The Enjoyment of Elementary Geometry Li Zhou ... - MAA Sections

The Enjoyment of Elementary Geometry
Li Zhou
Polk Community College
lzhou@polk.edu
February 20 - 21, 2004
A Well-known Exercise
(a) As appeared in a trigonometry textbook (Larson, Hostetler, & Edwards):
A photographer is taking a picture of a 4-foot-tall painting hung in an art gallery. The camera lens is
1 foot below the lower edge of the painting. The angle " subtended by the camera lens B feet from
the painting is
%B
B& #
" œ arctan , B !.
Use a graphing utility to graph " as a function of B. Move the cursor along the graph to
approximate the distance from the picture when " is maximum.
(b) As appeared in a calculus textbook (Larson, Hostetler, & Edwards):
A photographer is taking a picture of a 4-foot painting hung in an art gallery. The camera lens is 1
foot below the lower edge of the painting. How far should the camera be from the painting to
maximize the angle subtended by the camera lens?
B
Solution. Let " and Bbe as above. Then " œ arccot arccot B. So
&
. " #
"Î& " & " %Ð&B Ñ
.B
œ
" "B
œ
#&B
"B
œ
B #
# # # Ð#&B# ÑÐ"B#
Ñ
#&
.
Setting " œ! and applying the First Derivative Test, you can conclude that Bœ È&
yields a
.B
maximum value of ".
.
The Origin of the Problem
In 1471 Regiomontanus posed the following problem in a letter to Christian Roder, a Professor at
Erfurt:

At what point on the ground does a perpendicularly suspended rod appear largest (i.e., subtends
the greatest visual angle)?
Regiomontanus
- Born in Unfinden, near the town of Königsberg in Franconia, in 1436. Died in 1476 in Rome.
He was a man known by many names:
Johann Müller
Johannes Germanus (because he was a German)
Johannes Francus (because Franconia was known as Eastern France)
Johann von Kunsperk (after the town of Königsberg)
Regio Monte (Latin translation of Königsberg, "the royal mountain")
Joannes de Monte Regio
Regiomontanus
He is best remembered for the book: De triangulis omnimodis ( On Triangles of Every Kind), written
in 1464, printed in 1533. We quote a wonderful paragraph from the introduction of this book:
"You, who wish to study great and wonderous things, who wonder about the movement of the
stars, must read these theorem about triangles. ... For no one can bypass the science of triangles
and reach a satisfying knowledge of the stars. ... A new student should neither be frightened nor
despair. ... And where a theorem may present some problem, he may always look down to the
numerical examples for help."
A Geometric Solution
Suppose that the upper and lower edges Eand F of the painting are + and , feet above the
horizontal line 6 at the level of the camera lens. Draw the circle through EßF and tangent to 6, then
it is easy to see that " is maximum at the tangency point. By the power of a point, this optimal
distance is Bœ È+,
.

VIP Seats in a Multiplex Stadium-seating Movie Theater
Once in my class, I phrased the question as finding the best parking spot in a drive-in theater to
maximize the (vertical) viewing angle. It is then only natural to ask:
Where are the best seats in a multiplex stadium-seating theater to maximize the viewing angle of
your favorite Hollywood stars, car chases, explosions and other special effects?
The geometric solution is still valid, with 6 changed to the tilted line joining the audience's eye-
levels.
A Ruler-compass Construction
+, # +, #
Note that +, œ Ð Ñ Ð Ñ . So B œ È
+, +,
+, œ ÉÐ Ñ# Ð Ñ#
, which can be
# # # #
constructed by Pythagorean Theorem.
Another Ruler-compass Construction from the CRUX
Problem 2822 [2003, 114]. Proposed by Peter Y. Woo, Biola University, CA
Suppose that C is a parallelogram with sides of lengths #+ and #, and with acute interior angle !,
and that J and J w are the foci of the ellipse A that is tangent to the four sides of C at their mid-
points.
(a) Find the major and minor axes of A in terms of +, , , and !.
(b) Find a straight-edge and compass construction for J and J w .
Solution.
# #
# #
? @
+ Ð, sin! Ñ
(a) Let I be the ellipse œ " and V the rectangle tangent to I at its ?ß@-intercepts.
Under the affine change of coordinates
” ? • ” " cot! cos) •” sin)
œ
•” B
@ ! " sin) cos)
C
•
#
# # #
# #
+, cos# !
#
B C
, sin# !
E F
with cot #) œ , V is transformed to C and I to A with equation œ ", where
Eσ+ # , # G #
Fσ+ # , # G #
Gœ È%
, , and +
%
#+,
# #
cos #!
,
%.
# #

È # #
1
#
!
(b) By the law of cosines, :œ + , #+, cos Ð Ñ and ;œ
È+ #
,#
1
#+, cos Ð ! Ñ can be constructed. Now, we notice that G œ È:; œ
#
:; :;
# # # #
ÉÐ Ñ# Ð Ñ# G E œ É+ #
,
#
G
#
F œ É+ #
,
#
G
#
. So we can construct , thus and as
well.
w
Finally, let # , in order to satisfy nTHE œ nTFG œ %& °. Therefore, JM Ÿ JEand
LN Ÿ LG, where M and N are the feet of the perpendiculars from J and L to GK and EI
respectively. If GK²IE, then JMŸJEœLN ŸLG œJM, which forces all the
equalities to hold. Thus MœHand NœF, and EFGHis a square. Otherwise, without loss of
generality, we may extend IE beyond E to intersect GK beyond K at O. Then JE LN Ÿ
LG, and therefore JM LN œ LG JE ÐLG JEÑsin nGOI LG JE. This
contradiction completes the proof.

Erdös-Mordell-Barrow Inequality
The case of 8œ$ of the above MONTHLY problem is also true. In fact, it is an immediate
consequence of the famous Erdös-Mordell-Barrow Inequality:
Let T be a point in the interior of a triangle EFG . Let < , = , and > be the distances from T to the
vertices, and let BC , , and D be the distances from T to the sides. Then
#ÐBCDÑ,
with equality if and only if EFG is equilateral.
Proof. Let I and J be the feet of the perpendiculars from T to GE and EF. By the law of
# # # # # # #
cosines, IJ œC D #CDcosnITJ œ C D #CDcosEœC D
# # #
#CDcosÐFGÑœÐCsinGDsinFÑ ÐCcosGDcosFÑ ÐCsinGDsin FÑ . Hence
EI EI IJ sinG sinF sinE sinG
sinnET I sinnEJ I sinE sinE sinE sinF sinF
B C . By the AM-GM inequality #, etc., completing the proof.
Brocard Points
This special case of 8œ$ leads us further to the Brocard points. They are named after Henri
Brocard, a French army officer, who described them in 1875. However, they had been studied
earlier by Jacobi, and also by Crelle, in 1816. Crelle was led to exclaim:
"It is indeed wonderful that so simple a figure as the triangle is so inexhaustible in properties. How
many as yet unknown properties of other figures may there not be?"
For any triangle there is a unique angle = , the Brocard angle, and points H and H', the
w w w
Brocard
points, such that = œnHEFœnHFGœnHGEœnHEGœnHFEœnHGF.
A Dog Chase in a Triangle
To find the Brocard points, put three dogs at the vertices of the triangle and let them chase one
after the other at the same speed. Depending on the directions of chasing, clockwise or counterclockwise,
they will end up at the corresponding Brocard points.

The Most Important Fact about the Brocard Angle =
cot= œ cotE cotF cotG
An Angle of Many Inequalities
For example,
(a) = Ÿ$! ° (equality iff EFGis equilateral);
(b) = $ Ÿ ÐE = ÑÐF = ÑÐG = Ñ;
(c)
# = Ÿ È $
EFG
(d) # Ÿ .
$
=
" " "
E F G
(the Yff inequality);
An IMO Problem
IMO Problem 1991/5. Let EFG be a triangle and T an interior point in EFG . Show that at least
one angles nT EF, nT FGß nT GE is less than or equal to $! °.
Solution. T must be in one of the three smaller triangles HEF, HFG, or H GE. Since = Ÿ $! °,
the proof is complete.
A Recent MONTHLY Problem
A recent MONTHLY problem offers a new inequality of the Brocard angle.
Problem 11017 [2003, 439]. Proposed by C. R. Pranesachar, Indian Institute of Science, India.
Let T and Ube the Brocard points of a non-equilateral triangle X , and let = be the Brocard angle
of X .
(a) Prove that the line through T and U passes through a vertex of X if and only if the sides of the
triangle are in geometric progression.
1 FG
' $
(b) Prove that = min Ò ß Ó, where F and G are the smallest angles of X.

FG
1
$ #
Solution to (b). It suffices to show that = if F G . Since cot=
œ cotE cotF
cot G and cot B is a decreasing function of B − Ð!ß Ñ, we only need to show that cot
FG
1
$
FG
1
$
_
Using the Euler product sin # B #
. "
81
, we obtain cot
.B
lnsin
B
8œ"
_ _ _ _
! " " " " ! ! " B 5 B 5 " ! #5"
81Ð B
"
B
"
Ñ œ
B
81ÒÐ 81Ñ Ð81Ñ Ó œ
B
= 5B
81 81
8œ" 8œ" 5œ! 5œ!
_
# FG " " %
5 Ð8 Ñ $ F G FG
8œ"
_
where H œ !
"
#5" #5" #5"
= 5ÒÐ"
$ #5" ÑÐF GÑ F G Ó !
5œ!
cotÐ F GÑ cotF cot G, or equivalently, cot cotF cotG cot ÐF GÑ.
BœB Ò"Ð ÑÓ Bœ Ð BÑœ
, where
= œ ! 1
#5# . Hence, cotF cotG cotÐF GÑ cot œ H,
FG # " " %
#
" "
F G FG
inequality, , thus
, completing the proof.
F
G
. Finally, by the AM-HM
Of course,
" " %
F
G FG
Back to the Trigonometric Exercise
is merely another expression of
# #
ÐF GÑ %FG œ ÐF GÑ !, with equality iff F œ G .
%B %B
B& # B& #
Recall that " œ arctan . It suffices to maximize .
%B % " B " B
Note that œ Ÿ , with equality iff œ , i.e. B œ È&
.
B& # &
B B & B &
B
This kind of non-calculus ideas seemed to be what Regiomontanus used to solved his problem. It
is not clear whether he actually provided a solution.