Dynamic Geometry: P96 Series

This note gives the common calculations in the solution for problems titled Dynamic Geometry: P96, 101, 106, 109, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 125, 127, 128, and 129 so far by @Valentin Duringer. .

Radius, Chord Length, and Coordinates

Let the chord dividing a circle with center at OO into two circular segments of heights 44 and 99 be ABAB. Diameter DEDE through OO cut ABAB perpendicularly at FF. Then DF=4DF = 4, FE=9FE = 9, and diameter DE=DF+FE=13DE = DF + FE = 13, therefore radius of the circle is 6.56.5. By intersecting chords theorem AFFB=DFFE=49    AF=FB=6    AB=12AF \cdot FB = DF \cdot FE = 4 \cdot 9 \implies AF = FB = 6 \implies AB=12 . If we set OO as the origin (0,0)(0,0) of the xyxy-plane with ABAB parallel to the xx-axis, then ABAB is along y=2.5y=2.5, and A=(6,2.5)A = (-6, 2.5) and B=(6,2.5)B=(6,2.5).

Angles of Moving Vertices

Label the upper variable triangle as ABCABC and the lower variable triangle ABCABC'. Since the inscribed angle ACB\angle AC'B and the central angle AOB\angle AOB both form the same arc ACBACB, ACB=12AOB=12(2tan1125)=tan1125\angle AC'B = \frac 12 \angle AOB = \frac 12 \left(2 \tan^{-1} \frac {12}5 \right) = \tan^{-1} \frac {12}5. Similarly, ACB=πtan1125\angle ACB = \pi - \tan^{-1} \frac {12}5.

Side Lengths of Triangle

Since the diameter of a circumcircle is given by the sine rule:

D=ACsinCBA=13    AC=13sinCBAD = \frac {AC}{\sin \angle CBA} = 13 \implies AC = 13\sin \angle CBA

Similarly, BC=13sinCABBC = 13 \sin \angle CAB, AC=13sinCBAAC' = 13\sin \angle C'BA, and BC=13sinCABBC' = 13 \sin \angle C'AB,

The Largest Rectangle Inscribed by a Triangle

Consider any ABC\triangle ABC with a height CD=hCD = h and a width of AB=wAB=w. Let the rectangle ABC\triangle ABC inscribes be KLMNKLMN and the height of CKL\triangle CKL be CE=αhCE = \alpha h, where 0α10 \le \alpha \le 1. Then the area of rectangle KLMNKLMN, A=KLLMA = KL \cdot LM. Since CKL\triangle CKL and ABC\triangle ABC are similar KL=αAB=αwKL = \alpha \cdot AB = \alpha w. And LM=CDCE=hαhLM = CD - CE = h - \alpha h. Then A=α(1α)whA = \alpha (1-\alpha) wh. By AM-GM inequality, 2α(1α)α+(1α)=12\sqrt{\alpha(1-\alpha)} \le \alpha + (1 - \alpha) = 1, and equality occurs when α=1α    α=12\alpha = 1 - \alpha \implies \alpha = \frac 12. Then the rectangle has a maximum area Amax=14whA_{\max} = \frac 14 wh, when the height and breadth of the rectangle are half of that of the triangle.

The Square Inscribed by a Triangle

Let us consider the relationship between the side length ss of a square KLMNKLMN to the height CD=hCD=h and width AB=wAB=w of the triangle ABC\triangle ABC that inscribes it. Note that KLC\triangle KLC and ABC\triangle ABC are similar. Let the height of KLC\triangle KLC be CECE, then CE=swhCE = \dfrac swh. From CD=CE+EDCD = CE + ED, we have

h=swh+s    s=h1+hw=whh+w=12hh+12Since w=AB=12h = \frac swh + s \implies s = \frac h{1+\frac hw} = \frac {wh}{h+w} = \frac {12h}{h+12} \quad \small \blue{\text{Since }w=AB=12}

Similarly for the square in ABC\triangle ABC', s=12hh+12s' = \dfrac {12h'}{h'+12}.

Note by Chew-Seong Cheong
3 months, 2 weeks ago

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Nice work sir !

Valentin Duringer - 3 months, 2 weeks ago

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Actually, the coordinate system I use is different, I use point A(0,0)

Valentin Duringer - 3 months, 2 weeks ago

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