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1Metka is standing \(60\) m east and \(80\) m south of the spot where Tine is standing. Both are the same distance from a linden tree in the town park, and the tree stands due east of Tine's spot. At the …Open2Polona wants to draw three lines through one common point, as in the picture, so that the angles between them satisfy \(\beta = 2\alpha\) and \(\alpha = 3\gamma\). How many degrees must the angle \(\alpha\) …3Five semicircles, all of different sizes, stand side by side on one straight segment: each semicircle has its diameter on the segment, consecutive semicircles touch, and the five diameters together fill …4The three presents in the picture are all boxes in the shape of a cuboid with edge lengths \(10\) cm, \(20\) cm and \(30\) cm. Reading from left to right, the ribbon tying them measures \(x\) cm, \(y\) …5Five brothers - Jure, Klemen, Luka, Miha and Nace - bought a bar of chocolate. When they unwrapped it, they found that it had snapped into the seven pieces shown below, so they shared those seven pieces …6In the convex quadrilateral \(ABCD\) the line through \(A\) and \(D\) and the line through \(B\) and \(C\) meet at a right angle, as the figure shows. Furthermore \(|CD| = 1\), \(|AC| = 2\) and \(|BD| = 3\). …7The horizontal segment in the picture is divided into six parts of equal length, and every triangle appearing in the picture is equilateral. The whole figure is shaded in two colours, light grey and dark …8Each of the five figures drawn below is bounded entirely by semicircular arcs, and the largest arc is the same in all five figures. Among the figures with the smallest perimeter, which one has the largest …9Peter covered a wound with two rectangular plasters, laid across each other as in the picture. The region covered by both plasters at once has area \(40 \ \mathrm{cm}^{2}\) and perimeter \(30 \ \mathrm{cm}\). …10In triangle \(ABC\) the angle at \(A\) is a right angle. Points \(D\), \(E\) and \(F\) are chosen on the sides \(AB\), \(BC\) and \(CA\) respectively, so that \[ |BD| = |BE| \qquad\text{and}\qquad |CF| = |CE| , \] …11Let \(E\) be a point on the side \(CD\) of a square \(ABCD\). The point \(F\) lies on the line \(AB\) but not on the segment \(AB\), and satisfies \(|BF| = |DE|\). Prove that the lines \(AC\) and \(EF\) …12In triangle \(ABC\) the side \(AB\) is twice as long as \(AC\), that is \(|AB| = 2|AC|\). The point \(D\) lies on the ray \(CA\) and satisfies \(|CD| = 3|AC|\). Prove that the perpendicular dropped from …13Let \(AB\) be a segment and let \(P\) be any point of it other than \(A\) and \(B\). On the hypotenuses \(AP\) and \(PB\) erect isosceles right triangles \(APQ\) and \(PBR\), with the right angles at the …14A circle \(k\) has diameter \(AB\). A point \(M\) of \(k\) is chosen, different from \(A\) and from \(B\). Let \(k_1\) be the circle with centre \(M\) that touches the diameter \(AB\). The line \(AB\) …15Let \(E\) and \(F\) be the midpoints of the sides \(AD\) and \(DC\) of a rectangle \(ABCD\), and let \(G\) be the point where the segments \(AF\) and \(EC\) cross. Prove that \[ \angle CGF = \angle FBE . \] …16A square \(ABCD\) is given, together with points \(E\) and \(F\) lying outside the square such that the triangles \(BEC\) and \(CFD\) are equilateral. Prove that the triangle \(AEF\) is equilateral as …17Let \(ABC\) be a triangle. Points \(D\) and \(E\) lie on the rays \(CA\) and \(CB\) respectively, but not on the sides of the triangle \(ABC\), and are chosen so that \[ |AD| = |BE| = |AB| . \] Let \(G\) …18A disc \(K\) of radius \(R\) is cut into three circular sectors in such a way that the areas of the two smaller sectors add up to the area of the largest sector, while the difference of the areas of the …19Three squares are inscribed in a right triangle \(ABC\) with the right angle at \(B\), arranged as in the picture: the largest one has two sides lying on the legs of the triangle, and each of the two smaller …20The triangle \(ABC\) has side lengths \(|AB| = 15\) cm, \(|BC| = 14\) cm and \(|CA| = 13\) cm. Let \(D\) be the foot of the altitude drawn from \(A\) to the side \(BC\), and let \(E\) be the point of that …21Let \(ABCD\) be a convex quadrilateral and let \(E\) and \(F\) be points on the sides \(AB\) and \(AD\) respectively, chosen so that \(EF \parallel BD\). The segment \(CE\) meets the diagonal \(BD\) at …22In a rectangle \(ABCD\) with \(|AB| > |BC|\), the perpendicular bisector of the diagonal \(AC\) meets the side \(CD\) at the point \(E\). The circle with center \(E\) and radius \(|AE|\) meets the side …23A triangle \(ABC\) has a point \(D\) on side \(AB\) and a point \(E\) on side \(AC\) such that \[ |AE| = |ED| = |DB| \qquad\text{and}\qquad |AD| = |DC| = |CB| . \] Determine the angles of triangle \(ABC\). …24A parallelogram \(ABCD\) satisfies \(|AB| = |BD|\). Let \(K\) be the point of line \(AB\), different from \(A\), with \(|KD| = |AD|\). Let \(M\) be the image of \(C\) under the half-turn about \(K\) (so …25The lines containing the two diagonals of a quadrilateral \(ABCD\) meet at an angle of \(60^{\circ}\). Each vertex of the quadrilateral is reflected in the line containing the diagonal that joins its two …26Natasa glued a square and an equilateral triangle of the same side length into a pentagon. Out of seven copies of that pentagon she built the figure shown on the right, which sits inside a large square. …27Three curves are drawn inside a square \(ABCD\): the quarter circle \(\mathcal{Q}\) centred at the vertex \(A\) and passing through \(B\) and \(D\); the semicircle \(\mathcal{P}\) centred at the midpoint …28Let \(M\) be the midpoint of the base \(AB\) of a trapezoid \(ABCD\), so that \(AB \parallel CD\). A point \(E\) is taken in the interior of the segment \(AC\) in such a way that the lines \(BC\) and \(ME\) …29Let \(I\) be the incenter of a triangle \(ABC\), and suppose that \[ |CA| + |AI| = |BC| . \] Determine the ratio of the sizes of the angles \(\angle BAC\) and \(\angle CBA\).30Let \(I\) be the centre of the inscribed circle of a triangle \(ABC\), and let \(A_1\), \(B_1\), \(C_1\) be the feet of the perpendiculars dropped from \(I\) to the sides \(BC\), \(AC\) and \(AB\). The …31Let \(D\) be the midpoint of side \(AB\) of an acute triangle \(ABC\). Points \(A'\) and \(B'\) are chosen on the segments \(AC\) and \(BC\) so that the triangles \(ADA'\) and \(DBB'\) are both isosceles …32In a hexagon \(ABCDEF\) the following hold: \(\angle BAF = 150^{\circ}\), \(\angle ACB = \angle ADC = 90^{\circ}\), \(|AC| = |BC|\), triangle \(ABC\) is similar to triangle \(ADE\), and triangle \(BCD\) …
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