Russian Math Olympiad Problem | A Very Nice Geometry Challenge

[ناصريناصر-سب]

According to the Pythagorean theorem AB = √2329, we put CD = 27x and BD = x√2329, from which 40 = x(27 + √2329), so x = 40/(27 + √2329), from which BD = (40√2329)/(27 + √2329), and from which the area of triangle ABD is (40√2329)*(27)/2(27 + √2429) = 27(2329-27√2329)/80

[christianaxel]

By bisector angle theorem AB/BD=AC/DC, then √2329/BD=27/(40-BD), solving for BD=40√2329/(√2329+27)=(2329-27sqrt√2329)/40, finally

[prossvay]

Let CD=x
Tan(a)=x/27 (1)
Tan(2a)=40/27 (2)
(2): 2tan(a)/1-tan^2(a)=40/27
2(x/27)/1-(x/27)^2=40/27
So x=27√2328/40_729/40
Area of triangle

[maochuancko]

The ratio of area ABD and ADC is AB : AC. Since we know the area of ABC = 40 x 27 / 2, and AB = square root of 40x40+27x27, we can get the area of ABD = (40x27/2) x square root of (40x40+27x27)/[square root of (40x40+27x27) + 27]

[michaeldoerr]

The area is 27/80(2329-27*sqrt(2329)). It has been awhile since I have found another video that makes use of the double angle identity. Such a refresher!!!

[imetroangola]

*Outro método:*

Por Pitágoras no ∆ABC:

AB²=BC²+ AC²→AB²= 40² + 27²→ AB=√2329.

Pelo teorema da bissetriz interna no ∆ABC:

BD/CD=AB/AC→ BD/CD=√2329/27. Logo,

*CD/BD=27/√2329 (1)*

Sejam S=[ABD] e W=[ADC]. Daí

S/BD=W/CD→ W=S×CD/BD. Usando (1), obtemos:

W=27S/√2329 . Assim,

S+W= [ABC]

S + 27S/√2329= 27×40/2

S(1+ 27/√2329)=27×20

S(√2329 + 27)/√2329=27×20


S= 27×20√2329/(√2329+27)

S=

S=27×20√2329(√2329-27)/1600

*S=27√2329(√2329-27)/80*

Ou

*S=27(2329-27√2329)/80*

ou


*S≈346, 27*

[denisrenaldo]

There is no need to compute tan theta, in which you find sqrt(2329) and that is the hypotenuse AB. The best way is to jump directly to the area S of ABD with S=AC*BD/2=27*BD/2. You need only to find BD by drawing the perpendicular DH to AB (AH=AC=27) and using the analogy of BDH and BAC : BD/BA=BH/BC. So BD=BA*(BA-27)/40 and S=27*BA*(BA-27)/80 with BA=sqrt(2329). That’s it !

[shrikrishnagokhale]

Problem should be the one which does not need calculator to find the exact answer.

[lasalleman]

Given the larger right triangle dimensions for sine and cosine of angle ABC, angle ABC is 34 degrees. Angle BAC us thus 56 degrees. Split angle BAC and get 28 degrees. Use 28 degrees located at the apex A for the smaller unshaded triangle.. Line DC of the smaller unshaded triangle is the sine of angle DAC. And line AC is the respective cosine. Going Pythagorean, we have Line DC (the sine) = 14.35. Thus area of the unshaded portion of the triangle is simply (27 * 14.35)/2 = 193.73

( Basic formula for area of a triangle)

So, 540 (area of total triangle) less area of unshaded portion of total triangle, 193.73 = 346.28. Which is damn close to the suggested solution.

The Boosterman 's explanation is more detailed, but takes too long. I prefer a short-cut way of finding answers. Short and sweet you might say. And, when you're doing an exam, time counts. Counts bigtime!


But I am a subscriber and enjoy working on problems as they appear.

[SGuerra]

A questão é muito bonita, mas os números não colaboram muito. Parabéns pela escolha!!

[marioalb]

tan 2θ = 40/27 --> θ=27.99°
tan θ = DC/27 --> DC=14, 3505 cm
BD = 40 - DC = 25, 6497 cm = b

A = ½ bh = ½ . 25, 6497 . 27
A = 346, 27 cm² ( Solved √ )

I can't believe what extremely complicated, video solution is !!!
When, can be done easier !!!
Without need of quadratic equations and trigonometric identity !!!

[fouadyoussef]

AB^2= 27^2+ 40^2. AB= 48.3. If BD= x then AB/AC= BD/DC, 48.3/27= X/ (40- x). x=25.7. Area= 0.5(25.7)(27)= 347.

[himo]

CD=x BD=40-x
27/AB=x/(40-x)
27²+40²=(1080/x-27)² 2329=1166400/x²-58320/x+729
1166400/x²-58320/x-1600=0 1166400-58320x-1600x²=0
20x²+729x-14580=0
x>0, x≒14.35　　　　
BD=40-14.35=25.65

area of triangle ABD = 25.65*27*1/2 = 346.275cm²

[giuseppemalaguti]

BD=a..DA=t=27/cosθ..a/sinθ=t/sin(90-2θ)..a=27tgθ/cos2θ, con tgθ=(40-a)/27...sostituisco nella a, risulta una equazione assurda 40a^2-4658a+93160=0..a=25, 65.. Ayellow=27a/2=27*25, 65/2=346, 27...mah, i calcoli mi sembrano strani