Solutions to the Problems for 11/5/2016
1. On a certain day in winter The Oshkosh Northwester listed the length of daylight as 10 hours and 24 minutes,
the sunrise as 6:57 AM, and the sunset as 8:15 PM. The length of daylight and sunrise were correct, but the
sunset was wrong. When did the sun really set?
Solution.
5:21 PM.
2. Peter’s family ordered a 12 slice pizza for dinner. Peter ate one slice and shared another slice equally with his
brother. What fraction of the pizza did Peter eat? Reduce the fraction to lowest terms.
Solution.
Now
Peter ate one slice, which is 1/12th of the pizza, plus half a slice, which is 1/24th of the pizza.
3
1
1
+ 124 =
= .
12
24
8
An alternative solution: The pizza has 24 half-slices, Peter ate 3 half slices, thus 3/24 = 1/8 of the pizza.
3. In the land of Heptakronia the day is divided into seven hours. The picture below shows a typical Heptakronian
clock. Notice that their day begins and ends at 0 hours.
Here Are some questions; hours mean Heptakronian hours. Clock means Heptakronian clock.
(a) If the clock shows 4, what time will it show in 3204 hours?
(b) If it is now 4 o’clock (in Heptakronia), what time was it 7244 hours ago?
(c) It turns out that a Heptakronian hour is divided into 70 minutes, and each minute into 80 seconds
(Heptakronians are weird!) If the time is now 5:37:44, what time will it be in 4 hours, 68 minutes and 40
seconds?
Solution.
(a) 3204 = 457 × 7 + 5, so the time will be the same as 5 hours after 4, so 9 − 7 = 2. The answer is 2.
(b) 7244 = 1034 × 7 + 6, so the time is the same as 6 hours ago, 4 − 6 = −2, −2 + 7 = 5. The answer is 5.
(c) If we add,keeping hours,minutes, seconds, separate, we get 9 : 105 : 84. Now 84 seconds is the same as1
minute, 4 seconds so we now correct to 9 : 106 : 04. Since 106 = 70 + 36, so one hour and 36 seconds, we
add the additional hour to the hours toget 10 : 36 : 04. Since 10 − 7 = 3, theanswer is 3 : 36 : 04.
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4. Heptakronians love to do what is called arithmetic modulo 7. Here is how it works. If two numbers differ by a
multiple of 7 they are considered not exactly equal, but equal modulo 7. So, for example, Heppi of Heptakronia
will say 6 × 5 = 16 (mod 7), and Heppi is right! The value in regular arithmetic is 30, but 30 − 16 = 14, and 14
is a multiple of 7. But what Heptakronians really like to when practicing arithmetic modulo 7 is to also reduce
the answer so it is a number in the range 0, 1, 2, 3, 4, 5, 6. So Peppi, Heppi’s friend calculates 6 × 5 = 2 (mod
7); much better. As mentioned, two numbers are equal modulo 7 if their difference is a multiple of 7. They are
also equal modulo 7 if the have the same remainder when divided by 7. Can you solve the following equations
mod 7? Your answer in every case should be an integer in the range 0–6.
(a) 5x = 3 (mod 7)
(b) x2 = 2 (mod 7)
(c) 3x2 − 1 = 0 (mod 7)
(d) x6 = 1 (mod 7)
Solution.
(a) x = 2 (mod 7).
(b) x = 3 and x = 4 (mod 7).
(c) There is no solution.
(d) x = 1, x = 2, x = 3, x = 4, x = 5, and x = 6 (mod 7).
5. What is the unit digit of 132016 ?
Solution. One can do a few computations until one sees a pattern emerging. A more sophisticated way is
to use modular arithmetic, noticing first that because 13 is equal to 3 modulo 10, the last digit of 132016 is the
same as that of 32 016. No finding a pattern is easier: We have, always reducing modulo 10 (i.e. keeping only
the last digit) so that all equalities below are modulo 10:
31 = 3,
, 32 = 9,
33 = 7,
34 = 3 × 7 = 1,
35 = 3 × 1 = 3,
and we are back at the starting point. Every 4th power, beginning with the first power, will have a last digit
of 3. So we will have a lst digit of 3 at the 9th power, the 13 th power, the 17th power; at every power that is
equal to 1 plus a multiple of 4. Now 2013 = 1 + 4 × 503, so the last digit of 3 (hence also of 13) raised to the
power 2013 will be 3, to the power 2014 will be 9, to the power 2015, 7, and to the power 2016 will be 1. The
answer is 1.
6. Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five digit numbers so that they
have the largest possible sum. Which of the following could be one of the numbers?
(A) 76531
(B) 86724
(C) 87431
(D) 96240
(E) 97403
Solution. To get the largest possible value one of the numbers must have a first digit of 9, the other one
has to start with 8. The second digits must be 7 in one number, 6 in the other. The third digits 5 and 4.
The fourth digits 3 and 2, the last digits 1 and 0. Of the numbers given above, only 87431 has one of the two
required digits in the required places. The answer is C.
7. Let ABCD be a quadrilateral that has an inscribed circle. If |AB| = 10, |BC| = 5, and |CD| = 3, what is the
length of AD?
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Solution.
3
Let P, Q, R, S be the points on the circle tangent to the sides of the quadrilateral, as shown.
We use the fact that if M is a point outside of a circle, if from M we draw the tangent lines to the circle and
these lines touch the circle at points K and L, respectively, then |M K| = |M L|. Because of this
|AS| = |AP |
|SD| = |DR|
|BQ| = |BP |
|QC| = |CR|
Adding left and right hand sides we get
|AS| + |SD| + |BQ| + |QC| = |AP | + |DR| + |BP | + |CR|.
Now |AS| + |SD| = |AD|, |BQ| + |QC| = |BC|, |AP | + |BP | = |AB|, |DR| + |CR| = |CD|. We thus got,
|AD| + |BC| = |AB| + |CD| (this equality is known as Pitot’s Theorem) so that |AD| = |AB| + |CD| − |BC| =
10 + 3 − 5 = 8. The answer is 8.
8. A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown.
What is the ratio of the area of the star figure to the area of the original circle? The answer should have the
a + bπ
, where a, b are integers, not necessarily positive.
form
π
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Solution.
The region colored red in the figure on the right
is a circular segment obtained by removing from
one quarter of the circle the triangle ABC. Since
the area of the circle is 22 π = 4π and the area of
triangle ABC is 12 (2×2) = 2, the circular segment
in question has area π − 2. The blue region is just
a mirror image of the red one, so blue and red
regions together have an area of 2(π − 2) = 2π − 4.
To get the star we are removing four regions of
area 2π − 4 from the circle; the area of the star is
thus 4π − 4(2π − 4) = 16 − 8π. The ratio of this to
16 − 4π
4−π
the area of the circle is
=
, which is
4π
4
the answer.
4
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9. Triangle ABC is an equilateral triangle. A point P inside the triangle is situated so that if Q, R, and S are the
feet of the perpendiculars of P to sides AB, BC, and CA respectively, then |P Q| = 1, P R = 2, and |P S| = 3.
What is |AB|?
Solution. Let x = |AB|. We compute the area of triangle ABC in two ways. First, the area of an equilateral
triangle ofsides of length x is |sqrt3x2 /4. Next,we join the vertices to the point P and this divides the triangle
into the three triangles AP B, AP C, and BP C. The segments P Q, P R, P S are the altitudes of these triangles,
respectively, if we take sides AB, AC, BC, as the bases. We thus have that the areas of thes triangles, in the
1 1
1
order listed, are x, 2x, 3x, respectively. Adding up these areas, we see that the area of the big triangle is
2 2
2
1
3
x + x + x = 3x.
2
2
Equating to the previously found expression for the area we get
√
3 2
x = 3x,
4
√
which solves to (since x 6= 0) x = 4 3, which is the answer.
Note: By a similar argument one can show that the sum of distances from any point in the interior of an
equilateral triangle equals the altitude. This implies that in the triangle in question, the altitude h equals 6
and x is as found.
10. This is a problem that appeared in a college journal. But the solution is based on a simple trick. All you need
to know is that the sum of the angles of a triangle add up to 180◦ .
A triangle is inscribed in a square, as shown in the picture. What is the measure of the angle with the question
mark?
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Solution 1. (Using very basic geometry and some folding) Mark the vertices of the square and triangle as
indicated.
Now let’s forget for a moment the side EF of the triangle.
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Next, reflect triangle ABE along its hypotenuse, so that G is the image of B under this reflection.
It may be time to do a bit of angle chasing. Since the angles in a triangle have to add up to 180 degrees, we
see that angle BAE must measure 180 − (65 + 90) = 25 degrees. Looking at the corner of the square labeled
A, We must have
∠BAE + ∠EAF + ∠F AD = 90◦ ,
since ∠BAE − 25◦ , ∠EAF = 45◦ , we see that ∠F AD = 20◦ . Because we reflected triangle ABE along line
AE, we also have that ∠EAG = ∠EAB = 25◦ . Thus
∠GAF = ∠EAF − ∠EAG = 45◦ − 25◦ = 20◦ .
We see that ∠GAF = ∠F AD. If we now reflect triangle F AD along its hypotenuse F A, this last equality of
angles tells us that side AD would land on top of the segment AG. Since |AD| = |AB| = |AG|, it means that
the reflected segment AD would coincide exactly with AG. The reflected triangle would also have a right angle
at G, and this means that G is on the segment EF . Our picture now looks like this:
We see that ∠AEG = ∠AEB = 65◦ , thus ∠BEF = ∠AEG + ∠AEB = 130◦ , ∠CEF = 180◦ − ∠BEF = 50◦ ,
and finally ∠EF C = 90◦ − ∠BEF = 40◦ . The answer is 40◦ .
Solution 2. (Using trigonometry) Assume the square ABCD has sides of length 1 and let θ be the angle to
be determined. A tiny bit of angle chasing shows that the picture looks like:
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Then
|DE| = tan 20◦ ,
|EC| = 1 − tan 20◦ ,
|BF | = tan 25◦ ,
|CF | = 1 − tan 25◦ .
Thus, recalling the formula for tangent of a difference:
◦
tan θ
=
=
20
1 − 1−tan
|CF |
1 − tan 25◦
1 − tan (45◦ − 20◦ )
1+tan 20◦
=
=
=
CE|
1 − tan 20◦
1 − tan 20◦
1 − tan 20◦
2 tan 20◦
= tan(20◦ + 20◦ ) = tan 40◦ .
1 − tan2 20◦