Sample Questions for Gauss Contests


Questions chosen from previous Gauss contests
Gauss contests are organized by the Centre of Education for
Math and Computing, University of Waterloo
Problem 1

In the addition shown, P and Q each represent single digits, and the sum is 1PP7. What is P + Q?

(A) 9 (B) 12 (C) 14 (D) 15 (E) 13

 

Problem 2

In the right-angled triangle PQR, we have that PQ = QR. The three segments QS, TU and VW are perpendicular to PR, and the segments ST and UV are perpendicular to QR, as shown. What fraction of triangle PQR is shaded?

(A) 3 ⁄ 16 (B) 3 ⁄ 8 (C) 5 ⁄ 16 (D) 5 ⁄ 32 (E) 7 ⁄ 32

 

Problem 3

A box contains a total of 400 tickets that come in five colours: blue, green, red, yellow, and orange. The ratio of blue to green to red tickets is 1 : 2 : 4. The ratio of green to yellow to orange tickets is 1 : 3 : 6. What is the smallest number of tickets that must be drawn to ensure that at least 50 tickets of the same colour have been selected?

(A) 50 (B) 246 (C) 148 (D) 196 (E) 115

 

Problem 4

Greg, Charlize, and Azarah run at different but constant speeds. Each pair ran a race on a track that measured 100 m from start to finish. In the first race, when Azarah crossed the finish line, Charlize was 20 m behind. In the second race, when Charlize crossed the finish line, Greg was 10 m behind. In the third race, when Azarah crossed the finish line, how many metres was Greg behind?

(A) 20 (B) 25 (C) 28 (D) 32 (E) 40

 

Problem 5

In right-angled, isosceles triangle FGH, segment FH = √̅8. Arc FH is part of the circumference of a circle with centre G and radius GH. The area of the shaded region is

(A) π – 2; (B) 4 π – 2 (C) 4 π – (1 ⁄ 2) √̅8 ; (D) 4 π – 4 (E) π – √̅8

Proof: Diameter is the shortest curve that bisects circular area

The picture below actually shows the proof.

The goal is to show the red line (which is supposed to bisect circular area) is longer than the diameter AB’.
We see it from:

Length of red curve AB is greater than: AE + EB = AE + EB’

which is certainly longer than AB.

Just one more minute. Let’s review the construction-proof process.

Connect the two endpoints A, B of the curve by a line segment. Draw diameter CD // AB.

Take O (the midpoint of CD), then passing A and O we get another diameter AB’.

The two arguments needed for completing proof are:

(1) Red line must have at least one intersection with diameter CD; (think like this: if the red line resides completely at only one side, then there is no point that it can evenly divide the circular area) Suppose the intersection is at point E;

(2) B and B’ are symmetric to the diameter CD therefore EB = EB’ (think on why? using the property of circles and parallel lines)

Now the rest of the proof is straight forward.

Complete Numbers in Fraction Equations

The formula on our face page of “amazing numbers” is rather interesting:
1 – (1 ⁄ 28) = (1 ⁄ 2) + (1 ⁄ 4) + (1 ⁄ 7) + (1 ⁄ 14)

The point of interest is that: if you look at all divisors of 28: they are 1,2,4,7,14,28; with the exception of 28 which is itself, all divisors have appeared in this formula, and they appear in the form of so-called “unit fraction”, where numerator is 1. So (1 ⁄ 2), (1 ⁄ 4), etc. are all unit fractions.

Indeed, we present a fraction equation to make it a bit unusual, but there is a low-pitch but straightforward ways to present number 28. We have that:
28 = 1 + 2 + 4 + 7 + 14
To get to the earlier fraction form, just divide every term by the number 28.

The smallest complete number is 6 (=1+2+3), 28 is the 2nd complete number, and after that, you will not see a complete number until 496. So complete numbers are rare among all positive whole numbers.

Complete numbers 6 also has a nice fraction form, as:
1 – (1⁄6) = (1⁄2) + (1⁄3)

3D objects with 3 views from top, front and side

For a 3D objects, given three views to you: one from top, one from front, and one from side, can you imagine what the original 3D objects looks like?

The question is not posed to a mechanic engineer, it would be trivial in that case. The question is raised to get a junior middle student to think a bit.

For a cylinder one of the three views is a circle, and the other two views are rectangles. For a cone one of the views is a circle, and the other two views are triangles. What if the three views given are a circle, a rectangle and a triangle? Can you figure out the original shape?

Find the Center of a Circle — Do you know how to do it?

Using a compass, you can draw a circle at any place, with any radius.

Now let’s reverse the problem. Given a circle, do you know how to find its center? (Of course, once the circle is found, there shall be no problem at all to tell its diameter, or radius.) You only see the circle itself, there is no explicit indication on where the center is.

With two set of restrictions on what kind of tools you can use, there are actually two questions. In general, you can find the center using any convenient method, including copy-and-paste the circle onto a paper, and then fold it. In particular (from classical Euclidean geometry), where it’s required to do so with a ruler (with which you are allowed to draw lines and line segments only) and a compass (with which you are allowed to draw circles only).

See the following article on how to do it in general.

How to find the center of a given circle?

If you attempt to solve this problem with a ruler and a compass, then you are required to know how to make a perpendicular bisector. This will be discussed in another posting.