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In summary, the conversation discusses finding the length of PQ in an isosceles triangle ABC, where P is on AC such that AP=3PC and Q is on CB such that CQ=3BQ. The conversation mentions using the cosine theorem but the final answer is obtained without it, using only Pythagoras' theorem. The method involves setting up equations and solving for the unknown lengths, and then using Pythagoras' theorem again to find PQ.

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BvU

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mfb

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You can use the cosine theorem and then replace that cosine by a different expression.

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BvU

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post your work -- it's required to get assistance in PFDank2 said:I know i can use the Cosine theorem

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Dank2

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Thanks, solved using cosine for the two triangles

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aheight

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mfb

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aheight

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Can I ask if the problem is solvable without invoking the cosine law strictly through geometric means?

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mfb

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Put B at the origin, BC to the right along the x direction, BA (length c) upwards/right. Let a be the length CB.

C=(0, 0)

B=(a, 0)

A=(a/2, ##\sqrt{c^2-a^2/4}##)

Therefore:

Q=(3a/4, 0)

P=(a/8, 1/4##\sqrt{c^2-a^2/4}##)

Using Pythagoras again, ##PQ = \sqrt{(5a/8)^2 + 1/4^2 (c^2-a^2/4)} = \frac{1}{4}\sqrt{6a^2 + c^2}##

I don't think you can avoid anything like that completely. Square roots typically mean you need Pythagoras or more.

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aheight

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mfb said:

Put B at the origin, BC to the right along the x direction, BA (length c) upwards/right. Let a be the length CB.

C=(0, 0)

B=(a, 0)

A=(a/2, ##\sqrt{c^2-a^2/4}##)

Therefore:

Q=(3a/4, 0)

P=(a/8, 1/4##\sqrt{c^2-a^2/4}##)

Using Pythagoras again, ##PQ = \sqrt{(5a/8)^2 + 1/4^2 (c^2-a^2/4)} = \frac{1}{4}\sqrt{6a^2 + c^2}##

I don't think you can avoid anything like that completely. Square roots typically mean you need Pythagoras or more.

Thanks mfb! Here's how I worked through your solution:

1. Basically, using similar purple and red triangles below, we need to determine ##x## and ##y## in the diagram below. So we have

$$

\begin{array}{c}

\frac{\sqrt{c^2-a^2/4}}{a/2}=y/x\\

(c/4)^2=x^2+y^2

\end{array}$$

Once we have ##x## and ##y## , we can then use ##(QP)^2=(QR)^2+y^2## although quite a bit of algebra to get from P to Q.

A triangle is a polygon with three sides and three angles. It is one of the basic shapes in geometry.

The three types of triangles are equilateral, isosceles, and scalene. An equilateral triangle has three equal sides and three equal angles. An isosceles triangle has two equal sides and two equal angles. A scalene triangle has three unequal sides and three unequal angles.

The formula for finding the area of a triangle is A = 1/2 * base * height. The base is the length of one side of the triangle, and the height is the perpendicular distance from the base to the opposite vertex.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem is often used to find missing side lengths in right triangles.

Two triangles are similar if they have the same shape but different sizes. This means that their corresponding angles are equal, and their corresponding sides are in proportion. To determine if two triangles are similar, you can compare their corresponding angles or use the side-angle-side or angle-angle-angle similarity theorems.

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