Sweden Math Olympiad | A Very Nice Geometry Problem | 2 Methods
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you can find out length DC using congruent triangles. DC is 2/3 of AP as its triangle’s hypotenuse is 2/3 of length AB
Draw a horizontal line from D to the right. Its intersection with AC let be Q. Then DQ = 10/3, QC = 2×13/3, from Pitagor DC =8. Next steps, as you did.
thx for the effort for making multiple solutions. make a third one next time!!!
AD=x BD=2x CD=2y EA=3y
CE=10*1/2=5 √[13²-5²]=√144=12=3y y=4 EA=12 CD=8
△ABC=10*12*1/2=60
△ACD=60*1/3=20
The are is 20 square units. I was glad that I got a refresher on what the first method was about. And I kind of hope that you make a playlist that problems that make use of this kind of first method.
Solution:
E = point perpendicularly below A on the extension of BC.
1st set of rays: BE/BC = BA/BD = 3/2 ⟹ BE = 3/2*BC = 3/2*10 = 15 ⟹ CE = BE-BC = 15-10 = 5 ⟹ AE = √(AC²-CE²) = √(13²-5²) = 12 2nd set of rays: DC/AE= BD/BA = 2/3 ⟹ DC = 2/3*AE = 2/3*12 = 8 ⟹ Area of triangle ACD = BE*AE/2-BC*DC/2-CE*AE/2 = BE*AE/2-CE*AE/2-BC*DC/2 = (BE-CE)*AE/2-BC*DC/2 = BC*AE/2-BC*DC/2 = BC/2*(AE-DC) = 10/2*(12-8) = 5*4 = 20
A matemática é o u'nico meio REAL e VERDADEIRO da COMUNICAÇAO .Nao SEI nada do que esse PROFESSOR DIZ...Más ,por outro LADO ,entendo TUDO que ele ENSINA... e, repasso para minha filha Giselly que esta' cursando no oitava SÉRIE do segundo GRAU,e, que ficou em vigésimo lugar em LETRAS do vestibular UECE de fortaleza,ceará neste ANO de 2024... NOTA 10 ...A Giselly já chama quadrado de SQUARE...
A musica erudita também é uma linguagem universal; Uma partitura é entendida em qualquer lugar do mundo por um musico que estudou musica erudita.
F es la proyección ortogonal de A sobre la alineación BC y E la de D sobreAF→ Si BD=2DA→DE=BC/2=5=CF→ AC²-CF²=AF²→ AF=√((13²-5²)=12→ AF/BF=DC/BC→ DC=12*10/15=8 → Área ACD =DC*DE/2=8*5/2=20 ud².
Gracias y un saludo cordial.
Very nice problem with two wonderful solutions.
Thanks for sharing!
Drop a perpendicular from A to P on BC.
By simple proportion AP=(3/2) DC and CP=5
By Pythagoras AP^2 + 5^2 = 13^2
(9/4) DC^2 = 144
(3/2) DC = 12
DC = 8
Area ACD = CP.DC / 2
= 8 x 5 / 2
= 20
Extend BC to E such that angle AEC = angle AED = 90.
Now DC is parallel to AE in triangles BDC and BAE. Therefore they are similar (by AA, angle B common and angle BCD = BEA = 90)
Therefore, BD/BA = DC/AE = BC/BE = 2/3 (as its given that BD = 2.AD).
Therefore, CE = 5
In right triangle AEC, AC (hypotenuse) = 13, CE (base) = 5, therefore AE (height) = 12 (Pythagoras).
Thus, DC = 2/3 of AE = 8.
Now, in Triangle ACD,
Base = CD = 8
Height = CE = 5
Area = 1/2.8.5 = 40/2 = 20
Other way,
[ACD] = [AEB] - [DCB] - [AEC]
(All triangles on RHS are right triangle, and we know base and height of all 3 by now)
[ACD]
= (1/2.BE.AE) - (1/2.BC.CD) - (1/2.CE.AE)
= (1/2.15.12) - (1/2.10.8) - (1/2.5.12)
= 90 - 40 - 30
= 20
Another method would be to reduce the areas of triangles ACP and DCP from the area of triangle ABP.
Is it practically true?
If true,please show by a figure with proper scale.Thanks.
A=(10+5)*12/2-10*8/2-5*12/2=90-40-30=20
(10)^2=100 (13)^2^=169,{100+169}=269 180°ABCD/260=1.80ABCD1.8^10 1.8^2^5 1.2^32^5 1^1.1^1^12^1 2^1 (ABCD ➖ 2ABCD+1).
I hope one day you can reveal this Mars math to all of us on this channel! 😂😂😂
Note x and 2x are not given in the Q'
When a mathlete sees 13 as the hypotenuse,they suspect a 5-12-13 right triangle. A line parallel to one side of a triangle divides the other two sides proportionally. This answer should take no more than two or three minutes for a beginning mathlete!!
BD=2AD is not given in original Q
20
This solution is so complicated! You don't need the 'h' & all that algebra! You need only need construct point P. From the similar triangles & the BD = 2AD, you can get CP = 5. Using AC = 13 & Pythagorean formula you get PA = 12. From similar triangles again, working backward, you get CD = 8. Area = (1/2)(8)(5) = 20.
To be fair, method 1 is only marginally more complicated than your method. There's not much in it at all.
@@Grizzly01-vr4pn:
Appropriate reply!
10*12/2=60,,,,,,60/3=20
This one was really easy...
20 sq units