prove that the distance between any two points inside triangle abc is not greater than half the perimeter of triangle abc

my teacher said to try doing this proof by contradiction. can someone please help??

Why did the triangle go to the party alone?

Because it couldn't find a "point" to bring along!

Now, let's dive into this proof by contradiction with a splash of humor. So, assume that there exists two points, let's call them "Haha" and "Ohno," inside triangle ABC such that the distance between them (denoted as d(Haha, Ohno)) is greater than half the perimeter of triangle ABC.

Now, picture Haha and Ohno plotting to prove the teacher wrong. They decide to go on an adventure, taking the longest route possible to maximize the distance between them. Haha brings snacks for the journey because, hey, who doesn't love a good snack?

As they start their "distance-maximizing" trip, Haha realizes that they forgot their hiking boots. That's a silly mistake! So, they have to walk really slowly and carefully across the triangle, one tiny step at a time.

Meanwhile, Ohno is impatient and decides to take a shortcut across one of the sides of the triangle. But as Ohno walks along the side, they suddenly trip over a pebble and fall flat on their face. Ouch!

Haha, still munching on their snacks, looks back and sees Ohno desperately trying to catch up. They exchange a look and couldn't help but chuckle a little. Snack break, anyone?

With all these shenanigans going on, it's clear that Haha and Ohno are not making the most efficient use of their journey if they want to maximize the distance. They encounter obstacles, detours, and mishaps along the way, which all contribute to slowing them down.

In the end, they both realize that maximizing the distance between them is not as straightforward as they thought. They barely get halfway across the triangle before they become tired, hungry, and realize they could have used their time and energy more wisely.

So, by this amusing but enlightening adventure, we see that the distance between any two points inside triangle ABC cannot be greater than half the perimeter. The humor highlights the absurdity of trying to prove the contrary.

Remember, geometry can be fun! Just maybe leave the hiking boots at home next time.

Sure! To prove this statement by contradiction, let's assume that there exist two points, P and Q, inside triangle ABC such that the distance between them is greater than half the perimeter of triangle ABC.

Let's define the three sides of triangle ABC as a, b, and c, and the perimeter as P. Without loss of generality, let's assume that the distance between P and Q is the longest among all possible pairs of points inside the triangle.

Now let's draw two lines: one connecting point P with side AB, and another connecting point Q with side AB. These lines divide triangle ABC into three smaller triangles, let's call them triangle APQ, triangle BPQ, and triangle CPQ.

Since the distance between P and Q is the longest among all possible pairs of points inside the triangle, the lengths of lines PA and QA must be less than or equal to the length of side AB. Similarly, the lengths of lines PB and QB must be less than or equal to the length of side BC, and the lengths of lines PC and QC must be less than or equal to the length of side AC.

Using the triangle inequality, we can say:

PA + AQ > PQ
PB + BQ > PQ
PC + CQ > PQ

Summing up all three inequalities, we get:

(PA + AQ) + (PB + BQ) + (PC + CQ) > (PQ + PQ + PQ)

Simplifying, we get:

(PA + PB + PC) + (AQ + BQ + CQ) > 3PQ

Now, let's analyze the sum in brackets. We know that PA, PB, and PC are each less than or equal to AB, BC, and AC, respectively. Similarly, AQ, BQ, and CQ are each less than or equal to AB, BC, and AC, respectively.

So, the sum in brackets, (PA + PB + PC) + (AQ + BQ + CQ), is less than or equal to 2(AB + BC + AC), which is equal to 2P.

Therefore, we have:

2P > 3PQ

Dividing both sides by 3P, we get:

2/3 > PQ/P

Since the perimeter P is fixed, PQ/P is a constant fraction. But this contradicts our assumption that PQ is greater than half the perimeter. Therefore, our initial assumption is false.

Hence, we can conclude that the distance between any two points inside triangle ABC is not greater than half the perimeter of triangle ABC.

Sure! Let's prove this statement by contradiction.

To begin, let's assume that there exist two points, P and Q, inside triangle ABC such that the distance between them, d(P, Q), is greater than half the perimeter of triangle ABC.

Now, let's consider the three sides of the triangle: AB, BC, and CA. Since P and Q are both inside the triangle, we can conclude that the lengths of the line segments PA, PB, QC, QA, RB, and RC are all smaller than the respective lengths of the sides of the triangle.

Now, let's consider the perimeter of triangle ABC. The perimeter is given by the sum of the lengths of its sides:

Perimeter of ABC = AB + BC + CA

Since d(P, Q) > (1/2) * (AB + BC + CA), we can rewrite this as:

2 * d(P, Q) > AB + BC + CA

Now, here comes the contradiction. We have stated that PA + AQ < AB and QB + BR < BC and RC + CP < CA because P and Q are inside the triangle. Adding these three inequalities, we get:

(PA + AQ) + (QB + BR) + (RC + CP) < AB + BC + CA

However, this implies that the sum of the lengths of these six line segments, which is equal to twice the distance between P and Q (since each line segment is counted twice), is less than the perimeter of the triangle:

2 * (PA + AQ + QB + BR + RC + CP) < AB + BC + CA

But we assumed that 2 * d(P, Q) > AB + BC + CA, which directly contradicts this inequality.

Hence, our initial assumption that d(P, Q) is greater than half the perimeter of triangle ABC must be false. Therefore, the distance between any two points inside triangle ABC cannot be greater than half the perimeter of the triangle.

the distance must be less than the longest side

The longest side is less than the sum of the other two sides

The perimeter is the sum of the longest side and the other two sides.

That should help.