Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution: If the first number is $6$ , then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If the first number is $5$ $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If the first number is $4$ , ....
4 6 _ _ _ _ $\implies$ 24 ways
4 _ 6 _ _ _ $\implies$ 24 ways
4 _ _ 6 _ _ $\implies$ 24 ways
4 _ _ _ 6 _ $\implies$ 5 must go between $4$ and $6$ , so there are $3 \cdot 3! = 18$ ways.
$24 + 24 + 24 + 18 = 90$ ways if 4 is first.
If the first number is $3$ , ....
3 6 _ _ _ _ $\implies$ 24 ways
3 _ 6 _ _ _ $\implies$ 24 ways
3 1 _ 6 _ _ $\implies$ 4 ways
3 2 _ 6 _ _ $\implies$ 4 ways
3 4 _ 6 _ _ $\implies$ 6 ways
3 5 _ 6 _ _ $\implies$ 6 ways
3 5 _ _ 6 _ $\implies$ 6 ways
3 _ 5 _ 6 _ $\implies$ 6 ways
3 _ _ 5 6 _ $\implies$ 4 ways
$24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84$ ways
If the first number is $2$ , ....
2 6 _ _ _ _ $\implies$ 24 ways
2 _ 6 _ _ _ $\implies$ 18 ways
2 3 _ 6 _ _ $\implies$ 4 ways
2 4 _ 6 _ _ $\implies$ 6 ways
2 5 _ 6 _ _ $\implies$ 6 ways
2 5 _ _ 6 _ $\implies$ 6 ways
2 _ 5 _ 6 _ $\implies$ 4 ways
2 4 _ 5 6 _ $\implies$ 2 ways
2 3 4 5 6 1 $\implies$ 1 way
$24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ ways
Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Answer:  461
Problem: Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution: If $6$ is the first number, then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$ , and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If $6$ is the third number, then we cannot have the following:
1 _ 6 _ _ _ $\implies$ 24 ways
2 1 6 _ _ _ $\implies$ 6 ways
$120 - 24 - 6 = 90$ ways
If $6$ is the fourth number, then we cannot have the following:
1 _ _ 6 _ _ $\implies$ 24 ways
2 1 _ 6 _ _ $\implies$ 6 ways
2 3 1 6 _ _ $\implies$ 2 ways
3 1 2 6 _ _ $\implies$ 2 ways
3 2 1 6 _ _ $\implies$ 2 ways
$120 - 24 - 6 - 2 - 2 - 2 = 84$ ways
If $6$ is the fifth number, then we cannot have the following:
_ _ _ _ 6 5 $\implies$ 24 ways
1 5 _ _ 6 _ $\implies$ 6 ways
1 _ 5 _ 6 _ $\implies$ 6 ways
2 1 5 _ 6 _ $\implies$ 2 ways
1 _ _ 5 6 _ $\implies$ 6 ways
2 1 _ 5 6 _ $\implies$ 2 ways
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 $\implies$ 3 ways
$120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71$ ways
Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Answer:  461
Problem: Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution: Note the condition in the problem is equivalent to the following condition: for each $k$ with $1 \le k \le 5$ , the first $k$ terms is not a permutation $(1, 2, \ldots, k)$ (since it would mean there must be some integer $x$ in the first $k$ terms such that $x \not \in \{1, 2, \ldots, k\} \implies x > k$ ). Then, let $a_n$ denote the number of permutations of $(1, 2, \ldots, n)$ satisfying the condition in the problem. We use complementary counting to find $a_n$ . Notice that in order to not satisfy the condition in the problem, there are $n-1$ cases: the first $1 \le k \le n-1$ (we don't include $k = n$ since the condition in the problem only holds up to $n-1$ ) terms are a permutation of $(1, 2, \ldots, k)$ , and for all $k+1 \le i \le n-1$ , the condition in the problem still holds. Then, for each of these cases, there are $k!$ ways to arrange the first $k$ terms, and then $a_{n-k}$ ways to arrange the $k + 1$ to $n$ terms (assume by contradiction that terms from $k+1$ to $i$ is a permutation of $(k+1, k+2, \ldots, i)$ . Then, since the first $k$ terms are already a permutation of $(1, 2, \ldots, k)$ , the first $i$ terms form a permutation of $(1, 2, \ldots, i)$ . This contradicts our assumption that for all $k+1 \le i \le n-1$ , the condition still holds. Thus, we can only include the $a_{n-k}$ permutations of these terms). Now, summing the cases up and subtracting from $n!$ , we have: $a_n = n! - \sum_{k=1}^{n-1} a_{n-k} k!$ . From this recursion, we derive that $a_1 = 1$ $a_2 = 1$ $a_3 = 3$ $a_4 = 13$ $a_5 = 71$ , and finally $a_6 = \boxed{461}$

Answer:  461
Problem: Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution: Let $A_i$ be the set of permutations such that there is no number greater than $i$ in the first $i$ places. Note that $\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}$ for all $1\le b_0 < b_1\cdots < b_k \le 5$ and that the set of restricted permutations is $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$
We will compute the cardinality of this set with PIE. \begin{align*} &|A_1| + |A_2| + |A_3| + |A_4| + |A_5|\\ = &120 + 48 + 36 + 48 + 120 = 372\\ \\ &|A_1 \cap A_2| + |A_1 \cap A_3| + |A_1 \cap A_4| + |A_1 \cap A_5| + |A_2 \cap A_3|\\ + &|A_2 \cap A_4| + |A_2 \cap A_5| + |A_3 \cap A_4| + |A_3 \cap A_5| + |A_4 \cap A_5|\\=&24+12+12+24+12+8+12+12+12+24=152\\ \\ &|A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_2 \cap A_5| + |A_1 \cap A_3 \cap A_4| + |A_1 \cap A_3 \cap A_5|\\ +& |A_1 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_5| +  |A_2 \cap A_4 \cap A_5| + |A_3 \cap A_4 \cap A_5|\\=&6 + 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6 = 48\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4| + |A_1 \cap A_2 \cap A_3 \cap A_5| + |A_1 \cap A_2 \cap A_4 \cap A_5| + |A_1 \cap A_3 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4 \cap A_5|\\=&2 + 2 + 2 + 2 + 2 = 10\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = 1 \end{align*} To finish, $720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}$

Answer:  461
Problem: Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution: Define the function $f(p,q)$ as the amount of permutations with maximum digit $q$ and string length $p$ that satisfy the condition within bounds. For example, $f(4,5)$ would be the amount of ways to make a string with length $4$ with the highest digit being $5$ . We wish to obtain $f(6,6)=f(5,6)$
To generate recursion, consider how we would get to $f(p,q)$ from $f(p-1,a)$ for all $a$ such that $p\le{a}\le6$ . We could either jump from the old maximum $a$ to the new $q$ by concatenating the old string and the new digit $q$ , or one could retain the maximum, in which case $a=q$ . To retain the maximum, one would have to pick a new available digit not exceeding $q$
In the first case, there is only one way to pick the new digit, namely picking $q$ . For the second case, there are $q-p+1$ digits left to choose, because there are $q$ digits between 1 and $q$ total and there are $p-1$ digits already chosen below or equal to $q$ . Thus, $f(p,q)=[\sum^{q-1}_{n=p}f(p-1,n)] + (q-p+1)f(p-1,q)$ . Now that we have the recursive function, we can start evaluating the values of $f(p,q)$ until we get to $f(6,6)=f(5,6)$
\[f(2,3)=3, f(2,4)=5, f(2,5)=7, f(2,6)=9\] \[f(3,4)=13, f(3,5)=29, f(3,6)=51\] \[f(4,5)=71, f(4,6)=195\] \[f(5,6)=461\] Our requested answer is thus $\boxed{461}$ ~sigma

Answer:  461
Problem: Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution: We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first $k$ terms is greater than $k$ for $1\leq$ $k$ $\leq$ $5$ . We can further simplify this method by approaching it through casework on the first $k$ terms.
Case 1: None of the first one terms is greater than 1
The first term must obviously be one. Since there are five spaces left for numbers, there are a total of $5!=120$ permutations for this case.
Case 2: None of the first two terms is greater than 2
The first two terms must be 1 and 2 in some order. However, we already counted all cases starting with 1 in the previous case, so all of the permutations in this case must begin with $12\cdots$ . Since there are four spaces left, there are a total of $4!=24$ permutations for this case.
Case 3: None of the first three terms is greater than 3
The first three terms must be 1, 2, and 3 in some order. However, the cases beginning with 1__ and 21_ have already been accounted for. There are now $3!-3 = 3$ ways to order the first three numbers of the permutation, and $3!$ ways to order the last three numbers, for a total of $3\times6 = 18$ permutations.
Case 4: None of the first four terms is greater than 4
You can see where the pattern is going - the first four terms must be 1, 2, 3, and 4 in some order. All cases starting with 1 (there are $3!=6$ ), the cases starting with 21 (there are $2!=2$ ), and the 3 cases from case 3 (there are $3\times 1! = 3$ ) have been accounted for, so there are a total of $(4!-6-2-3)2!=26$ permutations for this case.
Case 5: None of the first five terms is greater than 5
This is perhaps the hardest case to work with, simply because there are so many subcases, so keeping track is crucial here. Obviously, the first five terms must be 1, 2, 3, 4, and 5, meaning there are 120 ways to order them. Now, we count the permutations we have already counted in previous cases. $4!$ start with 1, $3!$ start with 2, $3\times2!=6$ start with 3, and $13\times1!=13$ start with 4. Subtracting, we get a total of $120-24-6-6-13=71$ permutations.
Now, we subtract the total number of permutations from our cases from the total number of permutations, which is $6!$ $720 - 120 - 24 - 18 - 26 - 71 = \boxed{461}$

Answer:  461
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: For reference, $2\sqrt{65} \approx 16$ , so $\overline{AD}$ is the longest of the four sides of $ABCD$ . Let $h_1$ be the length of the altitude from $B$ to $\overline{AC}$ , and let $h_2$ be the length of the altitude from $D$ to $\overline{AC}$ . Then, the triangle area equation becomes
\[\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP\]
What an important finding! Note that the opposite sides $\overline{AB}$ and $\overline{CD}$ have equal length, and note that diagonal $\overline{DB}$ bisects diagonal $\overline{AC}$ . This is very similar to what happens if $ABCD$ were a parallelogram with $AB = CD = 10$ , so let's extend $\overline{DB}$ to point $E$ , such that $AECD$ is a parallelogram. In other words, \[AE = CD = 10\] and \[EC = DA = 2\sqrt{65}\] Now, let's examine $\triangle ABE$ . Since $AB = AE = 10$ , the triangle is isosceles, and $\angle ABE \cong \angle AEB$ . Note that in parallelogram $AECD$ $\angle AED$ and $\angle CDE$ are congruent, so $\angle ABE \cong \angle CDE$ and thus \[\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB\] Define $\alpha := \text{m}\angle CDB$ , so $180^\circ - \alpha = \text{m}\angle ABD$
We use the Law of Cosines on $\triangle DAB$ and $\triangle CDB$
\[\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha\]
\[14^2 = 10^2 + BD^2 - 20BD\cos\alpha\]
Subtracting the second equation from the first yields
\[260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}\]
This means that dropping an altitude from $B$ to some foot $Q$ on $\overline{CD}$ gives $DQ = \frac{8}{5}$ and therefore $CQ = \frac{42}{5}$ . Seeing that $CQ = \frac{3}{5}\cdot BC$ , we conclude that $\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \frac{56}{5}$ . Then, the area of $\triangle BCD$ is $\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56$ . Since $AP = CP$ , points $A$ and $C$ are equidistant from $\overline{BD}$ , so \[\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56\] and hence \[\left[ABCD\right] = 56 + 56 = \boxed{112}\] -kgator

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: Denote $\angle APB$ by $\alpha$ . Then $\sin(\angle APB)=\sin \alpha = \sin(\angle APD)$ .
Using the formula for the area of a triangle, we get \[\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha ,\] so \[(AP-CP)(BP-DP)=0\] Hence $AP=CP$ (note that $BP=DP$ makes no difference here).
Now, assume that $AP=CP=x$ $BP=y$ , and $DP=z$ . Using the cosine rule for $\triangle APB$ and $\triangle BPC$ , it is clear that \[x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)\] or \begin{align}x^2+y^2=148\end{align}. Likewise, using the cosine rule for triangles $APD$ and $CPD$ \begin{align}\tag{2}x^2+z^2=180\end{align}. It follows that \begin{align}\tag{3}z^2-y^2=32\end{align}. Since $\sin\alpha=\sqrt{1-\cos^2\alpha}$ \[\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}\] which simplifies to \[\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.\] Plugging this back to equations $(1)$ $(2)$ , and $(3)$ , it can be solved that $x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}$ . Then, the area of the quadrilateral is \[x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}\] --Solution by MicGu

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$ , but it's an AIME problem, we can take $AP=CP$ , and assume the other choice will lead to the same result (which is true).
From $AP=CP$ , we have $[DAP]=[DCP]$ , and $[BAP]=[BCP] \implies [ABD] = [CBD]$ , therefore, \begin{align} \nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\ \Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A  \end{align} By Law of Cosines, \begin{align} \nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\ \Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &=  \sqrt{65}\cos A \tag{2} \end{align} Square $(1)$ and $(2)$ , and add them, to get \[\left(\frac 85\right)^2  + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65\] Solve, $\cos C = 3/5  \implies \sin C = 4/5$ \[[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}\] -Mathdummy

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: Either $PA=PC$ or $PD=PB$ . Let $PD=PB=s$ . Applying Stewart's Theorem on $\triangle ABD$ and $\triangle BCD$ , dividing by $2s$ and rearranging, \[\tag{1}CP^2+s^2=148\] \[\tag{2}AP^2+s^2=180\] Applying Stewart on $\triangle CAB$ and $\triangle CAD$ \[\tag{3} 5CP^2=3AP^2\] Substituting equations 1 and 2 into 3 and rearranging, $s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}$ . By Law of Cosines on $\triangle APB$ $\cos(\angle APB)=\frac{4\sqrt{65}}{65}$ so $\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}$ . Using $[\triangle ABC]=\frac{ab\sin(\angle C)}{2}$ to find unknown areas, $[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}$

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: Now we prove P is the midpoint of $BD$ . Denote the height from $B$ to $AC$ as $h_1$ , height from $D$ to $AC$ as $h_2$ .According to the problem, $AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2$ implies $h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2$ . Then according to basic congruent triangles we get $BP=DP$ Firstly, denote that $CP=a,BP=b,CP=c,AP=d$ . Applying Stewart theorem, getting that $100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c$ , denote $b=5x,c=3x$ Applying Stewart Theorem, getting $260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)$ solve for a, getting $a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}$ Now everything is clear, we can find $cos\angle{BPA}=\frac{4}{\sqrt{65}}$ using LOC, $sin\angle{BPA}=\frac{7\sqrt{65}}{65}$ , the whole area is $\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}$

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: $BP = PD$ as in another solutions.
Let $D'$ be the reflection of $D$ across $C$ .
Let points $E, E',$ and $H$ be the foot of perpendiculars on $AC$ from $D,D'$ , and $B$ respectively. \begin{align*} &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ \Rightarrow &\qquad \triangle ABH =  \triangle CDE =  \triangle CD'E' \\ \Rightarrow &\qquad \angle BAC = \angle ACD' \\ \Rightarrow &\qquad \triangle ABC =  \triangle AD'C \\ \Rightarrow &\qquad BC = AD'. \end{align*} The area of quadrilateral $ABCD$ is equal to the area of triangle $ADD'$ with sides $AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20$ .
The semiperimeter is $s = 17 + \sqrt{65},$ the area \[[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.\]

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: Use your favorite method to get that $P$ is the midpoint of one of the two diagonals (suppose it's the midpoint of $\overline{AC}$ ). From here, let $x=AP=PC, y=BP, z=PD, a=\cos\theta$ where $\theta$ is the angle that the diagonals make. Then we have a system of four equations:
\begin{align*} x^2+y^2+2xya &= 100 \\ z^2+x^2+2xza &= 100 \\ x^2+y^2-2xya &= 196 \\ x^2+z^2-2xza &= 260 \\ \end{align*}
From these equations we get that \begin{align*} xya &= -24 \\ xza &= -40 \\ x^2+y^2-48 &= 10 \\ x^2+z^2-80 &= 10 \end{align*}
From here we can see that $\frac{z}{y}={5}{3}, z^2-y^2=32,$ so $z=5\sqrt{2}, y=3\sqrt{2}.$ Furthermore, this implies $x=\sqrt{130}$ and $xa=-4\sqrt{2},$ which implies $a=\cos\theta=\frac{4}{\sqrt{65}}.$ Then note that the area of the quadrilateral is \[\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.\]

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: Note that $\angle APB = \angle CPD = 180-\angle APD = 180-\angle BPC.$ (All angles are in degrees)
Since $\sin(\theta)=\sin(180-\theta),$ we can use sine area formula to get the following(after some simplifying steps): \begin{align*} BP \times AP + CP \times DP = BP \times PC + AP \times PD. \end{align*} For convenience, let $AP=a, BP=b, CP=c,DP=d.$ The above equation simplifies to: \begin{align*} ab + cd = bc + ad \\ab-ad+cd-bc=0 \\a(b-d)-c(b-d)=0 \\(a-c)(b-d)=0 \end{align*} From here, we see that $a=c$ or $b=d$ . Without loss of generality, let $a=c$ . Since triangles $ABP$ and $CDP$ are obviously not congruent, we see that one triangle is obtuse and the other one is acute.(Refer to the diagram) However, if we drop perpendiculars from $B$ to $AC$ and $D$ to $AC$ , we do get congruent triangles. If the foot of the perpendicular from $B$ is $M$ , and the foot of the perpendicular from $D$ is $N$ , then right triangle $BMP$ is congruent to right triangle $DNP$ .
From here, we see that the altitudes of triangles $ABC$ and $ADC$ to $AC$ are equal. Since they share base $AC$ , their areas are equal. We can use Heron's formula. To not have any fractions, let $AC=2x.$ \begin{align*} \sqrt{(12+x)(12-x)(x+2)(x-2)}=\sqrt{5+\sqrt{65}+x)(5+\sqrt{65}-x)(5-\sqrt{65}+x)(\sqrt{65}-5+x)} \end{align*} Even though this looks bad at first, it actually isn't too complicated to simplify. Expanding the differences of squares and simplifying completely, we get $x^2=32.$ Plugging this $x$ back into the Heron's formula, we get that the area of $ABC$ (or $ADC$ ) is $56$ . Since these triangles have equal area, the area of the quadrilateral is $2 \times 56 = \boxed{112}$ , and we are done. $\blacksquare$

Answer:  112
Problem: Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution: Use any method to derive that $P$ is the midpoint of $A$ and $C$ . Now, either construct a parallelogram with as shown in the diagram to the right, or use law of sines. We will take the parallelogram route here, but the same thing can be done with law of sines on triangles $\triangle \textnormal{ABP}$ and $\triangle \textnormal{CPD}$ . Reflect $D$ across $P$ to get $D'$ . Since $CD = AD' = AB = 10$ $\triangle \textnormal{ABD'}$ is isosceles. Thus, $\angle AD'B = \angle ABD'$ , and because $ADCD'$ is a parallelogram (since $AP = PC$ and $DP = PD'$ ), $\angle AD'B = \angle BDC = \angle ABD'$ . So, $\angle ABD = 180 - \angle ABD' = 180 - \angle BDC$ . Now, apply law of cosines on $\triangle \textnormal{ABD}$ and $\triangle \textnormal{CDB}$ . We get:
\begin{align}
100 + BD^2 - 20BD \cos{\angle ABD} &= 100 + BD^2 - 20 BD \cos {(180 - \angle BDC)} = \\
100 + BD^2 + 20 BD \cos{\angle BDC} &= 260 \\
&\textnormal{and} \\
100 + BD^2 - 20 BD \cos{\angle BDC} &= 196 \\
\textnormal{summing }&\textnormal{and simplifying,} \\
BD &= 8\sqrt{2}
\end{align}
Then, applying law of cosines on $\triangle \textnormal{BCD}$ again, we obtain \[100 + 196 - 280 \cos{\angle BCD} = BD^2 = 128 \implies \cos{\angle BCD} = \frac{3}{5} \implies \sin{\angle BCD} = \frac{4}{5}\] Since $AP = PD$ $[ABD] = [BCD] \implies [ABCD] = [ABD] + [BCD] = 2[BCD]$ . Thus, $[ABCD] = 2[BCD] = 2 \cdot \frac{1}{2} \cdot 10 \cdot 14 \sin{\angle BCD} = 140 \cdot \frac{4}{5} = \boxed{112}$

Answer:  112