Tag: G-8

Proof: Diameter is the Shortest Curve that bisects circular area

Post author By admin
Post date June 26, 2026
Sticky post

As shown in the figure below: (please scroll down a bit to see the figure) AB’ is one diameter of the circle, and a blue curve from A to B is supposed to bisect the area of the circle. We see from the figure below:

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

which is certainly longer than AB’; AB’ is the diameter.

The blue curve is obvious the focus. (Note when blue curve AB is mentioned, we mean the “curved one”: it curves around, not going straight from A to B.) Besides, point E is where the blue curve intersects with CD; and CD is one diameter.

So we have proved the claim that any curve bisecting the circular area got to be longer than the diameter.

Stay with us for one more minute. Let the construction-proof process be revealed to you, as follows.

Referring back to the figure. Let us start from the circle and the blue curve only (imagine all other lines and points disappear; now we draw them step by step). Connect the two endpoints A, B of the blue curve by a line segment AB. Then draw the diameter CD // AB (i.e. line CD is parallel to line AB).

Take point O (the midpoint of CD), and then passing A and O, let another diameter AB’ be drawn.

For completing the proof , two arguments are required:

(1) The blue curve has at least one intersection point with diameter CD (thinking it: if the blue curve resides at only one side of CD, then that curve cannot divide the circular area evenly into two parts with equal area; therefore, any curves that bisects the circular area must intersects diameter CD). Now suppose the intersection point is E.

(2) B and B’ are symmetric to the diameter CD therefore EB = EB’ (trying justifying it using the property of circles and parallel lines).

The rest of the proof is straightforward (and intuitive).

Tags G 9-10, G-8

Sample Questions for Gauss Contests

Post author By jonah.luo
Post date October 14, 2014
Sticky post

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 colors: 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

Tags G-7, G-8

Geometry Math BASICS Numbers

Protected: The (3-4-5) Pattern for Pythagorean Triplet [Thinking-of sides of a right triangle]

Post author By jonah.luo
Post date August 25, 2014

Tags G 9-10, G-8

Numbers Uncategorized

Complete Numbers in Fraction Equations

Post author By jonah.luo
Post date April 24, 2014

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)

Tags G 9-10, G-8

Geometry

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

Post author By jonah.luo
Post date March 18, 2014
No Comments on 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?

Tags G 9-10, G-8

Geometry

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

Post author By jonah.luo
Post date March 11, 2014
No Comments on 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.