Olympiad Mathematics | A Very Nice Geometry Problem

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[@soli9mana-soli4953]

Once known that BC = 1+2+6+3+3 = 15 (BH and MC with proportions on right triangles) we can find AP knowing that ADE and ABC are similar, so
AP : AQ = DE : BC
AP : (AP+6) = 6 : 15
AP = 4
Area = 15*(6+4)*1/2 = 75

[@lasalleman6792]

Just go Pythagorean . Use triangle DJI to find the angle at B, which is 63.43 degrees. Use the Triangle EKL to find the angle at C which is 45 degrees. Thus the angle at A is 72.57 degrees . Then work out the various seperate hypotenuse for cosine lines, DE, IJ, KL, BH and MC. Add all the sums for a total for each lines AB and AC. I got for line AB = 11.21. Line AC came out to = 14.13. Then find the height from line BC to apex at A. Turns out to be 9.96. Go Pythagorean again to find the base and then the total area. Total area turns out to be 74.54, according to my calculations. Pretty close to the suggested solution.

[@devondevon4366]

75
a different approach
The triangles are similar on both the left and right sides of the large triangle.
On the left, the height is twice the base, and on the right, it is the same as the base.
Hence, the length of the base of the first unshaded triangle =1, and the last unshaded =3
Hence, the base = 1 + 3 + 2 + 6 + 3 = 15 !!!
Drop a line from the vertex of the large triangle to create two triangles, and both will have the same.
Since the area of the blue square = 36, its height is 6.
Hence, the height of the large is 6 + the height of the unshaded triangle.
Since the base = 6, and since on the left the base is one-half the height but on the right it is
the same, then let the height = n,
then on the left the base = 1/2 n, and on the right the base =n
Hence, n + 1/2n =6
3/2n =6
n = 2/3 * 6
n=4
Hence, the height of the large triangle = 10 (6 +4)
hence area = base * height * 1/2 = 15 * 10 * 1/2 = 75

[@santiagoarosam430]

Pendiente AB =2 ; pendiente de CA =1 ---> DE =a+(6-a)---> Distancia de A a DE =h =2a=6-a---> a=2---> h=4 ---> Área ABC =BC*h/2 =(1+2+6+3+3)*(6+4)/2 =15*10/2 = 75. ud².
Gracias y saludos

[@hongningsuen1348]

Once BP = 1 and MC = 3 are found.
Area of trapezium BDEC = (1/2)(6+15)(6) = 63.
Triangle ADE & triangle ABC are similar triangles (AAA)
Hence area of ADE: area of ABC = (6/15)^2 = 4/25 (area ratio = side ratio^2)
Hence area of ABC: area of trapezium BDEC = 25:(25-4) = 25:21.
Hence area of ABC = (25/21)(63) = 75

[@hongningsuen1348]

Correction: BH, not BP = 1.

[@wasimahmad-t6c]

11.18+14.14+15=40.32 permetar )(15×10=150 trengle area 75

[@shaozheang5528]

This is quite easy. solved with some algebraic inequalities with the congruent triangles method you used. Solved in 5 minutes!

[@michaeldoerr5810]

I agree that it is quite easy. The area is 75 units square!!!

[@hongningsuen1348]

Exactly how?

[@Antony_V]

Since LMC is an isosceles right triangle C=45°, AQ=QC --> AP=4 --> area=75 squ

[@hongningsuen1348]

AQ=QC --> AP=4 reason?

[@Antony_V]

@@hongningsuen1348Simply I skipped the step AP=2*DP, so 12-x (QC) = 6+2x (QA) --> AP=4

[@hongningsuen1348]

@@Antony_V Thanks for your reply.

[@Antony_V]

@@hongningsuen1348 👍❤️

[@SGuerra]

Que questão bonita! Parabéns pela escolha. Brasil - Outubro de 2024.

[@alexniklas8777]

S=15×10/2=75😊

[@wasimahmad-t6c]

5+10=15×10=150÷2=75

[@devondevon4366]

75

[@shaozheang5528]

Solved in 5 minutes!

[@charlie-r1x]

I'm actually stupid

[@nenetstree914]

75

[@giuseppemalaguti435]

75