Problem 3 - Complex numbers from A to...Z

Question

Let \(A B O\) be an equilateral triangle with center \(S\) and let \(A^{\prime} B^{\prime} O\) be another equilateral triangle with the same orientation and \(S \neq A^{\prime}, S \neq B^{\prime} .\) Consider \(M\) and \(N\) the midpoints of the segments \(A^{\prime} B\) and \(A B^{\prime} .\) Prove that triangles \(S B^{\prime} M\) and \(S A^{\prime} N\) are similar. $$ \left(30^{\mathrm{th}} \mathrm{IMO}-\right.\text { Shortlist) } $$

Step-by-Step Solution

Answer

Short Answer: Triangles \(SB'M\) and \(SA'N\) are similar because they have congruent angles, \(\angle SA'N = \angle SB'M\), and proportional sides, \(\frac{|SA'|}{|SN|} = \frac{|SB'|}{|SM|}\). This is established by using the properties of the equilateral triangles, midpoints and the given conditions.

Step 1: Draw the diagram and note down the given information

First, we need to draw a diagram of the given situation. Start by drawing equilateral triangles \(ABO\) and \(A'B'O\), with their centers \(S\) on them. Mark \(M\) and \(N\) as midpoints of segments \(A'B\) and \(AB'\), respectively. Since \(M\) and \(N\) are midpoints, we have \(|AM|=|MB'|\) and \(|AN|=|NB'|\).

Step 2: Show that triangles \(A'BS\) and \(AB'S\) are congruent

To show that triangles \(A'BS\) and \(AB'S\) are congruent, we will use the SAS congruence criterion. We know \(|A'B|=|AB'|\) because the triangles \(A'BO\) and \(AB'O\) are equilateral. We are given that \(|\angle A'BS|=|\angle AB'S|\) because they are corresponding angles of equilateral triangles. Now, we need to show that \(|AS|=|SB'|\). For this, notice that triangles \(ASO\) and \(SB'O\) are equilateral triangles with side \(AS = SO = SB' = B'S.\) Hence, \(|AS| = |SB'|\). Therefore, by SAS criterion, triangles \(A'BS\) and \(AB'S\) are congruent.

Step 3: Show that triangles \(SA'N\) and \(SB'M\) share an angle

We are trying to prove that triangles \(SA'N\) and \(SB'M\) are similar. Let's try to show that they share an angle, which will help us establish similarity later. From step 2, we have \(\angle A'BS = \angle AB'S.\) Also, we know that \(\angle A'BM = \angle AB'N,\) as both are \(180^{\circ} - 60^{\circ} = 120^{\circ}.\) Now, subtracting \(\angle A'BM\) and \(\angle AB'N\) from both sides of the equation \(\angle A'BS = \angle AB'S,\) we get \(\angle SA'N = \angle SB'M.\)

Step 4: Show that triangles \(SA'N\) and \(SB'M\) have proportions

Again, from step 2, we know that triangle \(A'BS\) is congruent to triangle \(AB'S\). Therefore, \(|SA'| = |SB'|\). Recall that in step 1, we found that \(|AM|=|MB'|\) and \(|AN|=|NB'|\). Since \(M\) and \(N\) are midpoints, we have \(\frac{|SA'|}{|SN|} = \frac{|SB'|}{|SM|}\). Hence, the ratios between the sides of triangles \(SA'N\) and \(SB'M\) are equal.

Step 5: Prove that triangles \(SB'M\) and \(SA'N\) are similar

In step 3, we showed that \(\angle SA'N = \angle SB'M,\) and in step 4, we found that triangles \(SA'N\) and \(SB'M\) have equal ratios between their sides. Therefore, we conclude that triangles \(SB'M\) and \(SA'N\) are similar.

Key Concepts

SAS Congruence Criterion

Understanding the SAS Congruence Criterion is vital when exploring the properties of equilateral triangles and their congruency. The criterion denotes that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent. In geometry, congruence means that all corresponding sides and angles of the triangles are equal, effectively making them the same in shape and size.

For instance, if we have two triangles, \triangle ABC\ and \triangle DEF\, where \(AB = DE\), \(BC = EF\), and the angle \(\angle ABC = \angle DEF\), these two triangles would be congruent by the SAS Congruence Criterion. This is a powerful tool for proving congruence, as it requires just three pieces of information about the triangles, simplifying many geometric proofs. Applying this to our exercise, it was used to prove that the triangles \(A'BS\) and \(AB'S\) were congruent, as they shared two equal sides and an included angle.

In practical applications, the SAS Congruence Criterion can be used for anything from ensuring construction elements are matching perfectly, to confirming the congruency of molecular structures in chemistry, and for creating harmonious designs in art.

Similar Triangles

The concept of similar triangles is a cornerstone in geometry that frequently appears in various mathematical problems. Triangles are considered similar if their corresponding angles are equal and the lengths of their corresponding sides are proportional. This means that while similar triangles will have the same shape, they may differ in size.

The common methods to prove similarity in triangles include Angle-Angle (AA), Side-Angle-Side (SAS), and Side-Side-Side (SSS). However, it's important not to confuse SAS similarity with SAS congruence; in similarity, the 'A' stands for an angle that is included between the two proportional sides.

In our specific exercise, we investigated the similarity between triangles by comparing their angles and side lengths. We demonstrated that the triangles \(SB'M\) and \(SA'N\) have two pairs of sides in proportion, as well as a pair of equal corresponding angles, thus confirming their similarity. Understanding similarity is not only essential for geometry but is also applied in fields such as physics, where scaling principles are fundamental, as well as in real-world scenarios like creating maps or models.

Midpoints in Geometry

Midpoints play a significant role in geometry, particularly in problems involving triangles and their properties. A midpoint is the exact center point on a line segment, meaning it is equidistant from both endpoints of the segment. When you have a line segment AB, the midpoint M would satisfy the relationship that \(AM = MB\). Midpoints are used to divide lines into equal parts, which can be useful in constructing bisectors, median lines, and for use in proofs.

During our exercise, midpoints facilitated the analysis of the triangles’ relationships to one another. By identifying the midpoints M and N of the segments \(A'B\) and \(AB'\) respectively, we were able to infer that \(AM\) was equal to \(MB'\) and that \(AN\) was equal to \(NB'\). This was instrumental when establishing the proportional sides needed to prove similarity between triangles \(SB'M\) and \(SA'N\).

Midpoints are also crucial in real-life applications; for example, architects may use them when designing to ensure balance and symmetry. In technology, understanding midpoints can be applied in graphics programming and digital design, where objects are often manipulated based on their center points.