3
$\begingroup$

In a convex quadrilateral $ABCD$ ,$AB\cap CD=O$, let $r_{1},r_{2},r_{3},r_{4}$ is the radius of inscribed circle in triangle $\Delta OAB,\Delta OAD,\Delta OBC,\Delta ODC$ respectively,such $$r_{1}=r_{2}=r_{3}=r_{4}$$ show that:

the convex quadrilateral $ABCD$ is rhombus.

enter image description here

I guess this problem can use Calculation method ,.and I think this problem exsit very nice geometry methods and exsit Algebra methods( I only have this)

follow is my Algebra methods:

let $$OA=a,OB=b,OC=c,OD=d,\angle AOB=x$$ since $$r_{1}=r_{2}=r_{3}=r_{4}$$ then we have $$\dfrac{ab\sin{x}}{a+b+AB}=\dfrac{bc\sin{(\pi-x)}}{b+c+BC}=\dfrac{cd\sin{x}}{c+d+CD}=\dfrac{da\sin{(\pi-x)}}{d+a+DA}$$ let $$\dfrac{a+b+AB}{ab}=k$$ then $$AB=kab-a-b,BC=kbc-b-c,CD=kcd-c-d,DA=kda-a-d$$ and use Cosine therom we have $$AB^2=a^2+b^2-2ab\cos{x}$$ so $$abk^2-2(a+b)k=-2-2\cos{x}$$ let $$p=ak,q=bk,r=ck,s=dk$$ then $$(p-2)(q-2)=2-2\cos{x}$$ simialer we have $$(q-2)(r-2)=2+2\cos{x},(r-2)(s-2)=2-2\cos{x},(s-2)(p-2)=2+2\cos{x}$$ so $$(2-2\cos{x})^2-(2+2\cos{x})^2=0$$ so $$x=\dfrac{\pi}{2}$$ so $$p=q=r=s=2+\sqrt{2}$$ so $$a=b=c=d,x=\dfrac{\pi}{2}$$ so convex quadrilateral $ABCD$ is rhombus

This problem is my student ask me,maybe this problem is Mathematical olympiad problem.

I hope see a nice geometry methos,

Thank you

$\endgroup$
2
$\begingroup$

Hints:

Putting $\;r:=r_i\;$ , we get that if you join the four tangency points of the circles with the two diagonals, then you get a rhombus of side $\;\sqrt2\,r\;$ (why?) $\;\color{red}{(**)}\;$, with part of two given diagonals as its diagonals, which are then perpendicular to each other. But this, of course, it then also true when we look at the original diagonals of the quadrilateral, which is then a rhombus.

$\;\color{red}{(**)}\;$ Things you must remember/prove:

(1) The three sides of a triangle are tangents to the triangle's incircle

(2) If two circles are tangent to each other, the line joining the circles' center passes through the tangency point.

(3) Pythagoras Theorem, in particular: what's the length of a square's diagonal.

(4) A convex quadrilateral is rhombus iff its diagonals bisect each other and are perpendicular to each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.