Minimum/Maximum Problems Related To Triangles

Suppose AB and AC are equal sides. Let us draw height CD of triangle ABC. The area of ABC is AB^.CD/2. If angle BAC is not equal 90 degrees, then the length of CD is less than that of AC because CD is a leg and AC is the hypotenuse of triangle ACD. Therefore, BAC of 90 degrees gives the largest area.

What angle between two equal sides of a given length will result in an isosceles triangle with the largest area?

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Let us build a circle for which hypotenuse AB is a diameter. Consider radius CD of this circle perpendicular to AB. Triangle ADB has the biggest area. For any other right triangle AFB whose hypotenuse is AB, its third vertex F lies on the circle built. Its height EF is shorter than CD because EF is a leg and CF that equals CD is the hypotenuse of CEF. Therefore, the area of AFB is smaller than that of ADB.

What right triangle with a given hypotenuse has the largest area? Please construct it.

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Let us draw straight line BE so that it is perpendicular to the given line and AD=DE, and then let us draw a straight line AE. C is its intersection with the given line. Consider any other point F on the given line. AF+FE > AC+CE because AE is a triangle side, and AF and FE its other two sides. Since triangles CBD and CDE are congruent, and so are triangles FBD and FDE, CB=CE and FE=FB. Therefore, AF+FB+BA > AC+CB+BA.

Two points A and B are located on one side of a straight line. Find point C on this line such that the perimeter of ABC is minimal.

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Since triangle sides are chords of its circumcircle, the diameter of the circumcircle is longer or equal than each of the sides. Therefore, the diameter is a at least. Let us draw a circle whose diameter is a and intersect the circle with a line segment of length b.

Please construct a triangle whose two side lengths are a and b (a>b) and whose circumcircle is the smallest. What is the diameter of this circumcircle?

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Let draw a line going through C and which is perpendicular to AB. D is the point where this line crosses the circle. ABD has the smallest area because DE is the height of ABD and for any other point F on the circle, the height FG is longer than DE. (The assumption that FG is less or equal than DE leads to a contradiction: CF is the hypotenuse of a triangle whose leg is longer than CD or equals it.) This problem has a soltion only if the circle does not cross and does not touch line AB. If it does, than for any triangle built on A, B and a point on the circle as its vertices, it is possible to build yet another triangle having even smaller area.

Let points A and B lie outside of a circle whose center is C. Please construct such a triangle ABD that D lies on this circle and its area is the smallest. Define the cases for which this problem does not have a solution.

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The ratio should be one, that is, these segments should be equal. Suppose line BC is such that OB = CO. Consider another line - FG - going through O. Let us overlap triangles OFB and CGO. Angles COG and FOB are equal. Angle OCG is equal to the sum of angles CAB and ABC, and thus, it is bigger than angle ABC. Triangle OFB is a part of triangle OGC. Hence, the area of OFB is smaller. The area of AFG is the area of rectangle AFOC plus the area of triangle OGC. The area of ABC is the area of rectangle AFOC plus the area of triangle OFB. Therefore, the area of ABC is smaller. The case of line DE is considered similarly. See the picture of triangles OBD and ECO overlapped.

There is an angle and a point within it. Consider all possible lines going through the point and intersecting both sides of the angle. What should be the ratio of the lengths of the segments of such line from this point to angle sides so that the area of the triangle formed by the angle sides and this line is minimal?

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It is the isosceles triangle touching the circle at the point where the angle bisectrix crosses the circle. Since the bisectrix is also a meadian, BG = GC. Suppose DE forms another triangle with the same circle inscribed in it. Let us draw an isosceles triangle whose one side is equal BC, and two equal angles are the same as angles DFB and CFE. Suppose KL = FC and LM = BF, angle KLP equals FCE. Note that KL > LM and KLP is an obtuse angle. Therefore, LP crosses MN. Triangles KLP and FCE are equal. So are triangles LQM and BFD because angle BDF equals FCA. The difference between the ADE and ABC perimeters is: DF + FE + CE - DB - BF - FC. This difference can be expressed as: MQ + KP + (LQ + QP) - LQ - LM - KL = MQ + KP + QP - KM. KM is a straight line, hence it is shorter than KPQM. Therefore, the perimeter of ADE is bigger.

A circle is inscribed in an angle. How to draw a line intersecting with both sides of the angle so that the sides of the angle and this line form a triangle of minimal perimeter in which the circle is inscribed?

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Here is a trick: this sum is the same for all points on AC. Let PQ be parallel to AB and CF be a height. PD = GF because they are opposite sides of a rectangular. Triangle PQC is similar to ABC, and hence PQ = QC. CG = PE because they are heights crossing equal sides of an isosceles triangle. Therefore, PD + PE = CF.

Find such point P on side AC of isosceles triangle ABC (AB = BC) that the sum of lengths of two heights from P to the equal sides is maximum.