ARML Practice Problems – December 2013 - Solutions
NAMES, NAMES, NAMES!
1. Feuerbach Circle (http://www.math.uh.edu/~jmorgan/Rice/Feuerbach/index.html)
– named for Karl Feuerbach, who rediscovered it in 1822. This circle was known to
Euler in 1765.
a. List the points that are contained in the Feuerbach Circle:
For a triangle: the three midpoints of its sides, the three feet of the altitudes, and
the three midpoints of the segments from the orthocenter to each vertex.
b. What is the minimum number of unique points that could be on the Feuerbach for
a particular triangle? In what type of triangle does this happen?
The minimum is four and that occurs in an isosceles right triangle.
c. What other options are there for the number of unique points for different types of
triangles?
5 (non-isosceles right triangle), 6 (equilateral), 8 (non-equilateral isosceles), 9
(scalene triangle)
d. Can any of the points on the Feuerbach circle occur in the exterior of the triangle?
If so, how many points?
I think the most possible is 5 for an obtuse scalene triangle.
2. Euler Line (http://www.mathopenref.com/eulerline.html) was named for Leonhard Euler
(1707 – 1783).
a. List the three points of concurrency and special center point on the Euler Line.
Center of Feuerbach Circle, circumcenter, orthocenter, centroid
b. Give the special location for two of the four points from part a.
With the circumcenter and orthocenter as endpoints, the center of the Feuerbach
Circle is at the midpoint and the centroid is located at a position that has a ratio
of 1:2 from the two endpoints!
3. Thales’ Theorem (http://www.mathopenref.com/thalestheorem.html).
Thales proved that an angle inscribed in a semicircle is a right angle.
Can you?
Draw OA. There are now two isosceles triangles. Investigate the angle
congruences and sums!
4. Pythagorean Theorem – you know it! It has
been proved over and over and over again! This
book, written in 1927 contains 370 different proofs.
If you wish to buy it, go to amazon.com!
A couple of fun problems from my 75 year-old
“Plane Geometry” book:
a. The radii of two concentric circles are 9 and 15 inches respectively. Find the
length of a chord of the larger circle that is tangent to the smaller circle.
A tangent is  to a radius at the point of tangency and a segment through the
center of a circle that is  to a chord bisects that chord. Along with the
Pythagorean Theorem (and a 3-4-5 triangle), you get 24.
b. To lay out the strongest rectangular beam (one that will
carry the heaviest load without breaking) that can be cut
from a circular log, the sawyer trisects the diameter of a
cross section. At C and D, the points of trisection, he
then draws CE and DF, each perpendicular to AB. Next
he draws BF, FA, AE, and EB.
i. Prove that AEBF is a rectangle.
ii. Prove that ACE AEB .
1
iii. Prove that AE 2  AB 2
3
2
iv. Prove that BE 2  AB 2
3
AE
1
v. Prove that

BE
2
c. Follow-up: Give the dimensions of the strongest beam cut from a log with
diameter of 15 inches.
The exact values are 5 3 and 5 6 , or about 8.66 by 12.25 inches.
5. Ptolemy’s Theorem (http://www.cut-the-knot.org/proofs/PtolemyTheoremPWW.shtml)
For cyclic quadrilateral ABCD, the sum of the products of the lengths of opposite sides
equals the product of the lengths of the diagonals, AD  BC  AB  CD  AC  BD
a. From AoPS, v2: Find the diagonal length of an isosceles trapezoid with bases of
length 8 and 20 and legs of length 10.
The diagonals are congruent and an isosceles trapezoid is cyclic, so
d 2  102  8  20, so d  2 65
6.
Brahmagupta’s Formula (http://www.storyofmathematics.com/indian_brahmagupta.html)
For a cyclic quadrilateral with side lengths a, b, c, and d and semiperimeter s, the area can be
found using: A  (s  a)(s  b)(s  c)(s  d )
a. From Jim Wilson’s website: If ABCD is a
cyclic quadrilateral that also has an inscribed
circle, show that its area is given by A  abcd
Look at congruent tangent segments!
b. BCDE is inscribed in circle O. CG
and EF are drawn perpendicular to
diagonal BD , which has a length of 7.
Find CG  EF .
The area of BCDE is 4 35 using
Brahmagupta’s Formula. Finding the area of
the two triangles using the diagonal as base:
1
1
7
4 35   7  CG   7  EF   CG  EF  so
2
2
2
8
 CG  EF   35
7
(BE CAREFUL – NOT HERON’S!)
7. Heron’s Formula (http://www.mathsisfun.com/geometry/herons-formula.html)
The area of a triangle with side of length a, b, and c and with a semiperimeter of s is found
by: A  s(s  a)(s  b)(s  c)
A few investigations from Jim Wilson:
a. Find triangles having integer area and integer sides.
One possibility: a 13-13-10 triangle
b. Find a triangle with perimeter 12 having integer area and integer sides.
A 3-4-5 triangle.
c. Find a triangle having integer sides and integer area that is not a right triangle.
Can you find others? Generalize.
One possibility: a 13-13-10 triangle – an isosceles triangle in which half the base
and the legs form a Pythagorean triple.
For later investigation:
d. Find the smallest perimeter for which there are two different triangles with integer
sides and integer area.
e. Find 5 triangles with perimeter of 100 units having integer area and integer sides.
f. Find all the triangles with perimeter of 84 having integer sides and integer area.
8. Pick’s Theorem (http://www.cut-the-knot.org/ctk/Pick.shtml)
Let P be a lattice polygon. Assume there are I(P) lattice points in the interior of P, and B(P)
B( P)
1 .
lattice points on its boundary. Let A(P) denote the area of P. Then A( P)  I ( P) 
2
Find the area of the shaded region:
Pentagon: A = 38 + 6/3 – 1 = 40
Triangle: A = 3 + 3/2 – 1 = 3.5
Shaded: 36.5
9. Descartes’ Circle Theorem (http://www.ams.org/samplings/feature-column/fcarc-kissing)
If four circles are tangent to each other at six distinct points and k i (for i = 1, 2, 3, 4) represents
the curvature of each circle, then the following equation is true. (Note: The curvature of a circle
is the reciprocal of its radius. If the curvature is negative, the circle will circumscribe the other
1
2
circles.) k12  k22  k32  k42   k1  k2  k3  k4 
2
For later:
The following figure was made on Geogebra. Circles A, C, and E were drawn with radii 2.30,
1.57, and 1.01 respectively. Use a calculator to find the radii for the two circles which are
mutually tangent to A, C, and E. (The answer is in the box to the right of the figure.)
The small circle has r = .228
and the larger one is r = 3.909
For now:
A special, and simpler case occurs when there are two circles and a line tangent to those two
circles. The line has an infinite radius and a curvature of zero. So, what if you have a circle
with a curvature of 25 tangent to a circle with curvature 81 and they are both tangent to a line
as shown below. Find the curvatures and radii for the two possible mutually tangent circles
and sketch where they would occur.
A small circle with k = 196 (r =
1/196) is tucked in between the two
circles and the line. A large circle
with k = 16 (r = 1/16) lies to the
right of the circles and above the
line.
10. Euler’s Formula for Polyhedra (http://www.ams.org/samplings/feature-column/fcarceulers-formula)
For any convex three-dimensional polyhedron with F faces, V vertices, and E edges,
F+V=E+2
The truncated icosi-dodecahedron, AKA
hexacontakaidigon, has 62 faces and 120
vertices. How many edges does it have?
This shape has 180 edges. For fun, see
if you can determine the number of
vertices and the number of edges from
the fact that this polyhedron is
constructed of 12 decagons, 20
hexagons, and 30 squares.