Follow The Fold

I intend to calculate the area of $CPNM$ by finding the area of trapezoid $CDNM$ and subtracting the area of $\Delta PDN$ . So my formula is essentially

$\frac{3}{2}(DN+CM)-\frac{1}{2}\cdot DP\cdot DN.$

If we let $BM=x$ and $CM=5-x$ , then the Pythagorean Theorem tells us that $1+x^2=(5-x)^2$ . Solving this yields $BM=x=\frac{12}{5}$ and $CM=\frac{13}{5}$ .

It is straightforward to show that $\Delta CBM \sim \Delta PAC \sim \Delta PDN$ . This tells us that

$PC=\frac{CM}{BM}\cdot AC = \frac{13}{12}\cdot 2 = \frac{13}{6},$

which means that $DP=3-\frac{13}{6}=\frac{5}{6}$ . Using triangle similarity again, we see that

$DN=\frac{BM}{CB}\cdot DP = \frac{12}{5}\cdot\frac{5}{6} = 2.$

(Incidentally, this means that $\Delta PAC \cong \Delta PDN$ .) So now I can plug everything into my formula to get the area of $CPNM$ :

$\frac{3}{2}\Big(2+\frac{13}{5}\Big) - \frac{1}{2}\cdot \frac{5}{6}\cdot 2 = \frac{69}{10}-\frac{5}{6} = \frac{91}{15}.$

So the answer to the problem is $91+15=\boxed{106}$ .

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$\Delta MCB$ is a right triangle. So, $CB^2+BM^2=MC^2 \implies 1^2+x^2=(5-x)^2 \implies x=\frac{12}{5}$

$\Delta MCB \sim \Delta CPA \implies \frac{AP}{BC}=\frac{CA}{MB} \implies AP=\frac{5}{6}$

$\Delta PAC$ is a right triangle. So, $PA^2+AC^2=PC^2 \implies (\frac{5}{6})^2+2^2=PC^2 \implies PC=\frac{13}{6}$

$CD=3 \implies PD=3-\frac{13}{6}=\frac{5}{6}$

But, $\Delta PAC \sim \Delta PDN \implies \Delta PAC \simeq \Delta PDN \implies PN=PC=\frac{13}{6}$

Area of $CPNM=Ar.(ABNM)-Ar. \Delta APC -Ar. \Delta BMC$ $= \frac{1}{2} \times 3 \times (\frac{12}{5}+\frac{5}{6}+\frac{13}{6})-\frac{1}{2}(2 \times \frac{5}{6})-\frac{1}{2}(1 \times \frac{12}{5})=\frac{91}{15}$

So, $91+15= \boxed{106}$

The idea is to compute $BM$ and $AP$ and hence areas of $\triangle CBM$ and $\triangle ACP$ and subtract it from area of rectangle $ABCD$ which is $15$ . We Write the $C$ on $AB$ as $C_1$ and that $D$ after folding as $D_1$

Now, We notice that if $BM=x$ then $C_1M$ = $5-x$ . Hence, Pythagorus Theorem in $\triangle BMC_1$ , we have

$BM^2 + C_1B^2 = C_1M^2$

$\implies$ $x^2 + 1^2 = (5-x)^2$ . Solving we get $BM=\dfrac{12}{5}$ and hence $[\triangle C_1BM]=\boxed{\dfrac{6}{5}}$

Moreover, if $\angle BC_1M=\theta$ , then, $\angle AC_1P = (90^{\circ} - \theta)$ . We have $\tan\theta = \dfrac{12}{5}$ and hence $\dfrac{AC_1}{AP} = \cot \angle AC_1P = \cot (90^{\circ} - \theta) = \tan\theta = \dfrac{12}{5}$ . Solving, we get $AP = \dfrac{5}{6}$ and $[\triangle AC_1P] =\boxed{\dfrac{5}{6}}$ .

Now, we need to find $DN$ . For that, consider the right angled $\triangle PND_1$ . Now, it is easy to observe that $D_1N \parallel C_1M$ and $AD \parallel BC$ and thus $\angle D_1NP = \angle BMC_1$ . If we take $DN = D_1= y$ then we have $PN= 5 - y - \frac{5}{6}$ . We have $\tan\angle BMC_1 = \dfrac{12}{13}$ and hence $\dfrac{D_1N}{PN} = \dfrac{12}{13}$ . Solving, we get $DN=2$ .

Now we are left with computing the area of trapezium $NDCM = \frac{1}{2} \times \text{sum of the parallel sides} \times \text{distance between them}$ .

We get,

$[NDCM] = \dfrac{1}{2} \times (2+\frac{13}{5}) \times 3 = \boxed{\dfrac{69}{10}}$ . Hence,

$[CPNM] = [ABCD] - [APC_1] - [C_1BM] - [NDCM] = 15 - \dfrac{5}{6} - \dfrac{6}{5} - \dfrac{69}{10} = \dfrac{91}{15}$ . Hence the answer is $91 + 15 = \boxed{106}$

Piece of ice cream!

$A_{CPNM}=\dfrac{3(5)-\dfrac{1}{2}(1)\left(\dfrac{12}{5}\right)-2\left(\dfrac{5}{6}\right)}{2}=\dfrac{91}{15}$

The desired answer is $m+n=91+15=$ $\boxed{106}$

Consider CN as a diameter, then A and D lies on this crcle. So, NDC and NAC are congruent triangle. So DPN and APC are congruent.

=> AP=5/6, CP=13/6

ACP and BMC are similar triangles and not congruent. Using BC=1(given).Find the area of BMC

All the areas are available now. Ans => 91/15 => 106

To start, I will be calculating some length sizes in order to help me calculate the areas.

First of all, take a look at the triangle $CBM$ . If $BM$ is $x$ , then $CM$ is $5 - x$ ; and since CBM is a right triangle, we can use Pythagoras' Theorem to find $x$ .

$x^{2} + 1 = (5 - x)^{2}$

$x^{2} + 1 = 25 - 10x + x^{2}$

$10x = 24$

$x = 2.4$

Thus, $BM = 2.4$ and $CM = 2.6$

Now, take a look at the triangle $APC$ . If $AB$ is a line and the angle $PCM$ is a right angle, then the angle $ACP$ is $90° - BCM$ and thus the angle $APC$ is equal to the angle $BCM$ , and the triangles $ACP$ and $BCM$ are similar. Then, we can write: $AP/AC = BC/BM$ . Then $AP/2 = 1/2.4$ , so $AP = 5/6$

We can find PC by applying once more the Pythagorean Theorem:

$AP^{2} + AC^{2} = PC^{2}$

$(5/6)^{2} + 2^{2} = PC^{2}$

$25/36 + 4 = PC^{2}$

$PC = 13/6$

Then $DP = 3 - 13/6 = 5/6$ .

Look now at triangle $DPN$ . The angles $DPN$ and $APC$ are opposite, and thus are equal; the angle $PDN$ is right, so the angle $PND$ equals angle $PCA$ , and so the triangles $PDN$ and $APC$ are similar; furthermore, since $PA = PD$ , the triangles are equal, thus $DN = AC = 2$ , and $PN = 13/6$ .

Seeing as the angles $NDC$ and $DCM$ are right, the quadrilateral $DCMN$ is a trapezium, and its area is given by $(DN + CM)*DC/2 = (2 + 2.6)*3/2 = 4.6*3/2 = 6.9$

The area of the triangle $DPN$ is given by $DP*DN/2 = 2*(5/6)/2 = 5/6$ .

The area of $CPNM$ equals $6.9 - 5/6 = (69/10 - 5/6) = (207/30 - 25/30) = 182/30 = 91/15$ . Thus, $m = 91$ , $n = 15$ , and the sum $m + n$ equals $106$ .

BM=x and CM=5-x using pythagorean theorem you can solve for x and it comes out to be 2.4. Triangles CBM and APC are similar by a ratio of 1.2 usinig this ratio you can calculate the sides of triangle APC. Then triangles APC and PDN are congruent. The length of side AN is 3 so extend line BM to make it length 3. Call the point to which you extend it R. Side RM is 3-2.4 which equals .6. Now take the sum of the areas of triangles RMN, CBM, and APC this gives you 88/30 subtract that from 270/30 and you get 91/15

Triangle CMB, Triangle PCA and triangle PND are similar. Starting with triangle CMB angle MCB = Theta We've got sideBC. The sum of sides BM + MC =5 Get Sin^{2}(theta) +Cos^{2}(theta)=1. Thus all sides of the triangle are known. For a level 4 it wont be difficult to proceed from here :D

O segredo é somar a área dos 3 triângulo, somá-las,depois subtrair da área do retângulo e dividir por 2. O motivo de dividir por dois é porque o papel foi dobrado.

• para achar a área dos triângulos temos que fazer semelhança entre eles, pois tem os mesmos ângulos.