Squares and Triangles | AIME I, 1999 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Squares and triangles.

Squares and triangles - AIME I, 1999

The two squares share the same centre O and have sides of length 1, The length of AB is \(\frac{43}{99}\) and the area of octagon ABCDEFGH is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

is 107
is 185
is 840
cannot be determined from the given information

Key Concepts

Squares

Triangles

Algebra

Check the Answer

Answer: is 185.

AIME I, 1999, Question 4

Geometry Vol I to IV by Hall and Stevens

Try with Hints

Triangle AOB, triangleBOC, triangleCOD, triangleDOE, triangleEOF, triangleFOG, triangleGOH, triangleHOA are congruent triangles

with each area =\(\frac{\frac{43}{99} \times \frac{1}{2}}{2}\)

then the area of all 8 of them is (8)\(\frac{\frac{43}{99} \times \frac{1}{2}}{2}\)=\(\frac{86}{99}\) then 86+99=185.

Squares and Triangles | AIME I, 2008 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Squares and Triangles.

Squares, triangles and Trapezium - AIME I, 2008

Square AIME has sides of length 10 units, isosceles triangle GEM has base EM, and the area common to triangle GEM and square AIME is 80 square units.Find the length of the altitude to EM in triangle GEM.

is 107
is 25
is 840
cannot be determined from the given information

Key Concepts

Squares

Trapezium

Triangles

Check the Answer

Answer: is 25.

AIME I, 2008, Question 2

Geometry Revisited by Coxeter

Try with Hints

let X and Y be points where the triangle intersects the square and [AXE]=[YIM]=\(\frac{100-80}{2}\)=10 then AX=YI=2 units then XY=10-4=6 units

triangle GXY is similar to triangle GEM where h=height of triangle GXY then by similarity \(\frac{h+10}{10}=\frac{h}{6}\)

then h=15 and h+10=25.

Squares and Triangles | AIME I, 1999 | Question 4