France Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
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- Опубліковано 30 гру 2023
- France Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
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The figure is MISLEADING. It leds us to conclue that 10 is the length of the segment D to the point intersecting the segment CB.
I found AB= 5rt5 and BC= 11.Then I found tan EBA and tan EAB, and used them to find the height of triangle EAB from base AB.This is (2rt5)/3. Hence area =25/3.
Nice quick way to solve the problem after you find AB and BC . I solved the problem before I watched the video .I started the same as you , finding AB and BC . But ; then , I took a longer route to find the area . Your method was quicker than mine or math booster .
Origo is A point. AD line is x/(10/√125)-y/(5/√125)=0; BC line is (x-√125)/(11/√125) + y/(2/√125)=0. E is the intersection of these lines. The y coordinate of E is 2√5/3, this is the height of the red triangle, so the area of the triangle is √125*2√5/3/2=25/3
We can find AE and ED via similar triangles..So the area of triangle AED is 1/2 AE,ED sine.EBut Sine E equals its supplement. Then the solution is quick.
After getting the system of 2 equations of A and X, you can put it in matrix form and find X by Cramer's rule. We get the same answer, X=25/3.
Suppose EB=x.
Because △AEC is similar to △BED, we have: AE/BE=AC/BD
i.e., AE/x=2/5=>AE=(2/5)x -- (i)
Because CB has been found to be 11, it means that CE=11-BE=11-x -- (ii)
AD=10 (given). AE=(2/5)x as in (i), thus ED=10-AE=10-(2/5x)=(50-2x)/5 -- (iii)
Note that ∠C and ∠D are both right-angles, and hence ABCD must be a cyclic quadrilateral. Thus, AD and BC are chords of a circle.
By intersecting chord theorem, we have AE.ED=BE.EC
i.e., (2/5)x.(50-2x)/5=(x)(11-x)
i.e., [(2x)(50-2x)]/25=(x)(11-x)
multiplying throughout by 25, we get:
100x-4x²=25(11x-x²)
i.e., 100x-4x²=275x-25x²
i.e., 21x²=175x => x=0 or 21x=175.
Because x≠0, it must be that 21x=175. i.e., x=175/21=25/3
Thus, BE=x=25/3
In triangle BEA (the shaded area), base BE=25/3. Altitude=AC=2. Hence its area=(1/2)(25/3)(2)=25/3 square units
If BE = 5a, then AE = 2a by Similarity of ∆BED and ∆AEC,
hence ED = 10-2a.
In the right ∆BED, applying Pythagoras theorem,
21a² +40a -125 =0, gives a= 5/3
Area ∆AEB = ½*EB*AC = 25/3
Nice solution . I did a similar solution using the circle with diameter AB and related theorems in the area of triangles.
It's a beautiful question. By use of trigonometry, Pythagoras theorem, concept of similar triangle & Heron's formula the area of triangle can be found out. It is (35 +15x root 5)x 5/6 square unit.
I let de=x and ae= 10-x.since triangles ace and dbc are similar, 5/2= x/de. So ce=2x/5. Then ae= sqroot [4(x)^2\ (25)]+4. ae also equals 10-x.So you end up with 21(x)^2-500x+2400=0. Using the quadratic formula you get x= 20/3 and x= 120/7. The valid x value is 20/3. Since x=20/3 then ce =40/15= 8/3.Area of triangle ace=8/3 and area of triangle abc=11 so area of triangle abe =25/3.
BC*BC +(2*2)=(10*10)+(5*5) pythogoras
BC *BC=121 and BC=11
You named intersection point as 'E' fine ... Triangle ACE is similar to triangle BDE because at point 'E' angles are opposite angles
AE =y ; CE=x
5/2 = (10-y)/x =(11-x)/y
2.5x+y=10
x+2.5y=11
2.5x +. 6.25y =27.5
y=17.5/5.25=10/3
Shaded area =area of triangle ABD -area of triangle BDE =(10*5/2)-((10-y)*5/2)=25-(50/3)= 50/6=25/3 sq units
Happy new year math booster !
Happy new year!
Happy New Year, dear math lovers! ❤
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Correction to solution offered. 5/2=x/ce not x/de
4 2/3 cannot be = to 8/3. Should be 14/3.
It was by far the easiest olympiad question i've ever come across to
Hey maths booster
We can find area by herons formula too
By just finding sides of it
Yes, you can find.
25/3.
Área ADB =5*10/2=25
AB²=5²+10² =125 → CB=√(125-2²)=11 → Área ACB =2*11/2=11
Solapo entre ADB y ACB =Área roja =r
Las zonas blancas de los triángulos ADB y ACB son triángulos rectángulos y su razón de semejanza s=2/5 → s²=4/25 → (25-r)(1+s²)+2r=25+11→ (25-r)(29/25)+2r=36 → r=25/3
Interesante acertijo. Gracias y saludos cordiales.
I took a path that might have been a bit different.
Basically, I used the slopes of BC and AD to form intersecting algebraic equations-of-the-lines
[1.1] f(𝒙) = (-tan θ)𝒙 + (0 at 5√5) … we know tan
[1.2] f(𝒙) = -2𝒙/11 + ²⁄₁₁ × 5√5
[1.3] f(𝒙) = -2𝒙/11 + 10√5/2
And, we also can figure g(𝒙) just as easily
[2.1] g(𝒙) = (tan φ)𝒙 + 0
[2.2] g(𝒙) = ⁵⁄₁₀ 𝒙 + 0
[2.3] g(𝒙) = 𝒙/2 + 0
So, now just make those equal, to find out mathematically where they cross
[3.1] -2𝒙/11+ 10√5/2 = 𝒙/2 + 0 … and rearrange things to solve for 𝒙
[3.2] 𝒙 = 4√5/3
Then substitute that into [2.3] since its the easy one to find HEIGHT of the shaded △
[4.1] 𝒉 = ( 4√5/3 )/2
[4.2] 𝒉 = 2√5/3
Last, figure the area
[5.1] Area = ½ base • height
[5.2] Area = ½ 5√5 ( 2√5/3 )
[5.3] Area = 5 (√5 √5 )/3 )
[5.4] Area = 5 × 5 ÷ 3
[5.5] Area = 25 ÷ 3 = 8⅓
And that'd be the solution to the problem.
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