Michael's triangle

$\bigtriangleup ABD$ has

$AB^2+AD^2=1+1=2=BD^2$ and $AB=AD$ ,

so $\bigtriangleup ABD$ is an isosceles and right triangle at $A$ .

$\Rightarrow \widehat{ABD}=45^\circ$

$\bigtriangleup CBD$ has $CB^2+CD^2=\frac{1}{2} +\frac{3} {2}=2=BD^2$ ,

so $\bigtriangleup CBD$ is a right triangle at C.

Besides, $\sin \widehat{CBD}=\frac{CD}{DB}=\frac{\sqrt{3}}{2}$

$\Rightarrow \widehat{CBD}=60^\circ$

We have

$\sin \widehat{ABC}=\sin (\widehat{ABD}+\widehat{CBD})$

$=\sin (45^{\circ}+60^{\circ})$

$=\sin 45^{\circ}\cos 60^{\circ}+\cos 45^{\circ}\sin 60^{\circ}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Hence,

$(ABC)=\frac{1}{2}.AB.BC.\sin \widehat{ABC}=\frac{1}{2}.1.\frac{1}{\sqrt{2}}.\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{8}$

so $a+b+c=12.$

First, let's look at $\triangle BCD$ . Using converse of the Pythagorean theorem as $BC^2 + CD^2 = BD^2$ , we know that $\triangle BCD$ is rectangular triangle.

In rectangular triangle we have $BC$ - cathetus, that have half of the hypotenuse lenght, therefore $\angle CDB = 30 ^ \circ$ and $\angle CBD = 60 ^ \circ$ .

Similarly, let's look at $\triangle ABD$ . Using converse of the Pythagorean theorem we know that $\triangle ABD$ is rectangular triangle. Also it is isosceles, so $\angle DBA = 45 ^ \circ$ .

Now let's use square formula: $Area ABC = \frac{1}{2}*AB*BC*\sin \angle ABC = \\ = \frac{1}{2}*1*\frac{\sqrt{2}}{2}*\sin (45 ^ \circ+60 ^ \circ) = \frac{\sqrt{2}}{4} * \sin (105 ^ \circ)$ .

Let's use sin sum formula: $\sin (105 ^ \circ) = \sin (45 ^ \circ+60 ^ \circ) = \sin (45 ^ \circ)*\cos(60 ^ \circ) +\\ + \sin (60 ^ \circ)*\cos(45 ^ \circ) = \frac{1}{\sqrt{2}} * (\frac{1+\sqrt{3}}{2})$ .

So $Area ABC = \frac{\sqrt{2}}{4} * \sin (105 ^ \circ) = \frac{1+\sqrt{3}}{2*\sqrt{2}} * \frac{\sqrt{2}}{4} = \frac{1+\sqrt{3}}{8}$ .

And the answer is $1+3+8 = 12$ .

we see that AB^2 + AD^2 = BD^2. Hence angle A is 90 By sine rule BD/sin90 = AD/sin@ @=45

we see that BC^2 + CD^2 = BD^2. Hence angle C is 90 By sine rule BD/sin90 = CD/sin# #=60

area=1/2(AB)(BC)[sin(45+60)] =1/2(1)(1/\sqrt{2})[(\sqrt{3}/2)(1/ \sqrt{2} )+(1/\sqrt{2})(1/2) =\frac{\sqrt{3}+1}{8}

3+1+8=12

We can see that the triangle ABD is an isosceles, right triangle, and that the triangle BDC is a right triangle with the angles of 30(<BDC), 60(<DBC) and 90(<BCD) degrees. The area of the triangle ABC is given by S=ABxBCxsin(<ABC)/2=1x(\sqrt{2}/2)x[sin(45+60)]/2=(\sqrt{2}/2)(sin45xcos60+sin60xcos45)/2=(\sqrt{2}+\sqrt{6})\sqrt{2}/16=(2+2\sqrt{3})/16=(1+\sqrt{3})/8 a=3; b=1; c=8... a+b+c=12.

Solution:

The triangle $ABD$ is notable of $45^\circ$ so $\angle ABD = 45^\circ$ . The triangle $DBC$ is notable of $30^\circ$ and $60^\circ$ so $\angle DBC = 60^\circ$ . Form the triangle rectangle $CBP$ where $P$ be the foot of the perpedicular of $C$ in the prolongation of $AB$ . By substraction we find that $\angle CBP = 75^\circ$ , Therefore the triangle $BPC$ is notable and the side $BP = \frac{ \sqrt{3} + 1}{4}$ . Then the area of triangle $CBP = \frac{ \sqrt{3} + 1}{8}$ . The value of $a + b + c = 12$ .

First, notice that the three lengths AB, AD, and BD form a 45-45-90 right triangle. Additionally, the triangle ACD is also right, with its angles being 30-60-90. Therefore, the area of ABC can be found by $\frac{1}{2}*AB*BC*\sin(B)$ . AB and BC are given, and angle B is $45+30 = 75$ from out previous observation. Hence, $\sin(B) = \frac{\sqrt{6}+\sqrt{2}}{4}$ from the sine summation formula. Thus, $[ABC] = \frac{1+\sqrt{3}}{8}$ and $a+b+c = 3+1+8 = \bf{12}$ .

Clearly, the values form pythagoros triplets, angle BAD=angle BCD=90 angle ABD=45, angle CBD=60 => angle ABC= 105 area = 1/2 AB BC*sin(ABC)

Triangle ABD is a 45–45–90 triangle and triangle BCD is a 30–60–90 triangle. \angle ABD = 45 ^ \circ and \angle CBD = 60 ^ \circ, so \angle ABC = 105 ^ \circ.

Area of triangle ABC is \frac {\overline{AB} \times \overline{BC} \times \angle ABC}{2} = 12