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Replace all stars with ”+” or ”×” signs so the equation holds: 123456=100 Extra brackets may be added if necessary. Please write down the expression into the answer box.

In how many ways can one change \pounds2 into coins worth 50p, 20p and 10p? One does not necessarily need to use all available coin types, i.e. having 5 coins of 20p and 10 coins of 10p is allowed.

In a lot of geometric problems the main idea is to find congruent figures. We call two polygons congruent if all their corresponding sides and angles are equal. Triangles are the easiest sort of polygons to deal with. Assume we are given two triangles ABC and A1B1C1 and we need to check whether they are congruent or not, some rules that help are:

  • If all three corresponding sides of the triangles are equal, then the triangles are congruent.

  • If, in the given triangles ABC and A1B1C1, two corresponding sides AB=A1B1, AC=A1C1 and the angles between them BAC=B1A1C1 are equal, then the triangles are congruent.

  • If the sides AB=A1B1 and pairs of the corresponding angles next to them CAB=C1A1B1 and CBA=C1B1A1 are equal, then the triangles are congruent.

At a previous geometry lesson we have derived these rules from the axioms of Euclidean geometry, so now we can just use them.

Consider two congruent triangles ABC and A1B1C1. We draw a point M on the side BC and a point M1 on the side B1C1 such that the ratio of lengths BM:MC is equal to the ratio of lengths B1M1:M1C1. Prove that AM=A1M1.
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We call a median the segment from the vertex of a triangle to the midpoint of the opposite side. Prove that in two congruent triangles, the corresponding medians are of equal length.

We call a bisector the segment from the vertex of a triangle to the opposite side which divides in half the angle next to the starting vertex. Prove that in two congruent triangles, the corresponding bisectors are of equal length.

7 identical hexagons are arranged in a pattern on the picture below. Each hexagon has an area of 8. What is the area of the triangle ABC?

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In the triangle ABC the bisector BD coincides with the height. Prove that AB=BC.

In the triangle ABC the median BD coincides with the height. Prove that AB=BC.

In the triangle ABC with BC=12, the median AE is perpendicular to the bisector BD. Find the length of AB.
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