China Math Olympiad | A Nice Geometry Problem | 2 Methods

China Math Olympiad | A Nice Geometry Problem | 2 Methods

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[devondevon]

3.75
Easy problem for an Olympiad

Using the Law of Cosines triangle ABE angles are 71.565, 18.435, and 90 degrees.
Since angle AEB = 71.565, angle DEC = 63.435 degrees [ 180 - (71.565 + 45)].

Since angle D= 90 degrees then angle C = 26.565 degrees [ 180- (63.435 + 90)

Now we have an Angle, Side, an an Angle (90 degrees, 3 (other side of the rectangle), and 26.565) and thus can employ
the Law of Sines. Hence DE = 1.5
Hence, the area of the triangle is 7.5 ( 1.5 x 3)
the area of triangle ABE = 1.5 ( 1 x 3 x 1/2)
The area of triangle CDE = 2.25 ( 3 x 1.5 x 1/2)
Hence, the combined area of both triangles is 3.75 ( 1.5 + 2.25)
Hence, the area of the shaded region is 3.75 ( 7.5 - 3.75) Answer
So, the shaded region is one-half the rectangle.

[timeonly]

I did similarly to your second method:
From upper triangle, tan(θ)=3/1=3.
Call angle DEC = α.
So, α = 180° - (45°+θ) = 135°-θ.
From lower triangle:
tan(α) = 3/x
By sum/difference angle formula for tangent:
tan(135°-θ) =
=
tan(135°-θ) = [(-1)-(3)]/[1+(-1)(3)]
tan(135°-θ) = (-4)/(-2) = 2
But tan(α) = 3/x.
So, 2 = 3/x.
And x = 3/2,
which gives base of triangle BEC:
b = 1+x = 1 + 3/2 = 5/2
Area of triangle BEC =
= (1/2)bh
= (1/2)(5/2)(3)
= 15/4
Done!

[guyhoghton]

We can simplify method 2 by use of the following trigonometric identity:
_tan(X) + tan(Y) + tan(Z) = tan(X)tan(Y)tan(Z)_ ... ①
if and only if _X + Y + Z = 180k°, k ∊ _*_ℤ_*
Since _∠AEB + ∠BEC + ∠DEC = 180°_
_3 + 1 + 3/x = (3)(1)(3/x)_ using ①
⇒ _4 + 3/x = 9/x_
⇒ _x = 3/2_
∴ *_[ΔBEC] = ½(3)(5/2) = 15/4_*

[sinsn]

Another method is shared:

Draw a circle that pass through BCE, whose center is O.
Because BEC=45°, BOC=90°. Also, BO=CO=OE=r, thus BC=r√2
Distance from O to CB is same as distance from O to AB, which is r/√2

Let M be the center of AD,
ME=(r/√2-1)
OM=(3-r/√2), which is perpendicular to AD
ME^2+OM^2=OE^2
r^2-8*r/√2+10=r^2
r=5√2/4, BC=2.5

[Indigo]

Add line EF parallel to AB, Find angle BEF using tan, find angle CEF by 45 - BEF, Find FC using tan CEF, now you have the base and height of the triangle.

[noiha]

EB=√(1²+3²)=√10. Let F lie on BE and angle BFC = 90°. Triangle FEC is right angled and angle FEC = 45° => EF = CF. Triangles EAB and BFC are similar => CF/BF = BA/EA = 3/1 = 3 => EF = CF = 3*BF => EB = EF + BF = 4*BF
=> CF = 3*BF = 3*(EB/4) = 3√10/4.
Area (BEC) = EB*CF/2 = √10*(3√10/4)/2 = 15/4

[jimlocke]

Variation on the second method: At 10:48, tan(Θ) = 3. Since its tangent is greater than 1 and Θ is acute, Θ must be greater than 45° (but less than 90°) and <AEC = (Θ + 45°) must be greater than 90° (but less than 135°). Drop a perpendicular from E to BC and label the intersection F. Note that EF is parallel to AB and CD and <FEC = <AEC - 90° = (Θ + 45°) - 90° = Θ - 45°. Apply the tangent difference of angles formula, tan(α - ß) = (tan(α) - tan(ß))/(1 + tan(α)tan(ß)), to tan(Θ - 45°) to find tan(Θ - 45°) = 1/2. By alternate interior angles, <ECD = <FEC. tan(<ECD) = tan(FEC) = tan(Θ - 45°) = 1/2, so ED/CD = 1/2 and x = ED = 3/2. Skip ahead to 13:38 to complete the calculation of the shaded area.

[karelfindejs]

Construct triangle ECF as symetrical image of ECB by axis EC
Construct rectangle ABGH such that F lies on GH
Angle BEF has 90° and so triangles ABE and EHB are congruent
So |BG| = |AH| = |AE|+|EH| = 1+3
|FG| = |HG|-|HF| = |AB|-|AE| = 3-1 = 2
|BC| = |CF| = x
Using Pythagoras theroem on triangle CFG:
|CF|^2 = |CG|^2 + |FG|^2
x^2 = (4-x)^2 + 2^2
x^2 = 16 - 8x + x^2 +4
8x = 20
x = 5/2
S = 1/2*|BC|*|AB| = 1/2*5/2*3 = 15/4

[AmirgabYT]

My method is quite similar to method 2 in this video, but I want to share it:
Let ĿAEB=α, then ĿABE=90°-α.
ĿDEC=180°-45°-α=135°-α



DC=AB=3

135°-α=90°-a+45°
BE=√(9+1)=√10
sin(90°-α)=1/√10=√10/10
cos(90°-α)=3√10/10




=8√5/20=2√5/5

sin(DEC)=2√5/5

=3√5/2
BE=√10
EC=3√5/2
ĿBEC=45°
S(BEC)=

=√100•3/8=10•3/8=3, 75

[franciscook]

I found this easiest ...
Construct a line parallel to AB and DC through E (meets BC at F).
This splits angle 45° into an upper angle α with tan α=1/3 and β with tan β = x/3
Given that β=45 - α; tan β=(tan45 +tan(-α))/(1-tan45
but tan β = x/3 so x=3/2
area=(1/2)(1+3/2)(3)=15/4

[Irtsak]

Another solution with Geometry . Let BP⊥EC (construction).
Obviously orthogonal triangle EBP is isosceles cause < CEB=45° => EP=BP.
Math Booster found that BE=√10. By Pythagoras in ΔEBP => BP=√10/2
Also EC=√(x²+9) by Pythagoras in right triangle EDC
Notice that right triangles BPC, EDC are similar cause <PBC=<ECD (acute angles with vertical sides) => BC/EC=BP/DC => (x+1)/√(x²+9)=√5/3 =>3x+3=√5 (√(x²+9)) =>
9x²+18x+9=5(x²+9) => 2 x²+9x-18=0 => x=3/2
At last area (BEC) = 15/4 as Math Booster found.

[Irtsak]

Α variation on my previous solution.
Let BP⊥EC (construction).
Obviously orthogonal triangle EBP is isosceles cause < CEB=45° => EP=BP.
Math Booster found that BE=√10. By Pythagoras in ΔEBP => BP=√10/2
Also EC=√(x²+9)by Pythagoras in right triangle EDC
*Εxpress the area of the triangle BEC in two ways* :
1/2 BC⋅DC=1/2 EC⋅BP => (x+1)⋅3=√(x²+9)⋅√5 =>
9x²+18x+9=5(x²+9) => 2 x²+9x-18=0 => x=3/2
At last area (BEC) = 15/4 as Math Booster found.

[jimleahy]

Noting that <ABE+<DCE=45 taking the tan of both sides leads very quickly to a simple linear equation for x=3/2 and the rest is simple.

[Irtsak]

And *My son's solution*
Let Dxy a rectangular coordinate system and ABCD a rectangle.
D(0, 0), A(0, α+1), B(3, α+1), C(3, 0), E(0, -1).
Let λ₁, λ₂ the the directivity coefficients of the lines ΕΒ, ΕC.
λ₁=(α+1-α)/(3-0)=1/3 and
λ₂ =(0-α)/(3-0)=(-α)/3

(λ₁-λ₂ /(1+λ₁⋅λ₂ )=tan⁡45=1
=> λ₁-λ₂= 1+1+λ₁⋅λ₂₂=>……….. => α=3/2
E(0, 3/2) and B(3, 5/2)
You can easily find the equations of the lines : EB, EC.
EB : y-3/2=1/3 x => y=1/3 x+3/2 =>
f(x)=1/3 x+3/2
EC : y-0=-1/2(x-3) => g(x)=-1/2 x+3/2
f(x)-g(x)=5/6 x
So Area = defined integral with limits from 0 to 3
= ∫(f(x)-g(x))dx =∫(5/6 x)dx =

[llwlgqx]

Extend the sun ВС beam to the point F so that <BCF = 90 degrees. Then BE / EF = BC /CF = 1/3,
...=> BC = 2x, => CF = 6x, => BF = 8x = ...= 10, => 2x = 5/2 and so on – as you have! Don't thank me! I was glad of your assignment!

[michaeldoerr]

This is the fifth or sixth time I understood both methods. I have noticed that the first method was kind of dependent on the give angle but only computationally and that BOTH methods made use of the obvious symmetry of the triangle's height being equal to the length of the rectangle which was 3. I think that if the angle were NOT 45 degrees, the symmetry would not be obvious and you would have to make use of the first method. I could be wrong. Would those methods be applicable even with angles other than 45 degrees?

[prime]

The plan is whats important not the nitti gritti.These are teaching moments!!We have a rectangle and a lots of90 degree angles. We can find one side of the triangle and all the trig functions of one of the right triangles. The triangle whose area we want has an altitude equal to 1.We can use two different formulas to find the area of the triangle. There are other methoids!!

[soccerstar]

You can also just use cosine rule on EBC and solve for x.

[santiagoarosam]

EB=√(1²+3²)=√10 → El punto "F" es la proyección ortogonal de B sobre EC y las proyecciones de F sobre BC y sobre la horizontal por E son los puntos "G" y "H" → EFB es triángulo rectángulo isósceles de lados (√10/√2=√5)/(√5)/(√10) → (√5)²=√(1²+2²) → Los triángulos FHE y FGB son congruentes, de lados (1)/(2)/(√5) → La razón de semejanza entre los triángulos CGF y FGB es s=1/2→ Los catetos de CGF miden (1) y (1/2) → BC=BG+GC=2+(1/2)=5/2 → Área rosa CBE =(5/2)*3/2 =15/4 =3, 75 ud².
Gracias y un saludo cordial.

[zdrastvutye]

how to apply basic knowledge:
10 print "mathbooster-china math olympiad-a nice geometry problem":dim x(2, 2), y(2, 2)
20
30
50 x(0, 0)=0:y(0, 0)=0:x(0, 1)=l1:y(0, 1)=0:x(0, 2)=0:y(0, 2)=l4*cos(rad(w5))
60 x(1, 0)=0:y(1, 0)=lbc-l2:x(1, 1)=l1:y(1, 1)=0:x(1, 2)=x(1, 1):y(1, 2)=lbc
70 x(2, 0)=0:y(2, 0)=lbc-l2:x(2, 1)=l1:y(2, 1)=lbc:x(2, 2)=0:y(2, 2)=y(2, 1)
80 masx=1200/l1:masy=850/lbc:if masx<masy then mass=masx else mass=masy
90 goto 110
100 xbu=x*mass:ybu=y*mass:return
110 for a=0 to 2:x=x(a, 0):y=y(a, 0):gosub 100:xba=xbu:yba=ybu:gcol a+8
120 for b=1 to 3:ib=b:if ib=3 then ib=0
130 x=x(a, ib):y=y(a, ib):gosub 100:xbn=xbu:ybn=ybu:goto 150
140 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
150 gosub 140:next b:next a
mathbooster-china math olympiad-a nice geometry problem
die gesuchte flaeche=3.75
>
run in bbc basic sdl and hit ctrl tab to copy from the results window