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The Exaggerated Triangle Inequality

Triangle Inequality is an exaggerated version of the Basic Idea of the Euclidean Plane. Let’s do some Triangle inequality Problems and Solutions.

Triangle Inequality is an exaggerated version of the Basic Idea of the Euclidean Plane, something we follow every day in our lives –

The shortest distance between two points is the straight-line distance between the two points.

It tells you, if you move from A to B via another point P, then the distance of travel will be larger than traveling along AB.

But, we can play with and nurture this idea to solve interesting puzzles. Let’s do some Triangle inequality Problems and Solutions. A single diagram is sufficient for the arguments.

Excited right? Let’s explore!

Problem 1: A, B, C, D are four points. Find a point P such that PA + PB + PC + PD is minimum.

Solution:

ABCD is rotated and translated in such a way ABCD -> CEFG, and A, C, and F lie on a straight line. H -> I.

Now, observe, that HA + HB + HC + HD = AH + HC + CI + IF, which is minimized if H lies on AC and I lies on CF. I lie on CF is the same as H lies on BD. So, it is minimized if H lies on AC and BD both, which is the intersection point of the diagonal.

Problem 3: A is a line. L and M are two points in the same plane on the opposite side of the line. Find a point N on A such that NL + NM is minimum.

Solution:

LO + OM is minimized if L, O, M lie on a straight line. Therefore, N is the required point.

Problem 2: A is a line. L and M are two points in the same plane on the same side of the line. Find a point N on A such that NL + NM is minimum.

Solution:

M’ is the reflected point of M along line A (JK). Hence, NM’ = NM. Hence, LO + OM = LO + OM’. Hence, it is minimized if L, O, M’ lie on a straight line. Therefore, N is the required point.

Problem 4: A is a line. L and M are two points in the same plane on the same side of the line. Find a point P on A such that |PL – PM| is minimum.

Solution:

LP = PM. Just draw the perpendicular bisector of LM, and see its intersection with line A(JK).

Problem 5: A is a line. L and M are two points in the same plane on the opposite side of the line. Find a point P on A such that |PL – PM| is minimum.

Solution:

LP = PM. Just draw the perpendicular bisector of LM, and see its intersection with line A(JK).

Problem 6: A is a line. L and M are two points in the same plane on the same side of the line. Find a point O on A such that |OL – OM| is maximum.

Solution:

Observe that |LP-PM| \( \leq \) LM. Hence, it’s maximum value is AB. This occurs for the position O, when O, L, and M lie on a straight line. O is the intersection of the line LM and JK (Line A).

Problem 7: A is a line. L and M are two points in the same plane on the opposite side of the line. Find a point O on A such that |OL – OM| is maximum.

Solution:

Observe that |LP-PM|\(\leq\) LM. Hence, its maximum value is AB. This occurs for the position O when O, L, and M lie in a straight line. O is the intersection of the line LM and JK (Line A).

Problem 8: An ant is caught on one corner of a cuboid with sides l, b, and h. It wants to reach the diagonally opposite corner, However, the ant can perform a walk only along with the faces of the cuboid. What is the least amount of distance that the ant needs to walk to reach the other corner?

Solution:

triangle inequality solution
Just open the cube up. You will find the minimum length path just along the straight line.
Source of the Diagram: Math Stack Exchange

More, similar problems to come. Stay tuned!

By Srijit Mukherjee

I Learn. I Dream. I Enjoy. I Share.

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