A Very Nice Geometry Problem | Maths Olympiad | 3 Different Methods to Solve
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- Опубліковано 15 жов 2024
- A Very Nice Geometry Problem | Maths Olympiad | 3 Different Methods to Solve
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There's one more method, very similar to the 1st and 2nd. Run a line segment from D to AC that's perpendicular to AC. Define the point these intersect at as point E. This creates a triangle CED that is similar to ABC. By using similar triangle calculations on ABC versus CED and using pythagorean on triangle CED you get the same quadratic in x that leads to x=2*sqrt(10)
You gave us 3 different solutions (third option is the best) and I want to propose another one via analytic geometry.
I started using coordinates of point D as origin, that is D=(0,0), then point C=(10,0). Segment DC is within the X-axis.
Since ADC is an obtuse triangle and angle DAC=45 degrees, I looked at the option of drawing the circumcenter of triangle ADC. The circumcircle must pass through points A, D and C. I designated the center of the circumcircle as point O.
The arc CD=90 degrees because it is the opposite of inscribed angle DAC=45. So the central angle DOC=90 degrees and the triangle DOC is right isosceles, because OD=OC=R (Radius of the circumcircle). The hypotenuse DC=10, so R=5sqrt2.
Coming back to the right isosceles triangle DOC, I drew the height OH, meaning that H belongs to segment DC. This segment OH is median, mediatrix and also bisector of angle DOC=90. As a result, OH=5. This gave me that the center of the circumcircle O=(5,5). As a reminder, Segment DC is within the X-axis.
Now I can calculate the equation of the circumcircle.
(x-5)^2 + (y-5)^2 = R^2=50
The problem gave me the height AB=6 and point A belong to the circumcircle. So I can use the coordinates of point A=(x,6). After substitution y=6 on the equation of the circumcircle, I found two values for x=12 and x=-2. x=12 is not possible, because according to the triangle ABC, and the coordinates that I started with point D=(0,0), point A should have a negative abscissa. So, point B=(-2,0) is the correct solution.
Looking at the right triangle ABD and using B=(-2,0), I can say that BD=2. Since AB=6, and applying Pythagoras, I also found AD=2sqrt10
For the 3rd method, just doing trig: Angle A: tan^-1(10/6)=59, 59-45=14, then cos(14)=6/x, and x=6.18
no, perchè che misura 10 è DC. Ciao
Let θ = ⦟BAD, and y = BD.
In ∆ABC, ⦟BCA = 90° - ⦟BAC = 90°- (45°+θ) = 45°-θ.
Applying the Law of Sines on ∆ADC, we have:
sin(45°-θ)/x = sin(45°)/10, so that
x sin(45°) = 10 sin(45°-θ) = 10 [ sin(45°)cos(θ) - cos(45°)sin(θ) ]
Since sin(45°) = cos(45°), we can divide both sides by sin(45°) & get:
x = 10 [ cos(θ) - sin(θ) ]
From ∆ABD, we get cos(θ) = 6/x & sin(θ) = y/x, so
x = 10 [ (6/x) - (y/x) ], and multiplying thru by x:
x² = 60 - 10y. But y = √(x²-6²) = √(x²-36).
Subbing in for y, & isolating the radical we get:
10 √(x²-36) = 60 - x²
Squaring both sides and gathering like terms:
100(x²-36) = 3600 -120x² + x⁴
x⁴ - 220x² + 7200 = 0. Factoring we get:
( x² - 180 )(x² - 40 ) = 0 so that
x² = 180 or x² = 40, and
x = 6√5 (~ 13.42) or x = 2√10 (~ 6.32...).
Since 45°-θ < 45° < ⦟ADC (obtuse), x needs to be smallest side of ∆ADC.
Which means it can't be 6√5 (>10)... therefore x = 2√10. Done!
Yep, this is what I ended up with also.
👍
[Edit] Oops... almost what you did, but not quite. It didn't occur to me to just substitute the values for sin α and cos α directly. I reapplied the law of sines to the left triangle (just a _little_ more complicated.)
Applying the tangent sum of angles formula was the natural approach for myself and I was glad to see it done in method #3!
That was the method I used too. But I also like seeing the geometric solutions, since I'm definitely weaker in geometry than trigonometry.
Well, it took some serious headstanding, but I come up with:
X = 2√10
?
Declaring segment BD = Y, and ∠BAD to be α (probably shoulda picked ∠BDA = α, would've made it easier), and then establishing ∠BCA = 45° - α, lets us use the Law of Sines to establish X from both △ADC and △ABD. Setting those two results equal, and then combining with the Pythagorean relationship between the sides of △ABD yields the value of Y. The rest is easy.
Hope that's right... watching the video now.
I did it your third way, since expanding tan(45+x) is fairly simple. I left the equation as a quadratic in tantheta, getting tantheta = 1/3 (or -2), thus y=6/3=2, and x = 2rt10.
thanks for the time you spend!
φ = 30°; ∆ ABC → AB = 6; BC = AD + CD = y + 10; CD = DF + CF = 6 + 4; sin(DFE) = sin(ADE) = 1; BD = EF = y EAD = DEA = 3φ/2; AD = DE = x = ? BCA = ϑ → tan(ϑ) = y/4 = 6/(10 + y) → y = 2 → x = 2√10 → AE = 2CE = 4√5
Posto BD=a..risulta arctg6/(a+10)+45+arctg a/6=90...applico tang all'equazione,risulta a=2..poi teorema sei seni x/arctg(6/12)=10/sin45..x=10*arctg(1/2)√2=6,557...ovviamente manca qualcosa,come sempre... è sin(arct...)...meglio fare x^2=36+4=40
wenn cos alpha = 6/x , dann ist x = 6/cos alpha ; alpha = 90 - 45 - arctan 6/10 ; arctan 6/10 = 30,9637 ... ; alpha = 90 - 45 - 30,9367... = 14,0362 ! cos alpha = 0,9701 .... ! x = 6/ cos alpha = 6/0,9701 ... = 6,1846 ... !
i get the answer same as you x= 1.5sqrt(17)=6.1846584384
no, perchè che misura 10 è DC. Ciao
Love it!!!!!!!!!!!!!
Thank you 😊
AD=36-X²,
X=6.
The third method is the best
y=6+6-10=2