Another construction problem. A pretty easy one this time.

Using a straightedge, (a ruler with no markings that you can use to join two points) find a way to draw a perpendicular through a given point X that is arbitrarily placed on a plane that contains a circle with a given diameter AB.

Also, state the conditions for point X so that your procedure is valid.

I am very pleased to say that I figured it out in the exam hall pretty quickly. And this is from today's / this year's CMI entrance exam paper.

Okay, here's how you do it:

Join X and A and X and B. Let XA meet the circle at Y and XB meet the circle at Z.

Now, join B and Y and A and Z. The meeting point of BY and AZ will be the orthocentre (W) of triangle XAB.

Join XW. Now, XW is perpendicular to AB.

QED.

But is this possible for any point X?

No.

There are three regions in the given plane:

  • inside of the circle
  • outside of the circle
  • circle.

If X lies in the first two regions, then there's no problem. But if X lies on the circle, then X would be the orthocentre as XAB would be a right angled triangle and X=W. So, you can't join X and W because they're the same point. And there are infinitely many lines that can be drawn through a point.

Note by Vishnu C
3 years, 2 months ago

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Ramesh Goenka - 3 years, 1 month ago

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What exactly do you mean by the crossed out/ "wrong" lines?

Vishnu C - 3 years, 1 month ago

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In the crossed out, or "wrong" lines that you've posted, the triangle XAB is an obtuse angled triangle. So if you produce XB to meet the circle at Z, then AZ would be perpendicular to XZ and if the point of intersection of XA and the circle is Y, then the meeting point of BY and AZ would give the orthocentre (W) again. So why is it wrong?

Vishnu C - 3 years, 1 month ago

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@Vishnu C dint considered that .. !! u r right :D

Ramesh Goenka - 3 years, 1 month ago

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@Michael Mendrin I thought you'll enjoy solving this one as well. Try not to make a messy figure this time : )

Vishnu C - 3 years, 2 months ago

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