Jubayer's circle

Geometry Level 2

In triangle $\triangle ABC$ , $AB = 37, AC = 33$ , and $BC = 39$ . A circle $\Gamma$ is drawn with center $A$ and radius $AC$ . Let $D$ be the point on $\Gamma$ that is furthest from $B$ . What is the distance $BD$ ?

This problem is posed by Jubayer N .

The answer is 70.

1 solution

Pranav Kirsur
Aug 12, 2013

The problem becomes simple if we find out where D is. We can prove that D is the point where ray BA meets Γ. If D is any other point on the circle,construct AD,which is equal to 33. Then BA+AD>AD, due to the triangle inequality. Clearly, if D lies on ray BA, BA+AD=BD. So, B is greatest in the latter. Then we have< BD=BA+AD=37+33=70 So,the answer is 70.

Moderator note:

Great answer!

Avoid using the same notation (in this case, D) to refer to two different things. It is better to say

Let X be any other point on the circle. Construct AX, which is equal to 33. Then $70 = BA + AX > BX$ .

If the answer tell us to find the longest line of BD, and the question tell us that A is the center of the circle with radius of AC. AC=33, in this case
B _ _ _ _ A _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ D-------------------------------------------------------------------------------------------37---33 (RADIUS)---------------------------------------------------------------------------------------- Lastly, we add up the length of BA and AD=BA+AD=37+33=70 Therefore, the answer is 70

Tips: The furthest of a point in a circle depends on the radius of a circle.

Jing Cai Chan - 7 years, 10 months ago

brilliant

ojas dhiman - 7 years, 10 months ago

lovely solution!

A Former Brilliant Member - 7 years, 10 months ago

Jubayer's circle

The answer is 70.

1 solution

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