Find the area of the rectangle | A Very Nice Geometry Problem | Math Olympiad

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[RAG]

Very nice, slick answer. Mine was more clunky. If BC is r + x, then x^2 = 4x9 = 36 so x =6. This was encouraging, so I used triangle EBC to give r^2 + (6+r)^2 = 9^2. This gave a slightly nasty answer of r = (3rt14 - 6)/2, so r+6 =(3rt14 + 6)/2, and then 2r x (r+6) = 45. Oh well!

[imetroangola]

*Solution:*

Let O be the center of the circle and we draw the segment OP, where P belongs to the side BC. Using power at a point C, we have:

PC² = CF × CE = 4×9 → *PC=6.*

Now, we draw the segment OT, where T belongs to AD, note that AEOT and EBPO are squares whose sides are equal to the radius R of the circle. Therefore, AE=EB=BP=R.

By Pythagoras in ∆BEC:

EC² = BE² + BC² = BE² + (BP+ PC)²

9² = R² + (6+R)² = R²+36+12R+R²

2R² + 12R = 81 - 36 = 45

2R(R + 6) = 45 .

note that, AB = 2R e BC = R + 6, consequently,

*AB × BC = 45*, This is the area of rectangle ABCD.

[michaeldoerr]

The area is 45 units square. Also I have noticed something that I should have thought about: delta EFP and delta EQC are similar because these triangles are NOT opposite to one another. This is different from one of the last two problems on this channel which showed HL congruence. I hope that this counts as a sufficient reason being that shows why and how similar triangles correspond to the area of the square.

[jimlocke]

Let the point of tangency of BC and the circle be point M. Then, applying the tangent secant theorem, (CM)² = (CF)(CE) = (4)(4 + 5) = 36 and CM = 6. Let the radius of the circle be r. Then, EB = MB = r. Apply the Pythagorean theorem to ΔBCE: (EB)² + (BC)² = (9)², r² + (r + 6)² = 81, 2r² + 12r - 45 = 0. The positive root, simplified, is r = √(31.5) - 3. AB = 2r = 2(√(31.5) - 3). BC = r + 6 = √(31.5) + 3. Area of ABCD = (AB)(BC) = 2(√(31.5) - 3)(√(31.5) + 3) = 2(31.5 - 9) = 2(22.5) = 45, as Math Booster also found.

Checking the comments, I find that RAG981 may have solved it the same way.

[MorgKev]

Drop perpendicular from O onto chord EF. The little triangle formed is similar to triangle EBC. Let NC = x. Then (r+x)/9 = 2.5/r. This gives r^2 + rx =22.5 and 2r^2 + 2rx = 45. But this is the area since the rectangle has sides (r+x) and 2r.

[marioalb]

Similarity of triangles:
cos α = 5/2R = b/(5+4)
A = b.h= b.2R= 5.(5+4)
A = 45 cm² ( Solved √ )

[zygmuntserafin]

A=AB*BC=2r*x. r^2+x^2=81. (x-r)^2=9*4=36. (x-r)^2=x^2 + r^2 - 2r*x = 81 -A = 36.
A = 45.

[ВерцинГеториг-чь]

АD=BC=a, AB=DC=2r . Area ADCD=2ar, theorem about tangent and secant lines coming from one point : (a-r)*2=CExCF, a*2+r*2-2ar=(5+4)4 (1) . From a triangle BCE according to the Pythagorean theorem - CE*2=r*2+a*2, r*2+a*2=9*2, substituting into eguation (1) - 9*2-2ar=9x4, 2ar=81-36=45 . Area ABCD=45 .

[xaviersoenen]

puissance du point C par rapport au cercle= 4(4+5)=36= OQ²
BC=6+r et BC² =EC²-r²=(6+r)²=9²-r²
2r²+12r+36=81
2r(r+6)=45= surface du rectangle

[alexnikola]

NC=6... r^2+(r+6)^2=81... 2r^2+12r=45... S=2r(r+6)=2r^2+12r=45

[SGuerra]

Uau! Que questão bonita. Eu encontrei uma solução um pouco mais rápida que a sua. Parabéns pela escolha! 👏👏👏 Brasil 9 de novembro de 2024.

[RealQinnMalloryu]

(5)^2 (4)^2rm={25+16}=41 90ABCD 3^30 3^3^10 1^3^2^5 1^3^2^1 3^2 (ABCD ➖ 3ABCD+2)

[caperider]

Is it given that the point E is touching the tangent?

[solimana-soli]

2r:9=5:base
2r=height
Base *height =9*5=45

[xz]

Too clumsy,

4*(4+5)= tangent line sq, so tangent line =6. Let r is rad, so rectangle area =(6+r)*2r= 12r+ (2r) sq. Use Gougu theorem, r sq + (6+r) sq =81, expand, 12r+ (2r) sq -45=0. NO NEED TO SOLVE THE EQ, ang the 1st 2 terms is the area of rectangle, so =45

[giuseppemalaguti]

di F)...risulta arcsin(5r/9/5)=arccos(5/2r)..r^2=81/2-9√14..b(base)=√(81-(81/2-9√14))=√(81/2+9√14)..Arett=b*2r=2√(40, 5^2-81*14)=2*22, 5=45