The Most Missed Questions in the July 2025 ACSIMC

The July 2025 ACSIMC has concluded, scores are available, the leaderboard is out, and now it’s time to discuss the most missed problems. We will discuss the top 3 most missed questions in all rounds (i.e. Archimedes, Stokes, and Rutherford) and solve them in ingenious ways. Each problem had its own twist. The tests were certainly not easy for many competitors; although the tests could have been more difficult, the writers feel that it provided enough of a challenge in order to push competitors to use their mathematical sense and logic.

Archimedes Round Top 3 Most Missed Questions

Geometry #2:

This was the most missed question on the Archimedes test. We can see that the question asks us to find the area of parallelogram $ABCD$. The area of a parallelogram is equal to its base multiplied by its height. Its base has a length of 8. One might assume the height was 5, and answer (C) 40, but that would not be correct. The height of the parallelogram is what we would get if we drew an altitude from point $B$ to line segment $\overline{AD}$. This would have the same length as line segment $\overline{CE}$, which is part of a right triangle. To find the length of $\overline{CE}$, we need the lengths of the other sides and then we can find the length of the last side using the Pythagorean Theorem. $\overline{CD}$ is just the other side of the parallelogram and thus is congruent to $\overline{AB}$ with a length of 5. The length of $\overline{DE}$ is 3 because $AD+DE = 11$ and $\overline{AD}$ is congruent to $\overline{BC}$ with a length of 8. This forms a 3-4-5 Pythagorean triple, which means that the missing length is 4 (since we have the other two sides measuring 3 and 5). So with our new height, we can multiply by the base to get our answer: (A) 32.

Geometry #3:

This was the second most missed question on the Archimedes test. Here, we are asked to find the area of an equilateral triangle. This problem can be solved easily if one knows the formula for the area of an equilateral triangle. But, many do not know it, so we will approach this question differently. Draw an altitude from point $Y$ to $\overline{XZ}$ and call the point of intersection $O$. This divides $\overline{XZ}$ into two halves (property of equilateral triangles), with each half having a length of 3. We now have two congruent triangles, $\triangle XYO$ and $\triangle YOZ$. We can find the area of one triangle and multiply it by 2 to find the area of both triangles and thus the whole figure. Let’s take the figure $\triangle XYO$. We know that $\overline{XO}$ has a length of 3. When we made the altitude $\overline{YO}$, we bisected the angle $\angle XYZ$, making each half of this angle have a measure of 30 degrees. Now, using the common 30-60-90 triangle rule, as $\overline{XO}$ has a length of 3 and is opposite $\angle XYO$ (with a measure of 30 degrees), we can deduce that $\overline{YO}$ has length $3\sqrt{3}$. Now, we can find the area of $\triangle XYO$ by multiplying the length of $\overline{XO}$ and $\overline{YO}$ and then dividing by 2 (because of the formula for the area of a triangle). After that, we multiply by 2 to get the area of both triangles, and thus the whole figure. Notice, then, we are multiplying the lengths $\overline{XO}$ and $\overline{YO}$ (as dividing by 2 and then multiplying by 2 will cancel out), which will get us our answer of (D) $9\sqrt{3}$.

Geometry #4:

This problem can be tackled in many ways, with the most efficient one being to realize that the diagonals of a rhombus are perpendicular and that the diagonals divide the rhombus into 4 congruent right triangles. Using this knowledge, we can easily find the measure of $\angle ABD$ to equal 67.5 degrees. Then, by congruence, $\angle ADB$ is congruent to $\angle ABD$. From this, we get the answer of (B) 67.5 degrees. 

Stokes Round Top 3 Most Missed Questions

Algebra #5:

This question asks us to find how many numbers, when taken the square root of and rounded down, equal 7. We have the obvious one of $x$ being 49 where $\sqrt{49} = 7$, so flooring doesn’t affect it. Now, we want all values from $7^2$ up until (but not including) $8^2$, since including $8^2$ and values above $8^2$ would give us values higher than 7 when floored. So as $8^2$ is equal to 64, the maximum integral value for $x$ is 63. That means our range of value for $x$ is 49 to 63 (inclusive), which gets us 15 integers; thus, our answer is (C) 15.

Geometry #2:

The area of a right triangle with sides that measure $a$ and $b$ is $\displaystyle{\frac{1}{2}\cdot ab}$. That means, according to the problem, $\displaystyle{\frac{1}{2}\cdot ab=2(a+b)}$. Let’s simplify this:

$$\frac{1}{2}\cdot ab=2(a+b)$$

$$\frac{1}{2}\cdot ab=2a+2b$$

$$ab=4a+4b$$

Adding 16 to both sides, we get the following:

$$ab-4a-4b+16=16$$

$$(a-4)(b-4)=16$$

The factors of 16 in $(a,b)$ form are $(1,16)$, $(2,8)$, $(4,4)$, $(8,2)$, and $(16,1)$. Therefore, $a-4$ must equal the first value in each of the ordered pairs in the previous sentence; the same is true for $b-4$: it must equal the second value in each of the ordered pairs. From this process, we get:

$$(1,16)\Rightarrow a=5, b=20$$

$$(2,8)\Rightarrow a=6, b=12$$

$$(4,4)\Rightarrow a=8, b=8$$

$$(8,2)\Rightarrow a=12, b=6$$

$$(16,1)\Rightarrow a=20, b=5$$

We have all the possible side lengths of the triangles that will satisfy our condition. Now, we just have to find the areas. The list of areas is (respectively) 50, 36, 32, 36, and 50. Now, we have 36 and 50 as repeats, so our final list is 50, 36, and 32, and adding all of these numbers gets our answer: (C) 118.

Geometry #6:

This question requires some knowledge of the Law of Cosines, with which the problem can easily be solved. For the sake of time, we will simply state it here without giving it proof. Please learn it at your own time to truly understand this formula and not memorize. If we had a triangle $\triangle ABC$, with the angles $\angle x$, $\angle y$, and $\angle z$ being opposite of the sides $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$ respectively, then $(AC)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC)\cos(z)$.  

Now, we are ready to tackle the problem. Observe that the question asks for jumping from every other vertex, so let’s go from vertex F to D (assuming the vertices follow a standard sequence), making line segment $\overline{FD}$. 

Note that the sum of all angles in an octagon is $(8-2) \cdot 180$, and so each angle is $135^{\circ}$. Now we know that $\angle FED$ is $135^{\circ}$. Note now that we have triangle $\triangle FED$, and we want to find the length of $\overline{FD}$. So, we will use our aforementioned Law of Cosines to find this length. This means that  $(FD)^2 = (FD)^2 + (FE)^2 - 2(FD)(FE)\cos(\angle FED) $. We know the lengths of $\overline{FE}$ and $\overline{ED}$ are both equal to 1 since they are the sides of a regular octagon with side lengths of 1 (mentioned in the problem). So, we can write our equation as $ (FD)^2 = 1^2 + 1^2 -  2\cos(135^{\circ}) $. $\cos(135^{\circ}) = - \cos(45^{\circ})$ (because 45 degrees is the reference angle and this angle is on the negative $x$-axis side of a unit circle (if this doesn’t make sense be sure to learn that first)). $\displaystyle{-\cos(45^{\circ}) = -\frac{\sqrt{2}}{2}} $. Multiply this by 2 (from our Law of Cosines) to get $-\sqrt{2}$. Now, substitute it back into our equation to have the following:

$$ (FD)^2 = 1 + 1 - (-\sqrt{2}) $$

$$ (FD)^2 = 2 - (-\sqrt{2}) $$

$$ (FD)^2 = 2+ \sqrt{2}$$

After all these steps, we can now take the square root of both sides of the equation to get our answer of (C) $\sqrt{2+\sqrt{2}}$.  

Rutherford Round Top 3 Most Missed Questions

Algebra #3:

This question requires some extremely basic knowledge of vectors. Subtracting the vectors easily produces the new vector: $(6 - (-9)) + (3i - (-2)i)$. Simplifying this expression, we get $15+5i$. Now, the following is probably the only new vector knowledge one may require, but $15+5i$ is essentially a vector that starts at the origin and goes right 15 in the real axis and up 5 in the imaginary axis. Just imagine it as $(15,5i)$ being the end of the line segment originating at $(0,0i)$. We could delve into seeing the slope easily as $\displaystyle{\frac{5}{15}}$ and then getting a linear equation if we wanted, but we can use some logic and see how the imaginary coordinate ($y$-coordinate) is $\displaystyle{\frac{1}{3}}$ of the real coordinate ($x$-coordinate). Since a line segment has a constant slope, we wish to look for the same pattern in our answer. That eliminates every option except A and D. Since the question asks what is on the line, we will look for an answer that is between 0 and 5 for our imaginary coordinates and 0 and 15 for our real coordinates. Noting this, our correct answer will be (A) $\displaystyle{\left(2,\frac{2}{3}i\right)}$.

Combinatorics #4:

We can break this problem down into two parts: how many possible pairs of 2 distinct faces are there? 2. For each pair, how many outcomes (out of 6 dice rolls) contain just these two distinct faces?

The first part is simple. It's simply the choose function $\displaystyle{\binom{6}{2}}$. The second part is a bit more complex. For each pair, the dice have to roll only to contain those two dice rolls. There are $2^6$ ways to do this (2 options per dice). However, 2 of these options are if all the dice are the same, which means that there aren't 2 distinct options. Therefore we need to subtract 2: $2^6-2$.

For each pair there are a total of $6^6$ combinations so we divide everything by that. Combining everything together we get answer choice (C) $\displaystyle{\binom{6}{2}\frac{2^6-2}{6^6}}$.

Geometry #2:

Since the measure of angle $ABC$ is 135 degrees and the measure of angle $A$ is 15 degrees, by the Triangle Sum Theorem, the measure of angle $ACB$ is 30 degrees. Then, using the Law of Sines and the unit circle, we can find the length of side $\overline{AC}$:

$$\frac{\sin 30^\circ}{3}=\frac{\sin 135^\circ}{AC}$$

$$3\sin 135^\circ=AC\sin 30^\circ$$

$$AC\cdot\frac{1}{2}=3\cdot\frac{\sqrt{2}}{2}$$

$$AC=3\sqrt{2}$$

Now, we use the formula for the area of a triangle given two sides and the measure of the angle formed by those two sides: $\displaystyle{A=\frac{1}{2}ab\sin C}$. Substituting $a=3$, $b=3\sqrt{2}$, and $C=15^\circ$, we get $\displaystyle{A=\frac{1}{2}(3)\left(3\sqrt{2}\right)\sin 15^\circ}$. Since 15 can be represented as the difference between 45 and 30, we use the angle subtraction formula for sine: $\sin(A-B)=\sin A\cos B-\sin B\cos A$. From this formula being substituted into the spot for $\sin 15^\circ$ using $A=45^\circ$ and $B=30^\circ$, we get that the area of one of these triangles is equal to $\displaystyle{\frac{9\sqrt{3}-9}{4}}$ after simplification. Since the question asks for both areas combined, we simply multiply this expression by 2 to get answer choice (D) $\displaystyle{\frac{9\sqrt{3}-9}{2}}$.

Written by Ayush K.

Edited by Alex M.