New questionMy practice

This saved animation is in Chinese. Read the English solution, or enter an English question to create an English lesson.

Rotation and shortest distance: a moving-point problem

?

Adjust available parameters to see how shapes and values change. When a lesson has no adjustable elements, you can still watch the animation or compare frames.

    0.00s
    Step 0 / 0Waiting to start

    Submit a question to see the current step explained here.

    1x
    Enlarged question image
    Original image
    WRITTEN SOLUTION

    Full answer

    Original question

    Enable narration

    Your browser needs permission to play audio. Select Enable audio to hear the explanation.

    Preparing explanationPlease wait while your lesson is prepared.

    Depending on the problem, an explanation usually takes 1–5 minutes.

    Solution steps

      Geometry · Rotation

      Question

      In rectangle ABCD, AB = 4 and AD = 8. E moves on the circle with centre A and radius 2. Construct the right isosceles triangle EBF to the right of EB, with its right angle at B. Find the minimum length of DF.

      Rectangle ABCD with moving points E and FAB = 4 and AD = 8. E moves on a circle with centre A and radius 2. BE and BF are equal and perpendicular. G is the midpoint of BC; F moves on the circle with centre G and radius 2.ABCDEFG48
      G is an auxiliary point and the centre of F's circular locus. The animation shows the construction.

      Step-by-step solution

      1. Relate E and F by a rotation

      Since BE = BF and ∠EBF = 90°, F is the image of E under a clockwise rotation of 90° about B. Rotate E's entire circle to find the locus of F.

      2. Find the new centre G

      Rotate A clockwise through 90° about B to get G. AB = 4 and BC = AD = 8, so BG = 4 and G is the midpoint of BC. Rotation preserves distance: GF = AE = 2. Thus F moves on the circle with centre G and radius 2.

      3. Calculate DG

      In right triangle DCG, DC = AB = 4 and CG = BC − BG = 4. By Pythagoras, DG = √(DC² + CG²) = √(4² + 4²) = 4√2.

      4. Find the minimum and check equality

      The triangle inequality gives DF + FG ≥ DG, hence DF ≥ 4√2 − 2. Since DG > 2, equality holds when D, F and G are collinear with F on segment DG. This point F corresponds to a point E on the original circle, so the minimum is attainable.

      The minimum value of DF is 4√2−2.

      Which method does this problem use?

      Use the right isosceles triangle to identify a 90° rotation. It turns F's locus into a circle. For a fixed point outside a circle, the shortest distance is the distance to the centre minus the radius. Check both the outside condition and the equality position.