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# 丁丑 Minimum of AP*BP = ___.?

Line segment [AB = 7] is on a plane.

Distance of point P and AB is 3.

Minimum of AP*BP = ___.

### 2 Answers

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- atsuoLv 63 years agoFavorite Answer
Think △APB , its base is AB .

The height of △APB is the distance between P and AB , so it is 3 .

Therefore , the area of △APB is S = (1/2)7*3 = 21/2 .

On the other hand , the area of △ABP must be S = (1/2)AP*BP*sin(∠APB) .

So

(1/2)AP*BP*sin(∠APB) = 21/2

AP*BP*sin(∠APB) = 21

AP*BP = 21 / sin(∠APB)

And 0 < sin(∠APB) ≦ 1 , therefore

AP*BP has the minimum value of 21 when ∠APB = 90° .

(The lengths of AP and BP are not requested .)

- Φ² = Φ+1Lv 73 years ago
Minimum of AP*BP = (√((7/2)² + 3²) )² = 49/4 + 9 = 85/4

- ...Show all comments
Naive intuition sees only one symmety, which is what caused this answer to go wrong. Symmetrry arguments can be formulated rigorously, and are mathematically valid.

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