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I have been playing the app Euclidea, I have been doing quite well but this one has me stumped.

"Construct a triangle whose perimeter is the minimum possible whose vertices lie on two side of the angle and the third vertex is at point a"

see image >> minimum perimeter

I think the solution may have something to the Problem of Apollonius, but I cant quite wrap my head around it.

Any help much appreciated.

Mike
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2 Answers2

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enter image description here

Reflect the point $A$ in each of the two lines to get points $A_1,A_2$. Let the line $A_1A_2$ meet the two lines at $B,C$. The required triangle is $ABC$.

almagest
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  • That great - I can see that it is correct. I just can understand why? – Mike May 10 '16 at 21:47
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    It is just that the straight line is the shortest distance between two points! If you took any positions for $B,C$, you would still have $AB=A_1B$ and $CA=CA_2$, so that the perimeter would be $A_1B+BC+CA_2$. The positions of $A_1$ and $A_2$ would not change, but the line would be jagged not straight. – almagest May 11 '16 at 06:29
  • What if $\rm A_1A_2$ lies to the left of the intersection of the given lines? – TryingHardToBecomeAGoodPrSlvr Nov 01 '17 at 22:31
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Using the drawing above, IF you draw A1B congruent to BA and have drawn A2C congruent CA then mA1A2= the perimeter of triangle ABC.

Given an angle and a point A, you are to tell EXACTLY how you would position point B and then point C to give the minimum perimeter for ABC.

Look at line segment AB. Suppose AB was longer. Would that effect A1B? So would A1A2 and therefore the triangle perimeter be longer?

Start at A. What is the shortest distance to the upper side of the given angle? Make THAT point B. (Remember something about given a point off a line, the shortest distance is the perpendicular through that point to that line?)

Think about the same thing moving from A to C. What is the shortest distance from A to C?

Now that you HAVE a point C, as well as a point B, decide what is the shortest path from B to C.

Continue line segment BC upward where mA1B=mBA. Mark your new point A1.

Continue line segment BC downward where mA2C=mCA. Mark this point A2.

Now you may use THIS A1A2 (equal to the perimeter of the triangle) against which to compare any other attempts.

Any other path to the upper side should be longer than the perpendicular distance. Any other path to the lower side should be longer than the perpendicular distance. Any other path from B to C should be longer than the straight path.

Note here that drawing a straight line from current A to A2 looks like that would be a shorter distance. If you have drawn your current AB perpendicular to the upper side of the angle, moving B would increase the distance of AB, like BC would move further out to the right making BC longer, A1B which is the same length as AB would become longer. A similar problem occurs if you try to move C to some position between A and A2. Do play around with it though and see how the perimeter is effected judging by your new A1A2 you create if you move points B and C.

Why is this question difficult? You may have stalled at minimizing the perimeter. What we did was break the question into parts. Can you minimize the distance between a point and a line, could you do that again with another point and line, and also minimize the distance between two points? Would doing that help minimize the perimeter as desired? If you can break a larger question into parts and attack any of the parts, that is a beginning.

Katherine Rogers
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