How show that $AB>BR$ in rhombus?

Question

Let $A,B$ be points on the side $QR$ and $C,D$ be points on the side $RS$ of the rhombus $PQRS$, such that $B$ is closer to $R$ than $A$ is and that $C$ is closer to $R$ than $D$ is. Suppose that the segments $PA$, $PB$, $PC$, $PD$ divide the angle $\angle SPQ$ into five equal angles. How show that $AB>BR$?

score 0 · Answer 1 · edited Jun 12 '20 at 10:38

enter image description here Some basic facts:-

(1) a, the “altitude” of a rhombus is fixed and calculate-able from the given conditions.

(2) By symmetry, the angle distribution is in the ratio h : 2h : 2h as shown.

(3) k, as marked in the figure should also be a known quantity.

(E1) …. W = a tan (2h + k)

(E2) …. W + v = a tan (4h + k)

(E3) …. W + u + v = a tan (5h + k)

(E2) – (E1): v = a[tan (4h + k) – tan (2h + k)]

(E3) – (E2): u = a[tan (5h + k) – tan (4h + k)]

It is then required to show v – u > 0.

That is, a[tan (4h + k) – tan (2h + k)] – [a[tan (5h + k) – tan (4h + k)]] > 0

i.e. 2tan (4h + k) – [tan (5h + k) + tan (2h + k)] > 0 (since a is non-zero positive)

Applying the multiple angle formula, we should be able to show the above is true.

Note also that the last inequality must subject to the restriction 10h + k = 90 degrees.

Well, how prove that $2\tan (4h + k) – [\tan (5h + k) + \tan (2h + k)] > 0?$ — piteer, Aug 05 '14 at 18:07
@piteer Letting j = 2h + k, it is further reduced to 2tan (2h + j) – [tan (3h + j) + tan (j)] > 0. Then, the inequality can be simplified more easier. It passed with some dry run data. — Mick, Aug 06 '14 at 02:06

score 0 · Accepted Answer · answered Aug 06 '14 at 01:05

connect $SR$, let $\angle BSR=x \implies \angle ASB=2x, \angle BRS=5x<\dfrac{\pi}{2} $

let $BR=p,AB=q,SR=a,SB=b$, apply sin law:

$\dfrac{p}{\sin{x}}=\dfrac{a}{\sin{6x}},\dfrac{q}{\sin{2x}}=\dfrac{b}{\sin{8x}},\dfrac{a}{\sin{6x}}=\dfrac{b}{\sin{5x}} \implies \dfrac{q}{p}=\dfrac{\sin{2x}\sin{5x}}{\sin{x}\sin{8x}}=f(x)$

$f(x)=\dfrac{2\sin{5x}\cos{x}}{2\sin{4x}\cos{4x}}=\dfrac{\sin{6x}+\sin{4x}}{2\sin{4x}\cos{4x}}=\dfrac{\sin{4x}\cos{2x}+\cos{4x}\sin{2x}}{2\sin{4x}\cos{4x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{2x}}{2\cos{4x}}+\dfrac{\sin{2x}}{2\sin{4x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{2x}}{2\cos{4x}}+\dfrac{1}{4\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{2\cos^2{2x}+1}{4\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{4x}+2}{4\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{1}{4\cos{2x}}+\dfrac{1}{2\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}$

note $2x<4x <5x <\dfrac{\pi}{2}, \dfrac{1}{\cos{2x}}$ and $\dfrac{1}{\cos{4x}}$ are mono increasing funtion and positive $\implies f(x)$ is mono increasing function $ \implies f_{min}=f(0)=\dfrac{5}{4} \implies \dfrac{q}{p}>\dfrac{5}{4}>1 $

How show that $AB>BR$ in rhombus?

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