Qi Huan's square

Consider $A,$ $B,$ $C$ and $D$ are points on cartesian coordinates with $A(0,0),$ $B(10,0),$ $C(10,10)$ and $D(0,10).$ Let point $P(x,y)$ (with $x$ and $y$ both positive) located inside square $ABCD.$ Because $PA$ $=$ $6$ and $PB$ $=$ $8,$

$x^2$ $+$ $y^2$ $=$ $6^2$

$(10 - x)^2$ $+$ $y^2$ $=$ $8^2$

Solve both equations as we get $x$ $=$ $\frac {18}{5}$ and $y$ $=$ $\pm \frac {24}{5},$ hence $P(\frac {18}{5},\frac {24}{5}).$

$PC^2$ $=$ $(10 - \frac {18}{5})^2$ + $(10 - \frac {24}{5})^2$

$PC^2$ $=$ $68$

$PC$ $=$ $2\sqrt{17}$

As a result, $a + b$ = $2 + 17$ = $19$

$APB$ is a right triangle due to the Pythagorean theorem. Let $H$ be the height from $P$ of the right triangle $APB$ . Due to the first euclidean theorem, $HB= \frac {32}{5}$ and for the second euclidean theorem, $HP=\frac {24}{5}$ . Since $ABCD$ is a square, $CP^2=(BC-HP)^2+HB^2$ . Hence, $CP=2\cdot \sqrt{17}$ , so $a+b=19$

Let $\angle ABP$ be called $\theta$ , and $\angle PBC$ be called $\delta$ .

Our goal is to find $\cos \delta$ . Once we find this, we can use the law of cosines to solve for the missing side $PC$ of $\Delta PBC$ .

Given that $\Delta ABP$ has sides of length 6, 8, and 10, we know that $\Delta ABP$ is a right triangle. Using right triangle identities, we find that $\sin \theta = \frac{opp}{hyp} = \frac{AP}{AB} = \frac{6}{10} = \frac{3}{5}$ .

Let us note that $\theta = 90^\circ - \delta$ . Here we can use an important trig identity:
$\sin (90^\circ - \delta) = \cos \delta$

$\therefore$ $\frac{3}{5} = \sin \theta = \sin (90^\circ - \delta) = \cos \delta$

Now let's look at the triangle $\Delta PBC$ . We know $PB = 8$ , $BC = 10$ , and $\cos \delta = \frac{3}{5}$ .

We can use the Law of Cosines to solve for the missing side $PC$ :

Law of Cosines: $Z^2 = X^2 + Y^2 - 2*X*Y*\cos z$ , where $\angle z$ is the angle opposite side $Z$

First let $PC$ be called $y$ . Then $y^2 = 8^2 + 10^2 - 2*8*10*\cos \delta$ . We know $\cos \delta = \frac {3}{5}$ , so $y^2 = 164 - 160*\frac{3}{5} = 164-96 = 68$

$\therefore y = \sqrt {68} = 2\sqrt {17}$ . Recall that $y = PC$ , so $PC = 2\sqrt {17}$ .

Now that we have $PC$ written in the appropriate form $a\sqrt{b}$ , we can see that $a + b = 2 + 17 = 19$

First, we note that 6, 8 and 10 are the measures of a special rectangle triangle (I call special those with integers measures). Lets draw a segment perpendicular to AD from P, intersecting in point M. We obtain another rectangle triangle. Since \angle PAM = \angle PBA , and \angle APM = \angle PAB , they are similar triangles. What this means is that \frac {10}{8} = \frac {6}{AM} , so AM = \frac {24}{5}. By the same way, we can get PM = \frac {18}{5}.

Now lets draw another segment from P, perpendicular now to CD, intersecting at N. Note that triangle PNC is a rectangle triangle with hypotenuse a\sqrt {b}. As the square has sides of lenght 10, then hick PN = 10 - \frac {24}{5} = \frac {26}{5}. Equally, hick NC = 10 - \frac {18}{5} = \frac {32}{5}. Applying the Pythagorean Theorem, \sqrt{{\frac {26}{5}}^2 + {\frac {32}{5}}^2} = \sqrt{68}. 68 = 4 * 17, so a = 2 and b = 17. The value of a + b = 2 + 17 = 19.

Let us draw in a perpendicular from $P$ to $AB$ , calling the point on $AB$ $E$ . We now have two similar right triangles, $APB$ and $PEB$ . We can set up the equation $\frac{AB}{PB}=\frac{AP}{PE}$ . Plugging in values for $AB$ , $PB$ , and $AP$ , we find that $PE=\frac{24}{5}$ . Let us now draw in a perpendicular from $P$ to $BC$ , calling the point on $BC$ $F$ . By the Pythagorean theorem, we can set up the equation $BF^2+PF^2=PB^2$ . Plugging in values for $PF$ and $BF$ , since $BF=PE$ , we find that $PF^2=\frac{1024}{25}$ . Since we know that $BC=BF+FC$ , we can plug in values for $BC$ and $BF$ , getting that $FC=\frac{26}{5}$ . We now apply the Pythagorean theorem again, getting that $FC^2+PF^2=PC^2$ . Plugging in values for $FC$ and $PF^2$ , we get that $PC=2\sqrt{17}$ . Therefore, $a+b=2+17=19$ .

As everyone should know, the triangle ABP is rectangle, 'cause its sides are a pythagorean triple (the right angle is between AP and BP). From the Euclidean theorem, we get that h_\overline{AB}= \frac {\overline{PA} \cdot \overline{PB}}{\overline{AB}} . Replacing, h_\overline{AB} = \frac {6 \cdot 8}{10} = 4.8 .

Let's call Q the intersection between $\overline{AB}$ and h_\overline{AB} . In the BPQ triangle, we can use the pythagorean theorem again, we get:

$\overline{PQ}^2+\overline{BQ}^2=\overline{BP}^2$

$4.8^2+\overline{BQ}^2=8^2$

$23.04+\overline{BQ}^2=64$

$\overline{BQ}^2=40.96$

$\overline{BQ}=|6.4|$

Now, if we draw a rectangle that has P and C as its opposite corners and sides along $\overline{CD}$ and $\overline{BC}$ , it's clear that the distance between P and $\overline{CD}$ is $10-4.8=5.2$ , and the distance between P and $\overline{BC}$ is $6.4$ . Connecting P and C we form the diagonal of said rectangle, and using the pythagorean theorem for the last time we have:

$5.2^2+6.4^2=\overline{CP}^2$

$68=\overline{CP}^2$

$\overline{CP}=|\sqrt{68}|=\sqrt{68}=\sqrt{4\cdot17}=2 \sqrt{17}$

It's clear that a=2 and b=17, thus the answer is 19.

Drop a perpendicular from P onto AB at L. Now using similarity of triangles or basic trigonometry PL/6=8/10 Therefore PL=4.8. Next we drop another perpendicular from P onto BC at M. Clearly MC=BC-BM=BC-PL=10-4.8=5.2 Again applying basic trigonometry in triangle PBL we get BL/8=8/10 Therefore BL=6.4=PM Therefore by Pythagoras theorem in triangle PCM we get PC=2

a=2 b=17 a+b=19

Draw a perpendicular from P on AB(at E) and CD(at F). Area of triangle ABP=1/2 AB EP=1/2 AP PB.Put the values of AP,PB,AB and you will get EP=4.8, Now EF=EP+PF=10. By putting the value of EP=4.8, we get PF=5.2. in triangle EBP by usinmg Pythagoras theorem find the value of EB: EP EP+EB EB=8 8. =>EB=6.4=FC(by putting the value of EP=4.8) Now use the same method on triangle PFC. you will get PC=root(68)=2 root(17) so, 2+17=19

Draw a line EF that passes through point P and touches sides AD and BC at E and F respectively. Let AE , ED , EP , PF be a , b , c , d , respectively. Thus, we have a ^2+ c ^2=36 and a ^2+ d ^2=64.

Subtracting the above equations we get

( d - c )( d + c )=28

Since d + c =10, d - c =14/5

Solving the above equation we get d =32/5 and c =18/5

Thus, a =6^2-(18/5)^2=24/5 and b =10-24/5=26/5

Therefore, PC ^2=(32/5)^2+(26/5)^2=1700/25 and thus,

PC =(10 \sqrt {17})/5=2 \sqrt {17}

Finally a + b =17+2=19

Drop perpendicular from P to AB and BC at Q and R. Let BQ=PR=x and PQ=BR=y. From Pithagora in PQA and PQB we have (10-x)^2+y^2=36 and x^2+y^2=64.From second y^2=64-x^2 puting in first we have 36=64-x^2+(10-x)^2=64-100+20x. 20x=72, x=72/20, and y= sqrt(64-(72/20)^2)=24/5. Now PC=sqrt(RC^2+x^2)=sqrt((10-y)^2+x^2)=sqrt(68)=2*sqrt(17). a+b=2+17=19

Drop a perpendicular from P onto AB at L. Now using similarity of triangles or basic trigonometry PL/6=8/10

Therefore ,PL=4.8 Next we drop another perpendicular from P onto BC at M.

Clearly MC=BC-BM=BC-PL=10-4.8=5.2

Again applying basic trigonometry in triangle PBL , we get , BL/8=8/10

Therefore ,BL=6.4=PM

Therefore by Pythagoras theorem in triangle PCM we get PC=2*(17)^1/2

So now, a=2 b=17 therefore :- a+b=19

P is the interior point of square ABCD. Now, join (A,P), (B,P), (C,P), (D,P). Drop perpendiculars PM on AB and PN on BC.

In triangle PAB, PA=6, PB=8 and AB=10. (6, 8, 10) form a Pythagorean triplet. So, it is easy to conclude that angle APB = 90 degrees.

Consider the two similar triangles PAM and PBM. We get \frac {8}{6} = \frac {MB}{PM} = \frac {PM}{10-MB}

Solving we have PM = \frac {24}{5} and hence MB = \frac {32}{5}

In triangle PCN, applying Pythagoras' Theorem, PC^2 = PN^2 + NC^2 = MB^2 + (10 - PM)^2.

Putting the values of MB and PM, we finally get PC = 2 \sqrt{17}.

Hence the answer is 2+17 = 19.

Let the angle in front of $PA = A, PB = B, PC= C$ Because $PA=6$ and $PB=8$ and the hypotenuse in front of those two sides is $10$ we know that the angle in front of $∠A = 37°$ and $∠B= 53°$ . And also $∠C$ is complementary with $∠A$ which means $∠C = 90°- 37° = 53°$ . By using cosinus rule :

$PC^2 = PB^2+BC^2 - 2.BC. PB. cos ∠C$

$PC^2 = 64+100 - 2.8.10. cos 53°$

$PC^2 = 164 - 160 . \ (\frac{3}{5}$ )

$PC = 2 \sqrt{17}$

Because $PC = a\sqrt{b}$ , therefore $a+b = 2+17 = 19$