Let $ABCDE$ be a convex pentagon such that $AB=AE=CD=1$, $\angle ABC=\angle DEA=90^\circ$ and $BC+DE=1$. Compute the area of the pentagon. Greece
Problem
Source: JBMO 1998, Problem 2
Tags: geometry, geometry unsolved
RaMlaF
26.12.2004 01:31
just a hint: take P on the ray DE such that EP = BC. Then you'll have DP=1, and the triangle EAP is equal to BAC, so the area of ACDP is equal to the area of the original pentagon, and DP=CD=1, AP=AC, and you complete the solution...
pilot
26.12.2004 02:16
Thanks for the hint. I understand it now, there are two congruent triangles with base 1 and height 1 so the answer is 1.
Jessen1111
21.09.2014 10:11
test test test
KRIS17
29.11.2018 08:23
Let $BC = a, ED = 1 - a$ Let angle $DAC$ = $X$ Applying cosine rule to triangle $DAC$ we get: $Cos X = (AC ^ {2} + AD ^ {2} - DC ^ {2}) / (2 * AC * AD )$ Substituting $AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1$ we get: $Cos^{2} X = (1 - a - a ^ {2}) ^ {2} / ((1 + a^{2})(2 - 2a + a^{2}))$ From above, $Sin^{2} X = 1 - Cos^{2} X = 1 / ((1 + a^{2})(2 - 2a + a^{2})) = 1/(AC^{2}.AD^{2})$ Thus, $Sin X * AC * AD = 1$ So, $Area$ of triangle $DAC$ = $(1/2)*Sin X * AC * AD = 1/2$ Let $AF$ be the altitude of triangle DAC from A. So $1/2*DC*AF = 1/2$ This implies $AF = 1$. Since $AFCB$ is a cyclic quadrilateral with $AB = AF$, traingle $ABC$ is congruent to $AFC$. Similarly $AEDF$ is a cyclic quadrilateral and traingle $AED$ is congruent to $AFD$. So $area$ of triangle $ABC$ + $area$ of triangle $AED$ = $area$ of Triangle $ADC$. Thus $area$ of pentagon $ABCD$ = $area$ of $ABC$ + $area$ of $AED$ + $area$ of $DAC$ = $1/2 + 1/2 = 1$
abduzadr
23.06.2020 10:02
площадь треугольника APD=1/2. В четырехугольнике APDC AP=BC и PD=DC поэтому треугольники APD и ADC равны следовательно их площадь равна Sapd=Sadc=1/2 поэтому Sapd+Sadc=1/2+1/2=1=Sapdc=Sabcde
math31415926535
25.08.2021 02:42
Let $BC=x,$ so $DE=1-x,$ then we have that$$[AED]+[ABC]=\frac{x+(1-x)}{2}=\frac{1}{2}.$$So, now we just want to find the area of $\triangle{ACD}.$ Now, let the altitude from $A$ to $CD$ be $h,$ and can the foot of the altitude $H,$ also let $CH=y,$ so $DH=30-y,$ now, we pythag bash. We have$$h^2+y^2=1+x^2, (1)$$$$h^2+(1-y)^2=1+(1-x)^2 . (2)$$The second equation simplifies to$$h^2+y^2-2=x^2+1-2x,$$substituting $h^2+y^2=1+x^2$ we get that $x=y,$ so $(1)$ can be simplified down to $h^2=1 \implies h=1.$ Therefore the area of $\triangle{ACD}$ is $\frac{1 \cdot 1}{2}=\frac{1}{2},$ so the desired area if $\frac{1}{2}+\frac{1}{2}=\boxed{1}.$
DottedCaculator
25.08.2021 03:27
[asy][asy]
unitsize(0.2cm);
pair A, B, C, D, E, X;
A=(20,30);
B=(-100/13,240/13);
C=(0,0);
D=(30,0);
E=(38,6);
X=(20,0);
draw(A--B--C--D--E--A--X--B--X--E);
label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, dir(0));
label("$X$", X, S);
[/asy][/asy]
Let $X$ be the point on $CD$ such that $DX=DE$ and $CX=CB$. Then, we have $\angle DXE=\angle DEX$ and $\angle CXB=\angle CBX$. Therefore, $\angle EXB=\angle AEX+\angle ABX$, so $\angle BAE=360^{\circ}-2\angle EXB$. Therefore, since $AB=AE$, we must have that $A$ is the circumcenter of $BXE$, so $AB=AE=AX=1$. Since $DE=DX$, $AE=AX$, and $AD=AD$, we get that $\triangle AED\cong\triangle AXD$, so $\angle AXD=\angle AED=90^{\circ}$. Therefore, the area of triangle $ACD$ is $\frac121\cdot1=1$, and the sum of the areas of triangles $ABC$ and $ADE$ is $\frac12BC+\frac12DE=\frac12$. Therefore, the area of $ABCDE$ is $\boxed{1}$.