Triangles with equal areas

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Source: Baltic Way 2014, Problem 13

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#1 h Nov 11, 2014, 2:42 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a square inscribed in a circle $\omega$ and let $P$ be a point on the shorter arc $AB$ of $\omega$ . Let $CP\cap BD = R$ and $DP \cap AC = S.$
Show that triangles $ARB$ and $DSR$ have equal areas.

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#2 h Nov 12, 2014, 8:29 PM • 1 Y

Y by Adventure10

Let $O$ be the intersection point of $AC$ and $BD$ . As $\angle SOD=90^\circ=\angle BPD$ , $\angle OSD=90^\circ-\angle SDO=\angle PBD$ . Furthermore, $\angle SAD=45^\circ=\angle BAC=\angle BPC$ . By the law of sines: $\frac {DA}{DS}=\frac {\sin \angle ASD}{\sin \angle SAD}=\frac {\sin \angle OSD}{\sin \angle SAD}=\frac {\sin \angle PBR}{\sin \angle BPR}=\frac {RP}{RB}\,.$ Calculating the power of $R$ WRT $\omega$ , we get: $DR \cdot RB=CR \cdot RP$ . Thus, $\frac {DA}{DS}=\frac {RP}{RB}=\frac {DR}{CR}$ and $CB \cdot CR=DA \cdot CR=DS \cdot DR$ . Noting that $\angle BCR=\angle BCP=\angle BDP=\angle RDS$ , we obtain that $\triangle CBR$ and $\triangle DSR$ have equal areas. But $\triangle CRB \cong \triangle ARB$ .

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#3 h Nov 21, 2014, 3:50 PM • 1 Y

Y by Adventure10

I met this problem few years ago and I have used a small trick to avoid trigo; it took me some time to recall it, but here it is:
Clearly $\triangle ARB\cong\triangle CBR\ (\ 1\ )$ . Adding to its area the area of $\triangle CDR$ we get:
$[CRB]+[CRD]=[BCD]=R^2\ (\ 2\ )$ . Adding to $[RDS]$ the same $[CDR]$ we get $[RDS]+[CDR]=[CDSR]$ . $CDSR$ having perpendicular diagonals, its area is $[CDSR]=\frac{CS\cdot DR}{2}\ (\ 3\ )$ , so we need $CS\cdot DR=2R^2$ . The last relation we get from $\triangle CDR\sim\triangle SCD\implies \frac{DR}{CD}=\frac{CD}{CS}$ , done.
Here $R$ is the circumradius the square.

Best regards,
sunken rock

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#4 h Nov 21, 2014, 7:45 PM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the center of $ABCD$ and let $D_{\infty}$ be the point at infinity of $AC$ (also point at infinity of the tangent of $\omega$ at D).

$C(O,B,R,D) \equiv C(A,B,P,D)=D(A,B,P,D_{\infty}) \equiv D(A,O,S,D_{\infty}) \Longrightarrow$

$\frac{OB}{OD} \cdot \frac{RD}{RB}=\frac{OA}{OS} \Longrightarrow OS \cdot RD=OA \cdot RB \Longrightarrow [DSR]=[ARB].$

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jayme

#5 h Nov 7, 2015, 2:06 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
there is a typo in this problem: P is the midpoint of the shorter arc AB...

Sincerely
Jean-Louis

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IstekOlympiadTeam

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IstekOlympiadTeam

#6 h Nov 7, 2015, 2:29 PM • 2 Y

Y by Adventure10, Mango247

jayme wrote:

Dear Mathlinkers,
there is a typo in this problem: P is the midpoint of the shorter arc AB...

Sincerely
Jean-Louis

Special Case

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jayme

#7 h Nov 7, 2015, 2:40 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
my figure after toying desn't give the result...
Make a figure and verify again... Perhaps I am wrong but make a tentative...

Sincerely
Jean-Louis

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erbed

#8 h Oct 3, 2016, 7:24 AM • 2 Y

Y by Adventure10, Mango247

Trigonometry handles this problem quite easily.
Let $\phi=\angle ACP$ , then $\angle ADP = \phi$ too. Also let $r$ be the length of the radius of the circle.
In the right-angled triangle $OCR$ we have $tan\phi = \frac{OR}{OC}=\frac{OR}{r}$
Thus $OR = r tan\phi$ . Consequently $BR = OB - OR = r - r tan\phi = r(1-tan\phi)$
So the area of triangle $ARB$ is: $[ARB] = \frac{1}{2} \cdot AB \cdot BR \cdot \sin(\angle ABR)=\frac{1}{2} r\sqrt{2}\cdot r(1-tan\phi) \frac{\sqrt{2}}{2}=\frac{1}{2}r^2(1-tan\phi)$ On the other hand, in the right-angled triangle $ODS$ we have: $tan(45^\circ-\phi) = \frac{OS}{OD}=\frac{OS}{r}$ , so $OS=rtan(45^\circ-\phi)$
Also $DR=OD+OR=r+r tan\phi = r(1+tan\phi)$
So the area of the triangle $DSR$ is: $[DSR]=\frac{1}{2}\cdot OS \cdot DR=\frac{1}{2}rtan(45^\circ-\phi)r(1+tan\phi) = \frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)$ Finally, using the trigonometrical formula:
$tan(45^\circ-\phi) = \frac{tan 45^\circ - tan\phi}{1+tan 45^\circ \cdot tan\phi} = \frac{1-tan\phi}{1+tan\phi}$ So the area of the triangle $DSR$ is:
$[DSR]=\frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)=\frac{1}{2}r^2\frac{1-tan\phi}{1+tan\phi}(1+tan\phi)=\frac{1}{2}r^2(1-tan\phi)$ Thus the areas of $ARB$ and $DSR$ are equal.

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KRIS17

#9 h Sep 14, 2019, 8:55 PM • 1 Y

Y by Adventure10

Let $M$ be the intersection point of lines $PD$ and $AB$ and let $N$ be the projection of $R$ onto line $AB$ .
Let $O$ be the circumcenter of circle $\omega$ . Now $W.L.O.G$ , assume that side length of square $ABCD$ = 1.

Since $\angle DPC$ = $\angle DAC$ = $45^\circ$ = $\angle ABC$ , $PBRM$ is cyclic.
So, $\angle DRM$ = $\angle DPB$ = $90^\circ$ , implying that $MR \parallel AC$ .

So $\triangle DSO \sim \triangle DMR \implies DO / DR = SO / MR \implies SO.DR = DO.MR \implies$ area of $\triangle DSR = 1/2.(1/\sqrt2).MR ---> (1)$

In $\triangle MRB$ , $\angle BMR = \angle BAC = 45^\circ = \angle MBR \implies \triangle MRB$ is an isosceles right triangle. So $RN$ = $MR / \sqrt2$
Hence area of $\triangle ARB = 1/2.(1).(MR / \sqrt2) ---> (2)$

From $(1)$ and $(2)$ above, $[DSR]=[ARB].$

Attachments:

This post has been edited 3 times. Last edited by KRIS17, Sep 14, 2019, 9:13 PM

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jayme

#10 h Nov 1, 2019, 8:15 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20III.pdf p. 18...

Sincerely
Jean-Louis

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1079 posts

#11 h Aug 13, 2021, 8:54 PM

Y by

Claim. $DC$ is tangent to $(PSC)$ .
Proof.
We have
$\measuredangle SPC=\measuredangle DPC=\measuredangle DBC=45^\circ=\measuredangle ACD=\measuredangle SCD,$ the claim follows.

Now,
$\begin{align*} S_{DSR}&=S_{ARB}\Longleftrightarrow \\ DS\cdot PR\cdot \sin{\angle DPR}\cdot \frac{1}{2}&=AB\cdot BR\cdot \sin{\angle ABR}\cdot \frac{1}{2}\Longleftrightarrow \\ DS\cdot PR&=AB\cdot BR\Longleftrightarrow \\ \frac{RB}{RP}&=\frac{DS}{DC}\Longleftrightarrow \\ \frac{\sin{\angle RPB}}{\sin{\angle PBR} }\cdot &=\frac{\sin{\angle DCS}}{\sin{\angle DSC}}\Longleftrightarrow \\ \sin{\angle PCD}&=\sin{\angle DSC}, \end{align*}$ which is true by the claim, we are done.

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#12 h Apr 6, 2025, 9:50 AM

Y by

Here is nicer solution almost trigonofree.
By some angle chase we can prove that $\triangle PAS \sim \triangle PRB$ . Since $\angle SPR+\angle APB=180^{\circ}$ . From here one can prove that $[PAB]=[PSR]$ . So $[DSRC]=[PDC]-[PSR]=[PDC]-[PAB]=[DBC]$ . Which implies that $[DSR]=[CRB]$ , and we're done.

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