Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres.
Distance and Sphere - AIME I, 1987
What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)?
is 107
is 137
is 840
cannot be determined from the given information
Key Concepts
Angles
Algebra
Spheres
Check the Answer
Answer: is 137.
AIME I, 1987, Question 2
Geometry Vol I to Vol IV by Hall and Stevens
Try with Hints
The distance between the center of the spheres is \(\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}\)
=\(\sqrt{14^{2}+18^{2}+21^{2}}\)=31
The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
Series and sum - AIME I, 1999
given that \(\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}\) where angles are measured in degrees, m and n are relatively prime positive integer that satisfy \(\frac{m}{n} \lt 90\), find m+n.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Angles and Triangles.
Angles and Triangles - AIME I, 2012
Let triangle ABC be a right angled triangle with right angle at C. Let D and E be points on AB with D between A and E such that CD and CE trisect angle C. If \(\frac{DE}{BE}\)=\(\frac{8}{15}\), then tan B can be written as \(\frac{mp^\frac{1}{2}}{n}\) where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime , find m+n+p.
is 107
is 18
is 840
cannot be determined from the given information
Key Concepts
Angles
Algebra
Triangles
Check the Answer
Answer: is 18.
AIME I, 2012, Question 12
Geometry Vol I to Vol IV by Hall and Stevens
Try with Hints
Let CD=2a,then with angle bisector theorem of triangle we have for triangle CDB \(\frac{2a}{8}\)=\(\frac{CB}{15}\) then \(CB=\frac{15a}{4}\)
DF drawn perpendicular to BC gives CF=a, FD=\(a \times 3^\frac{1}{2}\), FB= \(\frac{11a}{4}\)
then tan B = \(\frac{a \times 3^\frac{1}{2}}{\frac{11a}{4}}\)=\(\frac{4 \times 3^\frac{1}{2}}{11}\) then m+n+p=4+3+11=18.
Area of Triangle and Integer | PRMO 2019 | Question 29
Try this beautiful problem from the PRMO, 2019 based on area of triangle and nearest integer.
Area of triangle and integer - PRMO 2019
In a triangle ABC, the median AD (with D on BC) and the angle bisector BE (with E on AC) are perpendicular to each other, if AD=7 and BE=9, find the integer nearest to the area of triangle ABC
is 107
is 47
is 840
cannot be determined from the given information
Key Concepts
Angles
Triangles
Integer
Check the Answer
Answer: is 47.
PRMO, 2019, Question 29
Geometry Vol I to IV by Hall and Stevens
Try with Hints
let AD and BE meet at F angle ABF =angleFBD=\(\frac{B}{2}\) angle AFB=angle BFD=90 (in degrees), BF is common with triangles ABF and BFD then triangle ABF is congruent to triangle BFD,AF=FD=\(\frac{7}{2}\) where AF=FD
AB=BD then\(\frac{AB}{BC}=\frac{AB}{BD+CD}=\frac{AB}{2BD}=\frac{1}{2}\) where AB=BD \(\frac{AE}{EC}\)=\(\frac{AB}{BC}\)=\(\frac{1}{2}\)
\(\frac{area triangle ABC}{area triangle ABE}=\frac{3}{1}\) then \(area triangle ABC=3area triangle ABE\)=\((3)(\frac{1}{2} \times AF \times BE)=\frac{3}{2} \times \frac{7}{2} \times 9\)=47.25 then nearest integer=47.
Triangle and Trigonometry | AIME I, 1999 Question 14
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Triangle and Trigonometry.
Triangle and Trigonometry - AIME 1999
Point P is located inside triangle ABC so that angles PAB,PBC and PCA are all congruent. The sides of the triangle have lengths AB=13, BC=14, CA=15, and the tangent of angle PAB is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.
is 107
is 463
is 840
cannot be determined from the given information
Key Concepts
Triangles
Angles
Trigonometry
Check the Answer
Answer: is 463.
AIME, 1999, Question 14
Geometry Revisited by Coxeter
Try with Hints
Let y be the angleOAB=angleOBC=angleOCA then from three triangles within triangleABC we have \(b^{2}=a^{2}+169-26acosy\) \(c^{2}=b^{2}+196-28bcosy\) \(a^{2}=c^{2}+225-30ccosy\) adding these gives cosy(13a+14b+15c)=295
[ABC]=[AOB]+[BOC]+[COA]=\(\frac{siny(13a+14b+15c)}{2}\)=84 then (13a+14b+15c)siny=168
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Triangles and sides.
Triangles and sides - AIME I, 2009
Triangle ABC AC=450 BC=300 points K and L are on AC and AB such that AK=CK and CL is angle bisectors of angle C. let P be the point of intersection of Bk and CL and let M be a point on line Bk for which K is the mid point of PM AM=180, find LP
is 107
is 72
is 840
cannot be determined from the given information
Key Concepts
Angles
Triangles
Side Length
Check the Answer
Answer: is 72.
AIME I, 2009, Question 5
Geometry Vol I to IV by Hall and Stevens
Try with Hints
since K is mid point of PM and AC quadrilateral AMCP is a parallelogram which implies AM parallel LP and triangle AMB is similar to triangle LPB
then \(\frac{AM}{LP}=\frac{AB}{LB}=\frac{AL+LB}{LB}=\frac{AL}{LB}+1\)
from angle bisector theorem, \(\frac{AL}{LB}=\frac{AC}{BC}=\frac{450}{300}=\frac{3}{2}\) then \(\frac{AM}{LP}=\frac{AL}{LB}+1=\frac{5}{2}\)
Circles and Triangles | AIME I, 2012 | Question 13
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Circles and triangles.
Circles and triangles - AIME I, 2012
Three concentric circles have radii 3,4 and 5. An equilateral triangle with one vertex on each circle has side length s. The largest possible area of the triangle can be written as \(a+\frac{b}{c}d^\frac{1}{2}\) where a,b,c,d are positive integers b and c are relative prime and d is not divisible by the square of any prime, find a+b+c+d.
is 107
is 41
is 840
cannot be determined from the given information
Key Concepts
Angles
Trigonometry
Triangles
Check the Answer
Answer: is 41.
AIME I, 2012, Question 13
Geometry Revisited by Coxeter
Try with Hints
In triangle ABC AO=3, BO=4, CO=5 let AB-BC=CA=s [ABC]=\(\frac{s^{2}3^\frac{1}{2}}{4}\)
\(s^{2}=3^{2}+4^{2}-2(3)(4)cosAOB\)=25-24cosAOB then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}-6(3)^\frac{1}{2}cosAOB\)
of the required form for angle AOB=150 (in degrees) then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}+9\) then a+b+c+d=25+3+4+9=41.
Triangles and Internal bisectors | PRMO 2019 | Question 10
Try this beautiful problem from the PRMO, 2019 based on triangles and internal bisectors.
Triangles and internal bisectors - PRMO 2019
Let ABC be a triangle and let D be its circumcircle, The internal bisectors of angles A,B and C intersect D at \(A_1,B_1 and C_1\) the internal bisectors of \(A_1,B_1,C_1\) of the triangle \(A_1B_1C_1\) intersect D at \(A_2,B_2,C_2\). If the smallest angle of triangle ABC is 40 find the magnitude of the smallest angle of triangle \(A_2B_2C_2\) in degrees.
Try this beautiful problem from the PRMO, 2019 based on Lines and Angles.
Lines and Angles - PRMO 2019
On a clock, there are two instants between 12 noon and 1 pm when the hour hand and the minute hand are at right angles. the difference in minutes between these two instants written as a+\(\frac{b}{c}\) where a b c are positive integers with \(b\lt c\) and \(\frac{b}{c} \) in the reduced form. find a+b+c.
is 107
is 51
is 840
cannot be determined from the given information
Key Concepts
Lines
Algebra
Angles
Check the Answer
Answer: is 51.
PRMO, 2019, Question 7
Higher Algebra by Hall and Knight
Try with Hints
Minute hand turns 6 (in degrees) in 1 min and hour hand turns half (in degree) in 1 min
x min after 12 is 6x -\(\frac{x}{2}\)=90 or 270 (in degrees) then x =16\(\frac{4}{11}\) or 49\(\frac{1}{11}\)
Difference 32\(\frac{8}{11}\) then a+b+c=32+8+11=51.
