FORENSICS ยท ALLEY CONTRACT CHECK
Generated by docs/spec/forensics/check_forensics.py from alley.scene.json,
props.json and each case entry. Every line, number and finding below is computed from those files.
Scene check: PASS.
The Open Window
hard PASS PLAUSIBLEThe second-floor window was found open. A ladder lay in the alley below it, and two scuffs in the grime sit below the window. The department's ladder rule: to climb through a window, the ladder rests on the sill and at least 0.95 m of it has to stand past the sill. Was this ladder long enough?
Scuff marks on level ground are 2.0 meters from a vertical wall. A ladder footed at the marks rests on a windowsill 3.6 meters above the ground. To climb through the window, at least 0.9 meters of the ladder must extend past the sill. What is the minimum length of the ladder, to the nearest tenth of a meter?
Answer 5.1 m (geometry 5.0683, accepted ±0.05)
From the scuffs, a climber needs at least 5.1 m of ladder. The ladder in the alley is 5.4 m and stands 1.28 m past the sill: enough to climb in.
Measurements and givens
| Scuffs to wall | s | 2.000 | required |
| Sill height | h | 3.600 | required |
| Ladder length | L | 5.400 | required |
| Door width | w | 0.900 | decoy |
| Ladder past the sill | m | 0.95 | given |
Misses
| variant | value | fault | feedback |
|---|---|---|---|
| base {"extend": null} | 4.118 | M-CONDITION-DROP | That ladder only reaches the sill. Add the 0.9 m that has to stand past it. |
| base {"leg_b": "hm", "extend": null} | 4.97 | M-CONCEPT | The 0.9 m runs along the ladder, past the sill. It isn't extra height on the wall. |
| leg_sum | 6.55 | M-FORMULA | Adding the sides walks the ladder along the ground and then up the wall. The ladder cuts straight across: it's the long side of a right triangle. |
| leg_difference | 3.943 | M-SIGN | The ladder is the longest side, so its square is the sum of the other two squares, not the difference. |
| no_root | 17.91 | M-PARTIAL | That's the square of the length to the sill. Take the square root before adding the 0.9 m. |
| echo | 5.4 | M-WRONG-TARGET | That's the ladder found in the alley. The question asks for the shortest ladder that would work. |
The Open Window
hard PASS RULED OUTThe second-floor window was found open. A ladder lay in the alley below it, and two scuffs in the grime sit below the window. The department's ladder rule: to climb through a window, the ladder rests on the sill and at least 0.95 m of it has to stand past the sill. Was this ladder long enough?
Scuff marks on level ground are 2.0 meters from a vertical wall. A ladder footed at the marks rests on a windowsill 3.6 meters above the ground. To climb through the window, at least 0.9 meters of the ladder must extend past the sill. What is the minimum length of the ladder, to the nearest tenth of a meter?
Answer 5.1 m (geometry 5.0683, accepted ±0.05)
From the scuffs, a climber needs at least 5.1 m of ladder. The ladder in the alley is only 4.8 m and stands 0.68 m past the sill: too little to climb in.
Measurements and givens
| Scuffs to wall | s | 2.000 | required |
| Sill height | h | 3.600 | required |
| Ladder length | L | 4.800 | required |
| Door width | w | 0.900 | decoy |
| Ladder past the sill | m | 0.95 | given |
Misses
| variant | value | fault | feedback |
|---|---|---|---|
| base {"extend": null} | 4.118 | M-CONDITION-DROP | That ladder only reaches the sill. Add the 0.9 m that has to stand past it. |
| base {"leg_b": "hm", "extend": null} | 4.97 | M-CONCEPT | The 0.9 m runs along the ladder, past the sill. It isn't extra height on the wall. |
| leg_sum | 6.55 | M-FORMULA | Adding the sides walks the ladder along the ground and then up the wall. The ladder cuts straight across: it's the long side of a right triangle. |
| leg_difference | 3.943 | M-SIGN | The ladder is the longest side, so its square is the sum of the other two squares, not the difference. |
| no_root | 17.91 | M-PARTIAL | That's the square of the length to the sill. Take the square root before adding the 0.9 m. |
| echo | 4.8 | M-WRONG-TARGET | That's the ladder found in the alley. The question asks for the shortest ladder that would work. |
The Shortcut
hard PASS PLAUSIBLEThe service door's alarm logged it opening at 23:42:06.0. At 23:42:08.5 a patrol car's lights swept the alley mouth, and its dash camera shows the alley empty. The theory: someone came out the door, ran around the stacked cases and was gone before the lights hit. Running empty-handed, a person manages about 6.0 m/s over a distance this short.
From a doorway, a person runs a route in two straight legs. The first leg ends at a point 2.4 meters farther along the wall and 1.8 meters out from it. The second leg runs 5.6 meters parallel to the wall. The person covers the whole route in 2.5 seconds at a constant speed. What is that speed, in meters per second, to the nearest tenth?
Answer 3.4 m/s (geometry 3.4400, accepted ±0.05)
The route takes 3.4 m/s. Running empty-handed, a person manages about 6.0. Someone could have been gone before the lights hit.
Measurements and givens
| Door to footprints, along the wall | a | 2.400 | required |
| Door to footprints, out from the wall | b | 1.800 | required |
| Footprints to alley mouth | c | 5.600 | required |
| Door straight to alley mouth | direct | 8.200 | decoy |
| Door alarm to headlights | t | 2.5 | given |
| Top speed, empty-handed | vmax | 6.0 | given |
Misses
| variant | value | fault | feedback |
|---|---|---|---|
| base {"segments": ["a", "b", "c"]} | 3.92 | M-FORMULA | Adding 2.4 and 1.8 walks the first leg along the wall and then out. The runner cut straight across: that leg is the long side of a right triangle. |
| base {"segments": ["direct"]} | 3.28 | M-MODEL-MAP | The straight line runs through the stacked cases. Use the route the footprints show. |
| first_only | 1.2 | M-PARTIAL | That's only the first leg. The runner still had 5.6 m to go. |
| invert | 0.291 | M-INVERT | That's seconds per meter. Speed is distance divided by time. |
| product | 21.5 | M-UNIT | Meters times seconds isn't a speed. Divide the distance by the time. |
The Shortcut
hard PASS RULED OUTThe service door's alarm logged it opening at 23:42:06.0. At 23:42:08.5 a patrol car's lights swept the alley mouth, and its dash camera shows the alley empty. The theory: someone came out the door, ran around the stacked cases and was gone before the lights hit. Whoever it was carried the missing equipment case, 32 kg, and a person carrying that much manages about 3.0 m/s.
From a doorway, a person runs a route in two straight legs. The first leg ends at a point 2.4 meters farther along the wall and 1.8 meters out from it. The second leg runs 5.6 meters parallel to the wall. The person covers the whole route in 2.5 seconds at a constant speed. What is that speed, in meters per second, to the nearest tenth?
Answer 3.4 m/s (geometry 3.4400, accepted ±0.05)
The route takes 3.4 m/s, faster than the 3.0 m/s a person manages carrying the case. Whoever opened the door was still in the alley when the lights hit.
Measurements and givens
| Door to footprints, along the wall | a | 2.400 | required |
| Door to footprints, out from the wall | b | 1.800 | required |
| Footprints to alley mouth | c | 5.600 | required |
| Door straight to alley mouth | direct | 8.200 | decoy |
| Door alarm to headlights | t | 2.5 | given |
| Top speed, carrying the case | vmax | 3.0 | given |
Misses
| variant | value | fault | feedback |
|---|---|---|---|
| base {"segments": ["a", "b", "c"]} | 3.92 | M-FORMULA | Adding 2.4 and 1.8 walks the first leg along the wall and then out. The runner cut straight across: that leg is the long side of a right triangle. |
| base {"segments": ["direct"]} | 3.28 | M-MODEL-MAP | The straight line runs through the stacked cases. Use the route the footprints show. |
| first_only | 1.2 | M-PARTIAL | That's only the first leg. The runner still had 5.6 m to go. |
| invert | 0.291 | M-INVERT | That's seconds per meter. Speed is distance divided by time. |
| product | 21.5 | M-UNIT | Meters times seconds isn't a speed. Divide the distance by the time. |
The Missing Shipment
standard PASS PLAUSIBLEA painted panel arrived crated last week and was stored in the back room. Tonight the crate turned up in the alley, opened and empty, leaning on the wall. The crate was built to the panel's size. The back room's only way out to the alley is the service door. Could the panel have gone out through it?
A rectangular doorway is 0.9 meter wide and 2.1 meters tall. What is the length of the doorway's diagonal, to the nearest hundredth of a meter?
Answer 2.28 m (geometry 2.2847, accepted ±0.05)
The panel's shorter side is 2.2 m, too tall for the door upright. The doorway's diagonal is 2.28 m, so tilted corner to corner, the panel clears it.
Measurements and givens
| Door width | dw | 0.900 | required |
| Door height | dh | 2.100 | required |
| Crate width | cw | 2.500 | required |
| Crate height | ch | 2.200 | required |
Misses
| variant | value | fault | feedback |
|---|---|---|---|
| leg_sum | 3.0 | M-FORMULA | Adding the width and height runs along two edges of the doorway. The diagonal cuts across them: it's the long side of a right triangle. |
| leg_difference | 1.897 | M-SIGN | The diagonal is the longest side, so its square is the sum of the other two squares, not the difference. |
| no_root | 5.22 | M-PARTIAL | That's the diagonal squared. Take the square root. |
| echo | 2.1 | M-WRONG-TARGET | That's the door's height. Tilted corner to corner, the doorway is longer than that. |
The Missing Shipment
standard PASS RULED OUTA painted panel arrived crated last week and was stored in the back room. Tonight the crate turned up in the alley, opened and empty, leaning on the wall. The crate was built to the panel's size. The back room's only way out to the alley is the service door. Could the panel have gone out through it?
A rectangular doorway is 0.9 meter wide and 2.1 meters tall. What is the length of the doorway's diagonal, to the nearest hundredth of a meter?
Answer 2.28 m (geometry 2.2847, accepted ±0.05)
The panel's shorter side is 2.4 m, longer than the doorway's 2.28 m diagonal. No tilt gets it through this door. It left another way.
Measurements and givens
| Door width | dw | 0.900 | required |
| Door height | dh | 2.100 | required |
| Crate width | cw | 2.500 | required |
| Crate height | ch | 2.400 | required |
Misses
| variant | value | fault | feedback |
|---|---|---|---|
| leg_sum | 3.0 | M-FORMULA | Adding the width and height runs along two edges of the doorway. The diagonal cuts across them: it's the long side of a right triangle. |
| leg_difference | 1.897 | M-SIGN | The diagonal is the longest side, so its square is the sum of the other two squares, not the difference. |
| no_root | 5.22 | M-PARTIAL | That's the diagonal squared. Take the square root. |
| echo | 2.1 | M-WRONG-TARGET | That's the door's height. Tilted corner to corner, the doorway is longer than that. |