = 2sin@cos@
so 10sin@ = 12sin@cos@
there fore cos@ = 5/6 sin@ = root11/6
answer = 5cos@ x 10sin@ = 250 root11 / 36
Good explanation sir
Let AB = x and DA = y. From the diagram we can see the following:
sin(θ) = (y/2)/5 = y/10
cos(θ) = x/5
sin(2θ) = y/6
But we can also derive sin(2θ) from sin(θ) and cos(θ).
sin(2θ) = 2sin(θ)cos(θ)
y/6 = 2(y/10)(x/5)
y/6 = 2xy/50 = xy/25
y(1/6) = y(x/25)
x/25 = 1/6
x = 25/6
And we can determine y from the formula relating cos²(θ) and sin²(θ):
cos²(θ) + sin²(θ) = 1
(x/5)² + (y/10)² = 1
y²/100 = 1 - (5/6)² = 1 - 25/36
y² = (11/36)100 = 275/9
y = √(275/9) = 5√11/3
A = xy = (25/6)(5√11/3)
A = 125√11/18 ≈ 23.032
Extend EF and let P the intersection point of EF and DC.
Obviously orthogonal triangles BEF, FCP are equals => EF=FP
In triangle DEP : DF is median and angle bisector at the same time =>
DEP is isosceles => DF is height , so EF²=ED²-DF²=36-25=> EF²=11
Let CF=x=BF , CP=y so DC =6-y
x²+y²=FP² => x²+y²=11 (1)
In orthogonal triangle DFP , FC is heigh => FC²=DC*CP => x²=(6-y)y =>
x²+y²=6y => 6y=11 , cause (1) . So y=11/6
I can easily estimate x from (1) => x=(5√11)/6
So area of ABCD = DC*BC=(6-y)2x= …….. =(125√11)/18
(250/36)vʼ11=(125/18)/vʼ11 !!!
S = (5 * cos(th)) * (2 * 5 * sin(th))
IlIndicando con t l'angolo theta, il metodo più semplice consiste nel calcolare t con l'equazione 6* sin(2t)=2*5*sin (t); svolgendo con le formule di duplicazione emerge per t il valore di 30° per cui b vale 5 ed a= 5*cos (t) = 25/6; poiché 5*25/6 è diverso dal valore calcolato dall'espositore ci deve essere qualche errore di calcolo.
cos(t) = 5/6 , a = 25/6
sin(t) = {sqrt(11)}/6, b = 2*5* {sqrt(11)}/6.
I don't know where the 'valore di 30 [degrees]' is from, but I see that cos 30 =[sqrt(3)]/2 is close to 5/6;
arccos(5/6) = 33.5573 degrees approx.(according to calculator)
125*sqrt(11)/18
24,095 degrees!
Hi sir.I would like to ask if there is any way to send math problems to you?I hope our self-created questions can be published or discussed.
You can ask the question in the comment section. I will definitely solve the problem, if the problem have good thumbnail.
What if I have some geometry questions to ask?I think this is difficult to express through just typing.