Russian Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Different Methods

Russian Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Different Methods

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[Irtsak]

DP⊥AC construction
Right triangle APD is isosceles => *AP=PD=y*
Pythagoras theorem in right triangle ADP => AD²=AP²+PD² =>AD²=y²+y² => AD²=2y² (1)
Pythagoras theorem in right triangle ABD => AD²=BD²-AB² => AD²=x²-5² => AD²=x²-25 (2)
(1), (2) => *2y²=x²-5²* (3)
*the median of a triangle divides the triangle into two triangles with equal areas*
AD is the median in triangle ABC => Area (ABD) = area (ADC) => 1/2• (AB•AD)=1/2 (AC•DP)
=> 5y√2=AC•y => AC=5√2
Median theorem in triangle ABC => AB²+AC²= AD² + BC²/2 =>
5²+(5√2)²=2y²+(2x)²/2 =>
75=2y²+2x² => 75=2(x²-25) + 2x² cause (3)
=> 4x²=125 => x=5√5/2

[sergeyvinns]

Дополнительное построение показывает, что AD, является средней линией треугольника ВСЕ, СЕ=5, АD=5/2.
По теореме Пифагора, находим Х. Х^2=5^2+(5/2)^2=125/4. Х=5\/5/2!

[timc]

Thanks, in particular for the first method that I wouldn't have found for myself. An alternative method, similar to the 2nd method, but where the arithmetic turns out to be a little simpler:
Let AC = b, and angle BDA = t.
In triangle ABD: sin(t) = 5/X, Eq.1
In triangle ADC: [sin(180 - t)]/b = [sin(45)]/X, (sin rule) Eq. 2
In triangle ABC: [ sin(ABC]/(b) = [sin(135)]/(2X), (sin rule, ), or
[cos(t)]/b = [sin(45]/[2X], (sin(90 - t) = cos(t)) Eq.3
From Eq, 2), 3) : b = [Xsin(t)]/sin(45) = [2Xcos(t)/sin(45), so
sin(t) = 2cos(t), tan(t) = 2, and sin(t) = 2/[sqrt(5)], then from Eq.1:
X = [5(sqrt(5))/2.

[ВикторШеховцов]

Plot the points by coordinates: A(5;0), B(0;0), F(5;5), D(10:5), H(10;0). Then from the triangle BDH according to the Pythagorean theorem BD = 5*sqrt(5), and x = (5*sqrt(5))/2

[jonpress]

Another way, similar to method 2 and @timc5768: By the Law of Sines in triangle ABC, 2x/sin 135 = 5 / sin C, so sin C = 5 sqrt(2)/2x. By the Law of Sines in triangle ADC, AD/sin C = x/sin 45, so sin C = AD sqrt (2) / 2x. Set the two expressions for sin C equal to each other, and you get AD = 5/2. Now we have two legs of right triangle ADB, so we can find the hypotenuse, which is x -- 5 sqrt (5)/2.

[prime]

Draw a parallel line to AB through D intersecting AC at E.This is then very simple.Recall, a line parallel to one side of a triangle divides the other two sides proportionally. This makes DE i/2 of AB and the problem is easily solved!!

[santoshkumarpradhan]

5.59 he answer itina long samjhane ki kya jarurat he

[SanketMarathe-wirx]

Draw a parallel line to ab from d towards ac...

[pwmiles]

By sine rule (with some common simplifications)
AC=5 sin 90 / sin 45 = 5 sqrt(2)

By cosine rule
(2x)^2 = 25 + 50 - 50 sqrt(2) cos 135
4x^2 = 75 + 50 = 125
x=5sqrt(5)/2

[alexbayan]

By similarity of BAD and BEC AE=5. AE=CE so CE=5 By similarity again AD=5/2 By Pythagorian x^2=5^2+(5/2)^2.

[murvetaykac]

Acording to Thales rules 5/10=x/5 so x=5/2=AD. ABD tringle pisagor rules there fore 5^2+5/2^2=x=125/4

[michaeldoerr]

This seems really similar to other problems.

[alexbayan]

Draw a ray AD and from C drop a perpendicular line and name the foot intersection K then since the area(s) of ABD and ACD are the same CK has a length 5 and triangle ABD and KCD are conqurent. Angle ACK is 45 degree and AC is 5 * sqrt 5. Using the law of cosine on the triangle ADC we get x^2= (x^2 - 25) + 50 - sqrt 2 * sqrt( x^2-25) * 5 * sqrt 5 After simplifying and squaring both sides we get
x^2-25=25/4; x^2 = 125/4; x = (1/2) * 5 * sqrt 5

[LongZhao-bzte]

Third method : draw a n auxiliary line from point and be parallel to AB crossing AC by point E. So DA²=DE²=x²-5²=(5/2)², x²=(5/2)²+5²=25/4+25=125/4