Problem 31
Let C be the midpont of an arc AB of a circle. Let P be any point on the circle, such that P and C lie on the opposite sides of AB. Show that:
(PA + PB)/PC is constant, if P varies
Here’s a simulation that illustrates the problem.
Try dragging A or B. Then, try dragging P. Does (PA + PB)/PC remain constant? How can we prove the same?
Solution
Here are some tools that can help you solve the problem.
Sine Rule
In any triangle ABC,
BC/sinA = CA/sinB = AB/sinC
Here’s a simulation where you can explore this property.
You can drag the three vertices and observe the three ratios BC/sinA, CA/sinB, and AB/sinC. Are they always equal?
Angles Subtended By Equal Arcs
If two arcs of a circle are equal, then the angles they subtend at the circumference are also equal.
Here’s a simulation where you can explore this property.
Here, AB and CD are equal. Try dragging P and Q. You can also drag A and C to change the positions of these arcs. Two things to ensure:
- P and the circle’s center should be on the same side of AB.
- Q and the circle’s center should be on the same side of CD.
Are the angles APB and CQD always equal?
Now, to proceed further, try applying the sine rule in the triangles BPC and APC (in the original figure). See if you can obtain the ratios PA/PC and PB/PC in terms of some angles. Then, add these two ratios. Maybe you’ll get an expression that always remains a constant?
Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Telegram group.
Problem Source: CutTheKnot
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