Problem 17: Australian Mathematics Competition 2023 – Intermediate Year

Let's discuss a problem from the AMC 2023 Intermediate Category: Problem 17 which revolves around the area of the triangles.

Question

A $15 \mathrm{~cm} \times 15 \mathrm{~cm}$ square of origami paper is dark blue on top and pale yellow underneath. The top-left corner is folded down so that a crease is made from the top-right corner to a point $x \mathrm{~cm}$ above the bottom-left corner. Once folded, the visible regions of yellow and blue paper have equal areas. What is the value of $x$?

(A) $5$
(B) $6 \frac{2}{3}$
(C) $3 \sqrt{3}$
(D) $6$
(E) $4 \sqrt{2}$

Solution

Let's start this problem by joining the upper portion of the square.

Let the size of $AE$ be $y$cm. The side length of $AB$ is $15$ cm. The area of $\triangle AEB$, $\triangle BEF$ and $ EDCB$ is equal to $T$ $cm^2$. Thus the total area of the square is being divided into three equal parts.
So the area of the sqauare is = $15 \times 15$ $cm^2$ = $225$ $cm^2$.
Thus the area of the individual 3 parts is = $225 \div 3 = 75$ $cm^2$.

If the area of $\triangle AEB$ is $75$$cm^2$.
Then y is : $\frac {1}{2} \times base \times height = \frac{1}{2} \times 15 \times y = 75$
$ y = \frac {75 \times 2}{15} = 10$
Thus $y = 10$ cm.

The total side length of $AD = x + y$.

$x = AD - y = 15 - 10 = 5$.

Thus the length of $x$ be $5$ cm.

Check the strategy to solve problems in Olympiads:

Can you Work Backward?

Problem 24: Australian Mathematics Competition 2023 – Upper Primary

Let's discuss a problem from the AMC 2023 Upper Primary: Problem 24 which revolves around basic algebra.

Problem

I have 4 whole numbers that add up to 98. If I were to add 6 to the first number, subtract 6 from the second number, multiply the third number by 6 and divide the fourth number by 6, the four answers would all be the same. What is the sum of the largest two of my original four numbers?

(A) 72 (B) 86 (C) 88 (D) 90 (E) 94

Solution

Let the four whole numbers be $x, y, z, a$.

According to the given data, $x + y + z + a = 98$ ----------------------------(1)

Again, it's mentioned if we add $6$ to the first number : $x + 6$,

If we subtract $6$ from the second number : $y - 6$,

If we multiply $6$ with the third number : $ 6 \times z$,

If we divide the fourth number by$6$: $\frac {a}{6}$, then all these expressions are equal in value.

Thus, $x + 6 = y - 6 = 6 \times z = \frac {a}{6}$ ----------- (2)

From the equation (2) we can write, $6 \times z = \frac {a}{6}$.

$\therefore$ $a = 36 \times z$.

The value of $a$ can be either $36$ or $72$ as the sum of all the numbers equal to 98. So any one number can't exceed the value $98$.

If $a$ is 36 then $z$ has to be 1. If $z $ is 1 then y is $ 6 \times 1 + 6 = 12$.

If $y$ is 12 then $x$ is = $12 - 6- 6 = 0$.

Let's implement all the values in equation (1) -> $36 + 12 + 0 + 1 = 59$.

Now if $a$ is 72 then $z$ has to be 2. If $z $ is 2 then y is $ 6 \times 2 + 6 = 18$.

If $y$ is 18 then $x$ is = $18 - 6 - 6 = 6$.

Let's implement all the values in equation (1) -> $72 + 18 + 6 + 2 = 98$.

Thus, the two largest numbers are - $ 72 + 18 = 90$.

So the answer is option $D ) 90$.

What is AMC (Australian Mathematics Competition)?

The Australian Mathematics Competition (AMC) is one of Australia's largest and oldest annual mathematics competitions, aimed at fostering interest and excellence in mathematics among students.

Solution: Australian Mathematics Competition 2023 – Junior, Problem 29

Let's discuss a problem from AMC Junior 2023 that focuses on divisibility. From this solution, we will learn how to write two-digit numbers in terms of tenth place and 1's place.

Problem

A two-digit number has the property, that when it is divided by the sum of its digits the result is $4$ with remainder $3$. What is the sum of all two-digit numbers with this property?(AMC Junior 2023 Problem 29)

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Australian Mathematics Competition 2023 – Junior, Problem 29

Solution

Let's start this problem by representing the two-digit number as $10x + y$, where x represents the tenth-place digit and y represents the one's place digit.

In the next step, we have to divide this number by the sum of its digits: $\frac {10x + y}{x + y}$.

The answer will have to be 4 : $\frac {10x + y}{x + y} = 4$.

And we will also get a remainder: $ 10x + y = 4 (x + y) + 3$.

$\Rightarrow 10 x + y = 4x + 4y +3$.

$\Rightarrow 6x - 3y = 3$.

$\Rightarrow 2x - y = 1$.

$\Rightarrow2x - 1 = y$.

Making a table :

If we only have to consider two-digit numbers we will only consider till $9$ for $y$.
Thus the list of numbers are $11, 23, 35, 47, 59$.

Now let's divide the numbers by the sum of their digits and find the remainder :

$\frac{11}{2}and ans =5, rem =1$, $\frac {23}{5} and ans = 4, rem =3$, $\frac {35}{8} and ans = 4, rem = 3$, $\frac {47}{11} and ans = 4, rem = 3$, $\frac {59}{14} and ans = 4, rem = 3$.

As $11$ doesn't satisfy the condition of getting the remainder as 3 and the answer as 4, after dividing the number by the sum of its digits, we will ignore 11.

Thus the sum of the remaining two digit numbers is = $ 23 + 35 +47 +59 = 164$.

So the answer to this problem is 164.